Schreiber Introduction to Homological Algebra

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This page is a detailed introduction to homological algebra. Starting with motivation from basic homotopy theory, it introduces the basics of the category of chain complexes and then develops the concepts of derived categories and derived functors in homological algebra, with the main examples of Ext and Tor. The last chapter introduces and proves the fundamental theorems of the field.

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For the application to spectral sequences see at Introduction to Spectral Sequences.

For background/outlook on abstract homotopy theory see at Introduction to Homotopy Theory.

For generalization to stable homotopy theory see at Introduction to Stable homotopy theory.

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This text is a first introduction to homological algebra, assuming only very basic prerequisites. For instance we do recall in some detail basic definitions and constructions in the theory of abelian groups and modules, though of course a prior familiarity with these ingredients will be helpful. Also we use very little category theory, if it all. Where universal constructions do appear we spell them out explicitly in components and just mention their category-theoretic names for those readers who want to dig deeper. We do however freely use the words functor and commuting diagram. The reader unfamiliar with these elementary notions should click on these keywords and follow the hyperlink to the explanation right now.

Motivation

The subject of homological algebra may be motivated by its archetypical application, which is the singular homology of a topological space $X$. This example illustrates homological algebra as being concerned with the abelianization of what is called the homotopy theory of $X$.

So we begin with some basic concepts in homotopy theory in section 1) Homotopy type of topological spaces. Then we consider the “abelianization” of this setup in 2) Simplicial and abelian homology.

Together this serves to motivate many constructions in homological algebra, such as centrally chain complexes, chain maps and homology, but also chain homotopies, mapping cones etc, which we discuss in detail in chapter II below. In the bulk we develop the general theory of homological algebra in chapter III and chapter IV. Finally we come back to a systematic discussion of the relation to homotopy theory at the end in chapter V. A section Outlook is appended for readers interested in the grand scheme of things.

We do use some basic category theory language in the following, but no actual category theory. The reader should know what a category is, what a functor is and what a commuting diagram is. These concepts are more elementary than any genuine concept in homological algebra to appear below and of general use. Where we do encounter universal constructions below we call them by their category-theoretic name but always spell them out in components explicity.

1) Homotopy type of topological spaces

This section reviews some basic notions in topology and homotopy theory. These will all serve as blueprints for corresponding notions in homological algebra.

Definition

A topological space is a set $X$ equipped with a set of subsets $U \subset X$, called open sets, which are closed under

1. finite intersections
2. arbitrary unions.

Example

The Cartesian space $\mathbb{R}^n$ with its standard notion of open subsets given by unions of open balls $D^n \subset \mathbb{R}^n$.

Definition

For $Y \hookrightarrow X$ an injection of sets and $\{U_i \subset X\}_{i \in I}$ a topology on $X$, the subspace topology on $Y$ is $\{U_i \cap Y \subset Y\}_{i \in I}$.

Definition

For $n \in \mathbb{N}$, the topological n-simplex is, up to homeomorphism, the topological space whose underlying set is the subset

$$\Delta^n \coloneqq \{ \vec x \in \mathbb{R}^{n+1} | \sum_{i = 0 }^n x_i = 1 \; and \; \forall i . x_i \geq 0 \} \subset \mathbb{R}^{n+1}$$

of the Cartesian space $\mathbb{R}^{n+1}$, and whose topology is the subspace topology induces from the canonical topology in $\mathbb{R}^{n+1}$.

Example

For $n = 0$ this is the point, $\Delta^0 = *$.

For $n = 1$ this is the standard interval object $\Delta^1 = [0,1]$.

For $n = 2$ this is the filled triangle.

For $n = 3$ this is the filled tetrahedron.

Definition

A homomorphisms between topological spaces $f : X \to Y$ is a continuous function:

a function $f:X\to Y$ of the underlying sets such that the preimage of every open set of $Y$ is an open set of $X$.

Topological spaces with continuous maps between them form the category Top.

Definition

For $n \in \mathbb{N}$, $\n \geq 1$ and $0 \leq k \leq n$, the $k$th $(n-1)$-face (inclusion) of the topological $n$-simplex, def. , is the subspace inclusion

$$\delta_k : \Delta^{n-1} \hookrightarrow \Delta^n$$

induced under the coordinate presentation of def. , by the inclusion

$$\mathbb{R}^n \hookrightarrow \mathbb{R}^{n+1}$$

which “omits” the $k$th canonical coordinate:

$$(x_0, \cdots , x_{n-1}) \mapsto (x_0, \cdots, x_{k-1} , 0 , x_{k}, \cdots, x_{n-1}) \,.$$

Example

The inclusion

$$\delta_0 : \Delta^0 \to \Delta^1$$

is the inclusion

$$\{1\} \hookrightarrow [0,1]$$

of the “right” end of the standard interval. The other inclusion

$$\delta_1 : \Delta^0 \to \Delta^1$$

is that of the “left” end $\{0\} \hookrightarrow [0,1]$.

Definition

For $n \in \mathbb{N}$ and $0 \leq k \lt n$ the $k$th degenerate $(n)$-simplex (projection) is the surjective map

$$\sigma_k : \Delta^{n} \to \Delta^{n-1}$$

induced under the barycentric coordinates of def. under the surjection

$$\mathbb{R}^{n+1} \to \mathbb{R}^n$$

which sends

$$(x_0, \cdots, x_n) \mapsto (x_0, \cdots, x_{k} + x_{k+1}, \cdots, x_n) \,.$$

Definition

For $X \in$ Top and $n \in \mathbb{N}$, a singular $n$-simplex in $X$ is a continuous map

$$\sigma : \Delta^n \to X$$

from the topological $n$-simplex, def. , to $X$.

Write

$$(Sing X)_n \coloneqq Hom_{Top}(\Delta^n , X)$$

for the set of singular $n$-simplices of $X$.

As $n$ varies, this forms the singular simplicial complex of $X$. This is the topic of the next section, see def. def. .

Definition

For $f,g : X \to Y$ two continuous functions between topological spaces, a left homotopy $\eta : f \Rightarrow g$ is a commuting diagram in Top of the form

$$\array{ X \\ {}^{\mathllap{(id,\delta_0)}}\downarrow & \searrow^{\mathrlap{f}} \\ X \times \Delta^1 &\stackrel{\eta}{\to}& Y \\ {}^{\mathllap{(id,\delta_1)}}\uparrow & \nearrow_{\mathrlap{g}} \\ X } \,.$$

Remark

In words this says that a homotopy between two continuous functions $f$ and $g$ is a continuous 1-parameter deformation of $f$ to $g$. That deformation parameter is the canonical coordinate along the interval $[0,1]$, hence along the “length” of the cylinder $X \times \Delta^1$.

Proposition

Left homotopy is an equivalence relation on $Hom_{Top}(X,Y)$.

The fundamental invariants of a topological space in the context of homotopy theory are its homotopy groups. We first review the first homotopy group, called the fundamental group of $X$:

Definition

For $X$ a topological space and $x : * \to X$ a point. A loop in $X$ based at $x$ is a continuous function

$$\gamma : \Delta^1 \to X$$

from the topological 1-simplex, such that $\gamma(0) = \gamma(1) = x$.

A based homotopy between two loops is a homotopy

$$\array{ \Delta^1 \\ \downarrow^{\mathrlap{(id,\delta_0)}} & \searrow^{\mathrlap{f}} \\ \Delta^1 \times \Delta^1 &\stackrel{\eta}{\to}& X \\ \uparrow^{\mathrlap{(id,\delta_1)}} & \nearrow_{\mathrlap{g}} \\ \Delta^1 }$$

such that $\eta(0,-) = \eta(1,-) = x$.

Proposition

This notion of based homotopy is an equivalence relation.

Proof

This is directly checked. It is also a special case of the general discussion at homotopy.

Definition

Given two loops $\gamma_1, \gamma_2 : \Delta^1 \to X$, define their concatenation to be the loop

$$\gamma_2 \cdot \gamma_1 : t \mapsto \left\{ \array{ \gamma_1(2 t) & ( 0 \leq t \leq 1/2 ) \\ \gamma_2(2 (t-1/2)) & (1/2 \leq t \leq 1) } \right. \,.$$

Proposition

Concatenation of loops respects based homotopy classes where it becomes an associative, unital binary pairing with inverses, hence the product in a group.

Definition

For $X$ a topological space and $x \in X$ a point, the set of based homotopy equivalence classes of based loops in $X$ equipped with the group structure from prop. is the fundamental group or first homotopy group of $(X,x)$, denoted

$$\pi_1(X,x) \in Grp \,.$$

Example

The fundamental group of the point is trivial: $\pi_1(*) = *$.

Example

The fundamental group of the circle is the group of integers $\pi_1(S^1) \simeq \mathbb{Z}$.

This construction has a fairly straightforward generalizations to “higher dimensional loops”.

Definition

Let $X$ be a topological space and $x : * \to X$ a point. For $(1 \leq n) \in \mathbb{N}$, the $n$th homotopy group $\pi_n(X,x)$ of $X$ at $x$ is the group:

• whose elements are left-homotopy equivalence classes of maps $S^n \to (X,x)$ in $Top^{*/}$;

• composition is given by gluing at the base point (wedge sum) of representatives.

The 0th homotopy group is taken to be the set of connected components.

Example

For $n = 1$ this reproduces the definition of the fundamental group of def. .

The homotopy theory of topological spaces is all controled by the following notion. The abelianization of this notion, the notion of quasi-isomorphism discussed in def. below is central to homological algebra.

Definition

For $X, Y \in$ Top two topological spaces, a continuous function $f : X \to Y$ between them is called a weak homotopy equivalence if

1. $f$ induces an isomorphism of connected components

   $$\pi_0(f) \colon \pi_0(X) \stackrel{\simeq}{\to} \pi_0(Y)$$

   in Set;

2. for all points $x \in X$ and for all $(1 \leq n) \in \mathbb{N}$ $f$ induces an isomorphism on homotopy groups

   $$\pi_n(f,x) \colon \pi_n(X,x) \stackrel{\simeq}{\to} \pi_n(Y,f(x))$$

   in Grp.

What is called homotopy theory is effectively the study of topological spaces not up to isomorphism (here: homeomorphism), but up to weak homotopy equivalence. Similarly, we will see that homological algebra is effectively the study of chain complexes not up to isomorphism, but up to quasi-isomorphism. But this is slightly more subtle than it may seem, in parts due to the following:

Proposition

The existence of a weak homotopy equivalence from $X$ to $Y$ is a reflexive and transitive relation on Top, but it is not a symmetric relation.

Proof

Reflexivity and transitivity are trivially checked. A counterexample to symmetry is the weak homotopy equivalence between the stanard circle and the pseudocircle.

But we can consider the genuine equivalence relation generated by weak homotopy equivalence:

Definition

We say two spaces $X$ and $Y$ have the same (weak) homotopy type if they are equivalent under the equivalence relation generated by weak homotopy equivalence.

Remark

Equivalently this means that $X$ and $Y$ have the same (weak) homotopy type if there exists a zigzag of weak homotopy equivalences

$$X \leftarrow \to\leftarrow \dots \to Y \,.$$

One can understand the homotopy type of a topological space just in terms of its homotopy groups and how they act on each other. (This data is called a Postnikov tower of $X$.) But computing and handling homotopy groups is in general hard, famously so already for the seemingly simple case of the homotopy groups of spheres. Therefore we now want to simplify the situation by passing to a “linear/abelian approximation”.

2) Simplicial and singular homology

This section discusses how the “abelianization” of a topological space by singular chains gives rise to the notion of chain complexes and their homology.

Above in def. we saw that to a topological space $X$ is associated a sequence of sets

$$(Sing X)_n \coloneqq Hom_{Top}(\Delta^n, X)$$

of singular simplices. Since the topological $n$-simplices $\Delta^n$ from def. sit inside each other by the face inclusions of def.

$$\delta_k : \Delta^{n-1} \to \Delta^{n}$$

and project onto each other by the degeneracy maps, def.

$$\sigma_k : \Delta^{n+1} \to \Delta^n$$

we dually have functions

$$d_k \coloneqq Hom_{Top}(\delta_k, X) : (Sing X)_n \to (Sing X)_{n-1}$$

that send each singular $n$-simplex to its $k$-face and functions

$$s_k \coloneqq Hom_{Top}(\sigma_k,X) : (Sing X)_{n} \to (Sing X)_{n+1}$$

that regard an $n$-simplex as beign a degenerate (“thin”) $(n+1)$-simplex. All these sets of simplices and face and degeneracy maps between them form the following structure.

Definition

A simplicial set $S \in sSet$ is

• for each $n \in \mathbb{N}$ a set $S_n \in Set$ – the set of $n$-simplices;

• for each injective map $\delta_i : \overline{n-1} \to \overline{n}$ of totally ordered sets $\bar n \coloneqq \{ 0 \lt 1 \lt \cdots \lt n \}$

  a function $d_i : S_{n} \to S_{n-1}$ – the $i$th face map on $n$-simplices;

• for each surjective map $\sigma_i : \overline{n+1} \to \bar n$ of totally ordered sets

  a function $\sigma_i : S_{n} \to S_{n+1}$ – the $i$th degeneracy map on $n$-simplices;

such that these functions satisfy the simplicial identities.

Definition

The simplicial identities satisfied by face and degeneracy maps as above are (whenever these maps are composable as indicated):

1. $d_i \circ d_j = d_{j-1} \circ d_i$ if $i \lt j$,

2. $s_i \circ s_j = s_j \circ s_{i-1}$ if $i \gt j$.

3. $d_i \circ s_j = \left\{ \array{ s_{j-1} \circ d_i & if \; i \lt j \\ id & if \; i = j \; or \; i = j+1 \\ s_j \circ d_{i-1} & if i \gt j+1 } \right.$

It is straightforward to check by explicit inspection that the evident injection and restriction maps between the sets of singular simplices make $(Sing X)_\bullet$ into a simplicial set. We now briefly indicate a systematic way to see this using basic category theory, but the reader already satisfied with this statement should jump ahead to the abelianization of $(Sing X)_n$ in prop. below.

Definition

The simplex category $\Delta$ is the full subcategory of Cat on the free categories of the form

$$\begin{aligned} [0] & \coloneqq \{0\} \\ [1] & \coloneqq \{0 \to 1\} \\ [2] & \coloneqq \{0 \to 1 \to 2\} \\ \vdots \end{aligned} \,.$$

Remark

This is called the “simplex category” because we are to think of the object $[n]$ as being the “spine” of the $n$-simplex. For instance for $n = 2$ we think of $0 \to 1 \to 2$ as the “spine” of the triangle. This becomes clear if we don’t just draw the morphisms that generate the category $[n]$, but draw also all their composites. For instance for $n = 2$ we have_

$$[2] = \left\{ \array{ && 1 \\ & \nearrow && \searrow \\ 0 &&\to&& 2 } \right\} \,.$$

Proposition

A functor

$$S : \Delta^{op} \to Set$$

from the opposite category of the simplex category to the category Set of sets is canonically identified with a simplicial set, def. .

Proof

One checks by inspection that the simplicial identities characterize precisely the behaviour of the morphisms in $\Delta^{op}([n],[n+1])$ and $\Delta^{op}([n],[n-1])$.

This makes the following evident:

Example

The topological simplices from def. arrange into a cosimplicial object in Top, namely a functor

$$\Delta^\bullet : \Delta \to Top \,.$$

With this now the structure of a simplicial set on the singular simplices $(Sing X)_\bullet$, def. , is manifest: it is just the nerve of $X$ with respect to $\Delta^\bullet$, namely:

Definition

For $X$ a topological space its simplicial set of singular simplicies (often called the singular simplicial complex)

$$(Sing X)_\bullet : \Delta^{op} \to Set$$

is given by composition of the functor from example with the hom functor of Top:

$$(Sing X) : [n] \mapsto Hom_{Top}( \Delta^n , X ) \,.$$

Remark (aside)

It turns out that homotopy type of the topological space $X$ is entirely captured by its singular simplicial complex $Sing X$ (this is the content of the homotopy hypothesis-theorem).

Now we abelianize the singular simplicial complex $(Sing X)_\bullet$ in order to make it simpler and hence more tractable.

Definition

A formal linear combination of elements of a set $S \in$ Set is a function

$$a : S \to \mathbb{Z}$$

such that only finitely many of the values $a_s \in \mathbb{Z}$ are non-zero.

Identifying an element $s \in S$ with the function $S \to \mathbb{Z}$, which sends $s$ to $1 \in \mathbb{Z}$ and all other elements to 0, this is written as

$$a = \sum_{s \in S} a_s \cdot s \,.$$

In this expression one calls $a_s \in \mathbb{Z}$ the coefficient of $s$ in the formal linear combination.

Remark

For $S \in$ Set, the group of formal linear combinations $\mathbb{Z}[S]$ is the group whose underlying set is that of formal linear combinations, def. , and whose group operation is the pointwise addition in $\mathbb{Z}$:

$$(\sum_{s \in S} a_s \cdot s) + (\sum_{s \in S} b_s \cdot s) = \sum_{s \in S} (a_s + b_s) \cdot s \,.$$

For the present purpose the following statement may be regarded as just introducing different terminology for the group of formal linear combinations:

Proposition

The group $\mathbb{Z}[S]$ is the free abelian group on $S$.

Definition

For $S_\bullet$ a simplicial set, def. , the free abelian group $\mathbb{Z}[S_n]$ is called the group of (simplicial) $n$-chains on $S$.

Definition

For $X$ a topological space, an $n$-chain on the singular simplicial complex $Sing X$ is called a singular $n$-chain on $X$.

This construction makes the sets of simplices into abelian groups. But this allows to formally add the different face maps in the simplicial set to one single boundary map:

Definition

For $S$ a simplicial set, its alternating face map differential in degree $n$ is the linear map

$$\partial : \mathbb{Z}[S_n] \to \mathbb{Z}[S_{n-1}]$$

defined on basis elements $\sigma \in S_n$ to be the alternating sum of the simplicial face maps:

(1)$$\partial \sigma \coloneqq \sum_{k = 0}^n (-1)^k d_k \sigma \,.$$

Proposition

The simplicial identity, def. part (1), implies that the alternating sum boundary map of def. squares to 0:

$$\partial \circ \partial = 0 \,.$$

Proof

By linearity, it is sufficient to check this on a basis element $\sigma \in S_n$. There we compute as follows:

$$\begin{aligned} \partial \partial \sigma & = \partial \left( \sum_{j = 0}^n (-1)^j d_j \sigma \right) \\ & = \sum_{j=0}^n \sum_{i = 0}^{n-1} (-1)^{i+j} d_i d_j \sigma \\ & = \sum_{0 \leq i \lt j \leq n} (-1)^{i+j} d_i d_j \sigma + \sum_{0 \leq j \leq i \lt n} (-1)^{i + j} d_i d_j \sigma \\ & = \sum_{0 \leq i \lt j \leq n} (-1)^{i+j} d_{j-1} d_i \sigma + \sum_{0 \leq j \leq i \lt n} (-1)^{i + j} d_i d_j \sigma \\ & = - \sum_{0 \leq i \leq j \lt n} (-1)^{i+j} d_{j} d_i \sigma + \sum_{0 \leq j \leq i \lt n} (-1)^{i + j} d_i d_j \sigma \\ & = 0 \end{aligned} \,.$$

Here

1. the first equality is (1);

2. the second is (1) together with the linearity of $d$;

3. the third is obtained by decomposing the sum into two summands;

4. the fourth finally uses the simplicial identity def. (1) in the first summand;

5. the fifth relabels the summation index $j$ by $j +1$;

6. the last one observes that the resulting two summands are negatives of each other.

Example

Let $X$ be a topological space. Let $\sigma^1 : \Delta^1 \to X$ be a singular 1-simplex, regarded as a 1-chain

$$\sigma^1 \in C_1(X) \,.$$

Then its boundary $\partial \sigma \in H_0(X)$ is

$$\partial \sigma^1 = \sigma(0) -\sigma(1)$$

or graphically (using notation as for orientals)

$$\partial \left( \sigma(0) \stackrel{\sigma}{\to} \sigma(1) \right) = (\sigma(0)) - (\sigma(1)) \,.$$

In particular $\sigma$ is a 1-cycle precisely if $\sigma(0) = \sigma(1)$, hence precisely if $\sigma$ is a loop.

Let $\sigma^2 : \Delta^2 \to X$ be a singular 2-chain. The boundary is

$$\partial \left( \array{ && \sigma(1) \\ & {}^{\mathllap{\sigma(0,1)}}\nearrow & \Downarrow^{\mathrlap{\sigma}}& \searrow^{\mathrlap{\sigma^{1,2}}} \\ \sigma(0) &&\underset{\sigma(0,2)}{\to}&& \sigma(2) } \right) = \left( \array{ && \sigma(1) \\ & {}^{\mathllap{\sigma(0,1)}}\nearrow & & \\ \sigma(0) } \right) - \left( \array{ && \\ & & & \\ \sigma(0) &\underset{\sigma(0,2)}{\to}& \sigma(2) } \right) + \left( \array{ && \sigma(1) \\ & & & \searrow^{\mathrlap{\sigma^{1,2}}} \\ && && \sigma(2) } \right) \,.$$

Hence the boundary of the boundary is:

$$\begin{aligned} \partial \partial \sigma &= \partial \left( \left( \array{ && \sigma(1) \\ & {}^{\mathllap{\sigma(0,1)}}\nearrow & & \\ \sigma(0) } \right) - \left( \array{ && \\ & & & \\ \sigma(0) &\underset{\sigma(0,2)}{\to}& \sigma(2) } \right) + \left( \array{ && \sigma(1) \\ & & & \searrow^{\mathrlap{\sigma^{1,2}}} \\ && && \sigma(2) } \right) \right) \\ & = \left( \array{ && \\ & & & \\ \sigma(0) } \right) - \left( \array{ && \sigma(1) \\ & & & \\ } \right) - \left( \array{ && \\ & & & \\ \sigma(0) && } \right) + \left( \array{ && \\ & & & \\ && \sigma(2) } \right) + \left( \array{ && \sigma(1) \\ & & & \\ && && } \right) - \left( \array{ && \\ & & & \\ && && \sigma(2) } \right) \\ & = 0 \end{aligned}$$

Definition

For $S$ a simplicial set, we call the collection

1. of abelian groups of chains $C_n(S) \coloneqq \mathbb{Z}[S_n]$, prop. ;

2. and boundary homomorphisms $\partial_n : C_{n+1}(S) \to C_n(X)$, def.

(for all $n \in \mathbb{N}$) the alternating face map chain complex of $S$:

$$C_\bullet(S) = [ \cdots \stackrel{\partial_2}{\to} \mathbb{Z}[S_2] \stackrel{\partial_1}{\to} \mathbb{Z}[S_1] \stackrel{\partial_0}{\to} \mathbb{Z}[S_0] ] \,.$$

Specifically for $S = Sing X$ we call this the singular chain complex of $X$.

This motivates the general definition:

Definition

A chain complex of abelian groups $C_\bullet$ is a collection $\{C_n \in Ab\}_{n}$ of abelian groups together with group homomorphisms $\{\partial_n : C_{n+1} \to C_n\}$ such that $\partial \circ \partial = 0$.

We turn to this definition in more detail in the next section. The thrust of this construction lies in the fact that the chain complex $C_\bullet(Sing X)$ remembers the abelianized fundamental group of $X$, as well as aspects of the higher homotopy groups: in its chain homology.

Definition

For $C_\bullet(S)$ a chain complex as in def. , and for $n \in \mathbb{N}$ we say

• an $n$-chain of the form $\partial \sigma \in C(S)_n$ is an $n$-boundary;

• a chain $\sigma \in C_n(S)$ is an $n$-cycle if $\partial \sigma = 0$

  (every 0-chain is a 0-cycle).

By linearity of $\partial$ the boundaries and cycles form abelian sub-groups of the group of chains, and we write

$$B_n \coloneqq im(\partial_n) \subset C_n(S)$$

for the group of $n$-boundaries, and

$$Z_n \coloneqq ker(\partial_n) \subset C_(S)$$

for the group of $n$-cycles.

Remark

This means that a singular chain is a cycle if the formal linear combination of the oriented boundaries of all its constituent singular simplices sums to 0.

Remark

More generally, for $R$ any unital ring one can form the degreewise free module $R[Sing X]$ over $R$. The corresponding homology is the singular homology with coefficients in $R$, denoted $H_n(X,R)$. This generality we come to below in the next section.

Definition

For $C_\bullet(S)$ a chain complex as in def. and for $n \in \mathbb{N}$, the degree-$n$ chain homology group $H_n(C(S)) \in Ab$ is the quotient group

$$H_n(C(S)) \coloneqq \frac{ker(\partial_{n-1})}{im(\partial_n)} = \frac{Z_n}{B_n}$$

of the $n$-cycles by the $n$-boundaries – where for $n = 0$ we declare that $\partial_{-1} \coloneqq 0$ and hence $Z_0 \coloneqq C_0$.

Specifically, the chain homology of $C_\bullet(Sing X)$ is called the singular homology of the topological space $X$.

One usually writes $H_n(X, \mathbb{Z})$ or just $H_n(X)$ for the singular homology of $X$ in degree $n$.

Remark

So $H_0(C_\bullet(S)) = C_0(S)/im(\partial_0)$.

Example

For $X$ a topological space we have that the degree-0 singular homology

$$H_0(X) \simeq \mathbb{Z}[\pi_0(X)]$$

is the free abelian group on the set of connected components of $X$.

Example

For $X$ a compact connected, orientable manifold of dimension $n$ we have

$$H_n(X) \simeq \mathbb{Z} \,.$$

The precise choice of this isomorphism is a choice of orientation on $X$. With a choice of orientation, the element $1 \in \mathbb{Z}$ under this identification is called the fundamental class

$$[X] \in H_n(X)$$

of the manifold $X$.

Definition

Given a continuous map $f : X \to Y$ between topological spaces, and given $n \in \mathbb{N}$, every singular $n$-simplex $\sigma : \Delta^n \to X$ in $X$ is sent to a singular $n$-simplex

$$f_* \sigma : \Delta^n \stackrel{\sigma}{\to} X \stackrel{f}{\to} Y$$

in $Y$. This is called the push-forward of $\sigma$ along $f$. Accordingly there is a push-forward map on groups of singular chains

$$(f_*)_n : C_n(X) \to C_n(Y) \,.$$

Proposition

These push-forward maps make all diagrams of the form

$$\array{ C_{n+1}(X) &\stackrel{(f_*)_{n+1}}{\to}& C_{n+1}(Y) \\ \downarrow^{\mathrlap{\partial^X_n}} && \downarrow^{\mathrlap{\partial^Y_n}} \\ C_n(X) &\stackrel{(f_*)_n}{\to}& C_n(Y) }$$

commute.

Proof

It is in fact evident that push-forward yields a functor of singular simplicial complexes

$$f_* : Sing X \to Sing Y \,.$$

From this the statement follows since $\mathbb{Z}[-] : sSet \to sAb$ is a functor.

Therefore we have an “abelianized analog” of the notion of topological space:

Definition

For $C_\bullet, D_\bullet$ two chain complexes, def. , a homomorphism between them – called a chain map $f_\bullet : C_\bullet \to D_\bullet$ – is for each $n \in \mathbb{N}$ a homomorphism $f_n : C_n \to D_n$ of abelian groups, such that $f_n \circ \partial^C_n = \partial^D_n \circ f_{n+1}$:

$$\array{ \vdots && \vdots \\ \downarrow^{\mathrlap{\partial^C_{n+1}}} && \downarrow^{\mathrlap{\partial^D_{n+1}}} \\ C_{n+1} &\stackrel{f_{n+1}}{\to}& D_{n+1} \\ \downarrow^{\mathrlap{\partial^C_n}} && \downarrow^{\mathrlap{\partial^D_n}} \\ C_{n} &\stackrel{f_{n}}{\to}& D_{n} \\ \downarrow^{\mathrlap{\partial^C_{n-1}}} && \downarrow^{\mathrlap{\partial^D_{n-1}}} \\ \vdots && \vdots } \,.$$

Composition of such chain maps is given by degreewise composition of their components. Clearly, chain complexes with chain maps between them hence form a category – the category of chain complexes in abelian groups, – which we write

$$Ch_\bullet(Ab)) \in Cat \,.$$

Accordingly we have:

Proposition

Sending a topological space to its singular chain complex $C_\bullet(X)$, def. , and a continuous map to its push-forward chain map, prop. , constitutes a functor

$$C_\bullet(-) : Top \to Ch_\bullet(Ab)$$

from the category Top of topological spaces and continuous maps, to the category of chain complexes.

In particular for each $n \in \mathbb{N}$ singular homology extends to a functor

$$H_n(-) : Top \to Ab \,.$$

We close this section by stating the basic properties of singular homology, which make precise the sense in which it is an abelian approximation to the homotopy type of $X$. The proof of these statements requires some of the tools of homological algebra that we develop in the later chapters, as well as some tools in algebraic topology.

Proposition

If $f : X \to Y$ is a continuous map between topological spaces which is a weak homotopy equivalence, def, , then the induced morphism on singular homology groups

$$H_n(f) : H_n(X) \to H_n(Y)$$

is an isomorphism.

(A proof (via CW approximations) is spelled out for instance in (Hatcher, prop. 4.21)).

We therefore also have an “abelian analog” of weak homotopy equivalences:

Definition

For $C_\bullet, D_\bullet$ two chain complexes, a chain map $f_\bullet : C_\bullet \to D_\bullet$ is called a quasi-isomorphism if it induces isomorphisms on all homology groups:

$$f_n : H_n(C) \stackrel{\simeq}{\to} H_n(D) \,.$$

In summary: chain homology sends weak homotopy equivalences to quasi-isomorphisms. Quasi-isomorphisms of chain complexes are the abelianized analog of weak homotopy equivalences of topological spaces.

In particular we have the analog of prop. :

Proposition

The relation “There exists a quasi-isomorphism from $C_\bullet$ to $D_\bullet$.” is a reflexive and transitive relation, but it is not a symmetric relation.

Proof

Reflexivity and transitivity are evident. An explicit counter-example showing the non-symmetry is the chain map

$$\array{ \cdots &\to& 0 &\to& \mathbb{Z} &\stackrel{\cdot 2}{\to}& \mathbb{Z} &\to& 0 &\to& \cdots \\ \cdots && \downarrow && \downarrow && \downarrow && \downarrow && \cdots \\ \cdots &\to& 0 &\to& 0 &\to& \mathbb{Z}/2\mathbb{Z} &\to& 0 &\to& \cdots }$$

from the chain complex concentrated on the morphism of multiplication by 2 on integers, to the chain complex concentrated on the cyclic group of order 2.

This clearly induces an isomorphism on all homology groups. But there is not even a non-zero chain map in the other direction, since there is no non-zero group homomorphism $\mathbb{Z}/2\mathbb{Z} \to \mathbb{Z}$.

Accordingly, as for homotopy types of topological spaces, in homological algebra one regards two chain complexes $C_\bullet$, $D_\bullet$ as essentially equivalent – “of the same weak homology type” – if there is a zigzag of quasi-isomorphisms

$$C_\bullet \leftarrow \to \leftarrow \cdots \to D_\bullet$$

between them. This is made precise by the central notion of the derived category of chain complexes. We turn to this below in section Derived categories and derived functors.

But quasi-isomorphisms are a little coarser than weak homotopy equivalences. The singular chain functor $C_\bullet(-)$ forgets some of the information in the homotopy types of topological spaces. The following series of statements characterizes to some extent what exactly is lost when passing to singular homology, and which information is in fact retained.

First we need a comparison map:

Definition

(Hurewicz homomorphism)

For $(X,x)$ a pointed topological space, the Hurewicz homomorphism is the function

$$\Phi : \pi_k(X,x) \to H_k(X)$$

from the $k$th homotopy group of $(X,x)$ to the $k$th singular homology group defined by sending

$$\Phi : (f : S^k \to X)_{\sim} \mapsto f_*[S_k]$$

a representative singular $k$-sphere $f$ in $X$ to the push-forward along $f$ of the fundamental class $[S_k] \in H_k(S^k)$, example .

Proposition

For $X$ a topological space the Hurewicz homomorphism in degree 0 exhibits an isomorphism between the free abelian group $\mathbb{Z}[\pi_0(X)]$ on the set of path connected components of $X$ and the degree-0 singular homlogy:

$$\mathbb{Z}[\pi_0(X)] \simeq H_0(X) \,.$$

Since a homotopy group in positive degree depends on the homotopy type of the connected component of the base point, while the singular homology does not depend on a basepoint, it is interesting to compare these groups only for the case that $X$ is connected.

Proposition

For $X$ a path-connected topological space the Hurewicz homomorphism in degree 1

$$\Phi : \pi_1(X,x) \to H_1(X)$$

is surjective. Its kernel is the commutator subgroup of $\pi_1(X,x)$. Therefore it induces an isomorphism from the abelianization $\pi_1(X,x)^{ab} \coloneqq \pi_1(X,x)/[\pi_1,\pi_1]$:

$$\pi_1(X,x)^{ab} \stackrel{\simeq}{\to} H_1(X) \,.$$

For higher connected $X$ we have the

Theorem

If $X$ is (n-1)-connected for $n \geq 2$ then

$$\Phi : \pi_n(X,x) \to H_n(X)$$

is an isomorphism.

This is known as the Hurewicz theorem.

This gives plenty of motivation for studying

1. chain complexes

2. chain homology

3. quasi-isomorphism

of chain complexes. This is essentially what homological algebra is about. In the next section we start to develop these notions more systematically.

Chain complexes

Chain complexes of modules with chain maps between them form a category, the category of chain complexes, which is where all of homological algebra takes place. We first construct this category and discuss its most fundamental properties in 3) Categories of chain complexes . Then we consider more interesting properties of this category: the most elementary and still already profoundly useful is the phenomenon of exact sequences and specifically of homology exact sequences, discussed in 4) Homology exact sequences. In 5) Homotopy fiber sequences and mapping cones we explain how these are the shadow under the homology functor of homotopy fiber sequences of chain complexes constructed using mapping cones. The construction of the connecting homomorphism obtained this way may be understood as a special case of the basic diagram chasing lemmas in double complexes, such as the snake lemma, which we discuss in 6) Double complexes and the diagram chasing lemmas.

This serves to provide a rich set of tools that is needed when in the next chaper Abelian homotopy theory we turn to the actual category of interest, which is not quite that of chain complexes and chain maps, but the localization of this at the quasi-isomorphisms: the derived category.

3) Categories of chain complexes

In def. we had encountered complexes of singular chains, of formal linear combinations of simplices in a topological space. Here we discuss such chain complexes in their own right in a bit more depth.

Also, above a singular chain was taken to be a formal sum of singular simplices with coefficients in the abelian group of integers $\mathbb{Z}$. It is just as straightforward, natural and useful to allow the coefficients to be an arbitrary abelian group $A$, or in fact to be a module over a ring. We have to postpone proper discussion of motivating examples for this step below in chapter III and chapter IV, but the reader eager to see a deeper motivation right now might look at Modules – As generalized vector bundles. See also the archetypical example below.

So we start by developing a bit of the theory of abelian groups, rings and modules.

Definition

Write Ab $\in$ Cat for the category of abelian groups and group homomorphisms between them:

• an object is a group $A$ such that for all elements $a_1, a_2 \in A$ we have that the group product of $a_1$ with $a_2$ is the same as that of $a_2$ with $a_1$, which we write $a_1 + a_2 \in A$ (and the neutral element is denoted by $0 \in A$);

• a morphism $\phi : A_1 \to A_2$ is a group homomorphism, hence a function of the underlying sets, such that for all elements as above $\phi(a_1 + a_2) = \phi(a_1) + \phi(a_2)$.

Among the basic constructions that produce new abelian groups from given ones are the tensor product of abelian groups and the direct sum of abelian groups. These we discuss now.

Definition

For $A$, $B$ and $C$ abelian groups and $A \times B$ the cartesian product group, a bilinear map

$$f : A \times B \to C$$

is a function of the underlying sets which is linear – hence is a group homomorphism – in each argument separately.

Remark

In terms of elements this means that a bilinear map $f : A \times B \to C$ is a function of sets that satisfies for all elements $a_1, a_2 \in A$ and $b_1, b_2 \in B$ the two relations

$$f(a_1 + a_2, b_1) = f(a_1,b_1) + f(a_2, b_1)$$

and

$$f(a_1, b_1 + b_2) = f(a_1, b_1) + f(a_1, b_2) \,.$$

Notice that this is not a group homomorphism out of the product group. The product group $A \times B$ is the group whose elements are pairs $(a,b)$ with $a \in A$ and $b \in B$, and whose group operation is

$$(a_1, b_1) + (a_2, b_2) = (a_1 + a_2 \;,\; b_1 + b_2) \,.$$

A group homomorphism

$$\phi : A \times B \to C$$

hence satisfies

$$\phi( a_1+a_2, b_1 + b_2 ) = \phi(a_1,b_1) + \phi(a_2, b_2)$$

and hence in particular

$$\phi( a_1+a_2, b_1 ) = \phi(a_1,b_1) + \phi(a_2, 0)$$

$$\phi( a_1, b_1 + b_2 ) = \phi(a_1,b_1) + \phi(0, b_2)$$

which is (in general) different from the behaviour of a bilinear map.

Definition

For $A, B$ two abelian groups, their tensor product of abelian groups is the abelian group $A \otimes B$ which is the quotient group of the free group on the product (direct sum) $A \times B$ by the relations

• $(a_1,b)+(a_2,b)\sim (a_1+a_2,b)$

• $(a,b_1)+(a,b_2)\sim (a,b_1+b_2)$

for all $a, a_1, a_2 \in A$ and $b, b_1, b_2 \in B$.

In words: it is the group whose elements are presented by pairs of elements in $A$ and $B$ and such that the group operation for one argument fixed is that of the other group in the other argument.

Remark

There is a canonical function of the underlying sets

$$A \times B \stackrel{\otimes}{\to} A \otimes B \,.$$

On elements this sends $(a,b)$ to the equivalence class that it represents under the above equivalence relations.

Proposition

A function of underlying sets $f : A \times B \to C$ is a bilinear function precisely if it factors by the morphism of through a group homomorphism $\phi : A \otimes B \to C$ out of the tensor product:

$$f : A \times B \stackrel{\otimes}{\to} A \otimes B \stackrel{\phi}{\to} C \,.$$

Proposition

Equipped with the tensor product $\otimes$ of def. Ab becomes a monoidal category.

The unit object in $(Ab, \otimes)$ is the additive group of integers $\mathbb{Z}$.

This means:

1. forming the tensor product is a functor in each argument

   $$A \otimes (-) : Ab \to Ab \,,$$

2. there is an associativity natural isomorphism $(A \otimes B) \otimes C \stackrel{\simeq}{\to} A \otimes (B \otimes C)$ which is “coherent” in the sense that all possible ways of using it to rebracket a given expression are equal.

3. There is a unit natural isomorphism $A \otimes \mathbb{Z} \stackrel{\simeq}{\to} A$ which is compatible with the asscociativity isomorphism in the evident sense.

Proof

To see that $\mathbb{Z}$ is the unit object, consider for any abelian group $A$ the map

$$A \otimes \mathbb{Z} \to A$$

which sends for $n \in \mathbb{N} \subset \mathbb{Z}$

$$(a, n) \mapsto n \cdot a \coloneqq \underbrace{a + a + \cdots + a}_{n\;summands} \,.$$

Due to the quotient relation defining the tensor product, the element on the left is also equal to

$$(a, n) = (a, \underbrace{1 + 1 \cdots + 1}_{n\; summands}) = \underbrace{ (a,1) + (a,1) + \cdots + (a,1) }_{n\; summands} \,.$$

This shows that $A \otimes \mathbb{Z} \to A$ is in fact an isomorphism.

The other properties are similarly direct to check.

We see simple but useful examples of tensor products of abelian groups put to work below in the context of example and then in many of the applications to follow. An elementary but not entirely trivial example that may help to illustrate the nature of the tensor product is the following.

Example

For $a,b \in \mathbb{N}$ and positive, we have

$$\mathbb{Z}_a \oplus \mathbb{Z}_b \simeq \mathbb{Z}_{LCM(a,b)} \,,$$

where $LCM(-,-)$ denotes the least common multiple, whereas

$$\mathbb{Z}_a \otimes \mathbb{Z}_b \simeq \mathbb{Z}_{GCD(a,b)} \,,$$

where $GCD(-,-)$ denotes the greatest common divisor.

Definition

Let $I \in$ Set be a set and $\{A_i\}_{i \in I}$ an $I$-indexed family of abelian groups. The direct sum $\oplus_{i \in I} \in Ab$ is the coproduct of these objects in Ab.

This means: the direct sum is an abelian group equipped with a collection of homomorphisms

$$\array{ A_j &&\cdots && A_k \\ & {}_{\mathllap{\iota_j} }\searrow &\cdots& \swarrow_{\mathrlap{\iota_{k}}} \\ && \oplus_{i \in I} A_i } \,,$$

which is characterized (up to unique isomorphism) by the following universal property: for every other abelian group $K$ equipped with maps

$$\array{ A_j &&\cdots && A_k \\ & {}_{\mathllap{f_j} }\searrow &\cdots& \swarrow_{\mathrlap{f_{k}}} \\ && K }$$

there is a unique homomorphism $\phi : \oplus_{i \in I} A_i \to K$ such that $f_i = \phi \circ \iota_i$ for all $i \in I$.

Explicitly in terms of elements we have:

Proposition

The direct sum $\oplus_{i \in I} A_i$ is the abelian group whose ements are formal sums

$$a_1 + a_2 + \cdots + a_k$$

of finitely many elements of the $\{A_i\}$, with addition given by componentwise addition in the corresponding $A_i$.

Example

If each $A_i = \mathbb{Z}$, then the direct sum is again the free abelian group on $I$

$$\oplus_{i \in I} \mathbb{Z} \simeq \mathbb{Z}[I] \,.$$

Proposition

The tensor product of abelian groups distributes over arbitrary direct sums:

$$A \otimes (\oplus_{i \in I} B_i) \simeq \oplus_{i \in I} A \otimes B_i \,.$$

Example

For $I \in Set$ and $A \in Ab$, the direct sum of ${\vert I\vert}$ copies of $A$ with itself is equivalently the tensor product of abelian groups of the free abelian group on $I$ with $A$:

$$\oplus_{i \in I} A \simeq (\oplus_{i \in I} \mathbb{Z}) \otimes A \simeq (\mathbb{Z}[I]) \otimes A \,.$$

Remark

Together, tensor product and direct sum of abelian groups make Ab into what is called a bimonoidal category.

This now gives us enough structure to define rings and consider basic examples of their modules.

Definition

A ring (unital and not-necessarily commutative) is an abelian group $R$ equipped with

1. an element $1 \in R$

2. a bilinear operation, hence a group homomorphism

   $$\cdot : R \otimes R \to R$$

   out of the tensor product of abelian groups,

such that this is associative and unital with respect to 1.

Remark

The fact that the product is a bilinear map is the distributivity law: for all $r, r_1, r_2 \in R$ we have

$$r \cdot (r_1 + r_2) = r \cdot r_1 + r \cdot r_2$$

and

$$(r_1 + r_2) \cdot r = (r_1 + r_2) \cdot r \,.$$

Example

• The integers $\mathbb{Z}$ are a ring under the standard addition and multiplication operation.

• For each $n$, this induces a ring structure on the cyclic group $\mathbb{Z}_n$, given by operations in $\mathbb{Z}$ modulo $n$.

• The rational numbers $\mathbb{Q}$, real numbers $\mathbb{R}$ and complex numbers are rings under their standard operations (in fact these are even fields).

Example

For $R$ a ring, the polynomials

$$r_0 + r_1 x + r_2 x^2 + \cdots + r_n x^n$$

(for arbitrary $n \in\mathbb{N}$) in a variable $x$ with coefficients in $R$ form another ring, the polynomial ring denoted $R[x]$. This is the free $R$-associative algebra on a single generator $x$.

Example

For $R$ a ring and $n \in \mathbb{N}$, the set $M(n,R)$ of $n \times n$-matrices with coefficients in $R$ is a ring under elementwise addition and matrix multiplication.

Example

For $X$ a topological space, the set of continuous functions $C(X,\mathbb{R})$ or $C(X,\mathbb{C})$ with values in the real numbers or complex numbers is a ring under pointwise (points in $X$) addition and multiplication.

Just as an outlook and a suggestion for how to think geometrically of the objects appearing here, we mention the following.

Remark

The Gelfand duality theorem says that if one remembers certain extra structure on the rings of functions $C(X, \mathbb{C})$ in example – called the structure of a C-star algebra, then this construction

$$C(-,\mathbb{C}) : Top \stackrel{\simeq}{\to} C^\ast Alg^{op} \stackrel{forget}{\to} Ring^op$$

is an equivalence of categories between that of topological spaces, and the opposite category of $C^\ast$-algebras. Together with remark further below this provides a useful dual geometric way of thinking about the theory of modules.

From now on and throughout, we take $R$ to be a commutative ring.

Definition

A module $N$ over a ring $R$ is

1. an object $N \in$ Ab, hence an abelian group;

2. equipped with a morphism

   $$\alpha : R \otimes N \to N$$

   in Ab; hence a function of the underlying sets that sends elements

   $$(r,n) \mapsto r n \coloneqq \alpha(r,n)$$

   and which is a bilinear function in that it satisfies

   $$(r, n_1 + n_2) \mapsto r n_1 + r n_2$$

   and

   $$(r_1 + r_2, n) \mapsto r_1 n + r_2 n$$

   for all $r, r_1, r_2 \in R$ and $n,n_1, n_2 \in N$;

3. such that the diagram

   $$\array{ R \otimes R \otimes N &\stackrel{\cdot_R \otimes Id_N}{\to}& R \otimes N \\ {}^{\mathllap{Id_R \otimes \alpha}}\downarrow && \downarrow^{\mathrlap{\alpha}} \\ R \otimes N &\to& N }$$

   commutes in Ab, which means that for all elements as before we have

   $$(r_1 \cdot r_2) n = r_1 (r_2 n) \,.$$

4. such that the diagram

   $$\array{ 1 \otimes N &&\stackrel{1 \otimes id_N}{\to}&& R \otimes N \\ & \searrow && \swarrow_{\mathrlap{\alpha}} \\ && N }$$

   commutes, which means that on elements as above

   $$1 \cdot n = n \,.$$

Example

The ring $R$ is naturally a module over itself, by regarding its multiplication map $R \otimes R \to R$ as a module action $R \otimes N \to N$ with $N \coloneqq R$.

Example

More generally, for $n \in \mathbb{N}$ the $n$-fold direct sum of the abelian group underlying $R$ is naturally a module over $R$

$$R^n \coloneqq R^{\oplus_n} \coloneqq \underbrace{R \oplus R \oplus \cdots \oplus R}_{n\;summands} \,.$$

The module action is componentwise:

$$r \cdot (r_1, r_2, \cdots, r_n) = (r \cdot r_1, r\cdot r_2, \cdots, r \cdot r_n) \,.$$

Example

Even more generally, for $I \in$ Set any set, the direct sum $\oplus_{i \in I} R$ is an $R$-module.

This is the free module (over $R$) on the set $S$.

The set $I$ serves as the basis of a free module: a general element $v \in \oplus_i R$ is a formal linear combination of elements of $I$ with coefficients in $R$.

For special cases of the ring $R$, the notion of $R$-module is equivalent to other notions:

Example

For $R = \mathbb{Z}$ the integers, an $R$-module is equivalently just an abelian group.

Example

For $R = k$ a field, an $R$-module is equivalently a vector space over $k$.

Every finitely-generated free $k$-module is a free module, hence every finite dimensional vector space has a basis. For infinite dimensions this is true if the axiom of choice holds.

Example

For $N$ a module and $\{n_i\}_{i \in I}$ a set of elements, the linear span

$$\langle n_i\rangle_{i \in I} \hookrightarrow N \,,$$

(hence the completion of this set under addition in $N$ and multiplication by $R$) is a submodule of $N$.

Example

Consider example for the case that the module is $N = R$, the ring itself, as in example . Then a submodule is equivalently (called) an ideal of $R$.

Definition

Write $R$Mod for the category or $R$-modules and $R$-linear maps between them.

Example

For $R = \mathbb{Z}$ we have $\mathbb{Z} Mod \simeq Ab$.

Example

Let $X$ be a topological space and let

$$R \coloneqq C(X,\mathbb{C})$$

be the ring of continuous functions on $X$ with values in the complex numbers.

Given a complex vector bundle $E \to X$ on $X$, write $\Gamma(E)$ for its set of continuous sections. Since for each point $x \in X$ the fiber $E_x$ of $E$ over $x$ is a $\mathbb{C}$-module (by example ), $\Gamma(X)$ is a $C(X,\mathbb{C})$-module.

Just as an outlook and a suggestion for how to think of modules geometrically, we mention the following.

Remark

The Serre-Swan theorem says that if $X$ is Hausdorff and compact with ring of functions $C(X,\mathbb{C})$ – as in remark above – then $\Gamma(X)$ is a projective $C(X,\mathbb{C})$-module and indeed there is an equivalence of categories between projective $C(X,\mathbb{C})$-modules and complex vector bundles over $X$. (We introduce the notion of projective modules below in Derived categories and derived functors.)

We now discuss a bunch of properties of the category $R$Mod which together will show that there is a reasonable concept of chain complexes of $R$-modules, in generalization of how there is a good concept of chain complexes of abelian groups. In a more abstract category theoretical context than we invoke here, all of the following properties are summarized in the following statement.

Theorem

Let $R$ be a commutative ring. Then $R Mod$ is an abelian category.

But for the moment we ignore this further abstraction and just consider the following list of properties.

Definition

An object in a category which is both an initial object and a terminal object is called a zero object.

Remark

This means that $0 \in \mathcal{C}$ is a zero object precisely if for every other object $A$ there is a unique morphism $A \to 0$ to the zero object as well as a unique morphism $0 \to A$ from the zero object.

Proposition

The trivial group is a zero object in Ab.

The trivial module is a zero object in $R$Mod.

Proof

Clearly the 0-module $0$ is a terminal object, since every morphism $N \to 0$ has to send all elements of $N$ to the unique element of $0$, and every such morphism is a homomorphism. Also, 0 is an initial object because a morphism $0 \to N$ always exists and is unique, as it has to send the unique element of 0, which is the neutral element, to the neutral element of $N$.

Definition

In a category with an initial object $0$ and pullbacks, the kernel $ker(f)$ of a morphism $f: A \to B$ is the pullback $ker(f) \to A$ along $f$ of the unique morphism $0 \to B$

$$\array{ ker(f) &\to& 0 \\ {}^{\mathllap{p}}\downarrow && \downarrow \\ A &\stackrel{f}{\to}& B } \,.$$

Remark

More explicitly, this characterizes the object $ker(f)$ as the object (unique up to unique isomorphism) that satisfies the following universal property:

for every object $C$ and every morphism $h : C \to A$ such that $f\circ h = 0$ is the zero morphism, there is a unique morphism $\phi : C \to ker(f)$ such that $h = p\circ \phi$.

Example

In the category Ab of abelian groups, the kernel of a group homomorphism $f : A \to B$ is the subgroup of $A$ on the set $f^{-1}(0)$ of elements of $A$ that are sent to the zero-element of $B$.

Example

More generally, for $R$ any ring, this is true in $R$Mod: the kernel of a morphism of modules is the preimage of the zero-element at the level of the underlying sets, equipped with the unique sub-module structure on that set.

Definition

In a category with zero object, the cokernel of a morphism $f : A \to B$ is the pushout $coker(f)$ in

$$\array{ A &\stackrel{f}{\to}& B \\ \downarrow && \downarrow^{\mathrlap{i}} \\ 0 &\to& coker(f) } \,.$$

Remark

More explicitly, this characterizes the object $coker(f)$ as the object (unique up to unique isomorphism) that satisfies the following universal property:

for every object $C$ and every morphism $h : B \to C$ such that $h \circ f = 0$ is the zero morphism, there is a unique morphism $\phi : coker(f) \to C$ such that $h = \phi \circ i$.

Example

In the category Ab of abelian groups the cokernel of a morphism $f : A \to B$ is the quotient group of $B$ by the image (of the underlying morphism of sets) of $f$.

Proposition

$R Mod$ has all kernels. The kernel of a homomorphism $f : N_1 \to N_2$ is the set-theoretic preimage $U(f)^{-1}(0)$ equipped with the induced $R$-module structure.

$R Mod$ has all cokernels. The cokernel of a homomorphism $f : N_1 \to N_2$ is the quotient abelian group

$$coker f = \frac{N_2}{im(f)}$$

of $N_2$ by the image of $f$.

The reader unfamiliar with the general concept of monomorphism and epimorphism may take the following to define these in Ab to be simply the injections and surjections.

Proposition

$U : R Mod \to Set$ preserves and reflects monomorphisms and epimorphisms:

A homomorphism $f : N_1 \to N_2$ in $R Mod$ is a monomorphism / epimorphism precisely if $U(f)$ is an injection / surjection.

Proof

Suppose that $f$ is a monomorphism, hence that $f : N_1 \to N_2$ is such that for all morphisms $g_1, g_2 : K \to N_1$ such that $f \circ g_1 = f \circ g_2$ already $g_1 = g_2$. Let then $g_1$ and $g_2$ be the inclusion of submodules generated by a single element $k_1 \in K$ and $k_2 \in K$, respectively. It follows that if $f(k_1) = f(k_2)$ then already $k_1 = k_2$ and so $f$ is an injection. Conversely, if $f$ is an injection then its image is a submodule and it follows directly that $f$ is a monomorphism.

Suppose now that $f$ is an epimorphism and hence that $f : N_1 \to N_2$ is such that for all morphisms $g_1, g_2 : N_2 \to K$ such that $f \circ g_1 = f \circ g_2$ already $g_1 = g_2$. Let then $g_1 : N_2 \to \frac{N_2}{im(f)}$ be the natural projection. and let $g_2 : N_2 \to 0$ be the zero morphism. Since by construction $f \circ g_1 = 0$ and $f \circ g_2 = 0$ we have that $g_1 = 0$, which means that $\frac{N}{im(f)} = 0$ and hence that $N = im(f)$ and so that $f$ is surjective. The other direction is evident on elements.

Definition

For $N_1, N_2 \in R Mod$ two modules, define on the hom set $Hom_{R Mod}(N_1,N_2)$ the structure of an abelian group whose addition is given by argumentwise addition in $N_2$: $(f_1 + f_2) : n \mapsto f_1(n) + f_2(n)$.

Proposition

With def. $R Mod$ composition of morphisms

$$\circ : Hom(N_1, N_2) \times Hom(N_2, N_3) \to Hom(N_1,N_3)$$

is a bilinear map, hence is equivalently a morphism

$$Hom(N_1, N_2) \otimes Hom(N_2,N_3) \to Hom(N_1, N_3)$$

out of the tensor product of abelian groups.

This makes $R Mod$ into an Ab-enriched category.

Proof

Linearity of composition in the second argument is immediate from the pointwise definition of the abelian group structure on morphisms. Linearity of the composition in the first argument comes down to linearity of the second module homomorphism.

Remark

In fact $R Mod$ is even a closed category, but this we do not need for showing that it is abelian.

Prop. and prop. together say that:

Corollary

$R Mod$ is an pre-additive category.

Proposition

$R Mod$ has all products and coproducts, being direct products and direct sums.

The products are given by cartesian product of the underlying sets with componentwise addition and $R$-action.

The direct sum is the subobject of the product consisting of tuples of elements such that only finitely many are non-zero.

Proof

The defining universal properties are directly checked. Notice that the direct product $\prod_{i \in I} N_i$ consists of arbitrary tuples because it needs to have a projection map

$$p_j : \prod_{i \in I} N_i \to N_j$$

to each of the modules in the product, reproducing all of a possibly infinite number of non-trivial maps $\{K \to N_j\}$. On the other hand, the direct sum just needs to contain all the modules in the sum

$$\iota_j : N_j \to \oplus_{i \in I} N_i$$

and since, being a module, it needs to be closed only under addition of finitely many elements, so it consists only of linear combinations of the elements in the $N_j$, hence of finite formal sums of these.

Together cor. and prop. say that:

Corollary

$R Mod$ is an additive category.

Proposition

In $R Mod$

• every monomorphism is the kernel of its cokernel;

• every epimorphism is the cokernel of its kernel.

Proof

Using prop. this is directly checked on the underlying sets: given a monomorphism $K \hookrightarrow N$, its cokernel is $N \to \frac{N}{K}$, The kernel of that morphism is evidently $K \hookrightarrow N$.

Now cor. and prop. imply theorem , by definition.

Now we finally have all the ingredients to talk about chain complexes of $R$-modules. The following definitions are the direct analogs of the definitions of chain complexes of abelian groups in Simplicial and singular homology above.

Definition

A ($\mathbb{Z}$-graded) chain complex in $R$Mod is

• a collection of objects $\{C_n\}_{n\in\mathbb{Z}}$,

• and of morphisms $\partial_n : C_n \to C_{n-1}$

$$\cdots \overset{\partial_3}{\to} C_2 \overset{\partial_2}{\to} C_1 \overset{\partial_1}{\to} C_0 \overset{\partial_0}{\to} C_{-1} \overset{\partial_{-1}}{\to} \cdots$$

such that

$$\partial_n \circ \partial_{n+1} = 0$$

(the zero morphism) for all $n \in \mathbb{N}$.

Definition

For $C_\bullet$ a chain complex and $n \in \mathbb{N}$

• the morphisms $\partial_n$ are called the differentials or boundary maps;

• the elements of $C_n$ are called the $n$-chains;

• for $n \geq 1$ the elements in the kernel

  $$Z_n \coloneqq ker(\partial_{n-1})$$

  of $\partial_{n-1} : C_n \to C_{n-1}$ are called the $n$-cycles

  and for $n = 0$ we say that every 0-chain is a 0-cycle

  $$Z_0 \coloneqq C_0$$

  (equivalently we declare that $\partial_{-1} = 0$).

• the elements in the image

  $$B_n \coloneqq im(\partial_n)$$

  of $\partial_{n} : C_{n+1} \to C_{n}$ are called the $n$-boundaries;

Notice that due to $\partial \partial = 0$ we have canonical inclusions

$$0 \hookrightarrow B_n \hookrightarrow Z_n \hookrightarrow C_n \,.$$

• the cokernel

  $$H_n \coloneqq Z_n/B_n$$

  is called the degree-$n$ chain homology of $C_\bullet$.

$$0 \to B_n \to Z_n \to H_n \to 0 \,.$$

Definition

A chain map $f : V_\bullet \to W_\bullet$ is a collection of morphism $\{f_n : V_n \to W_n\}_{n \in \mathbb{Z}}$ in $\mathcal{A}$ such that all the diagrams

$$\array{ V_{n+1} &\stackrel{d^V_n}{\to}& V_n \\ \downarrow^{\mathrlap{f_{n+1}}} && \downarrow^{\mathrlap{f_{n}}} \\ W_{n+1} &\stackrel{d^W_n}{\to} & W_n }$$

commute, hence such that all the equations

$$f_n \circ d^V_n = d^W_{n+1} \circ f_{n+1}$$

hold.

Proposition

For $f : C_\bullet \to D_\bullet$ a chain map, it respects boundaries and cycles, so that for all $n \in \mathbb{Z}$ it restricts to a morphism

$$B_n(f) : B_n(C_\bullet) \to B_n(D_\bullet)$$

and

$$Z_n(f) : Z_n(C_\bullet) \to Z_n(D_\bullet) \,.$$

In particular it also respects chain homology

$$H_n(f) : H_n(C_\bullet) \to H_n(D_\bullet) \,.$$

Corollary

Conversely this means that taking chain homology is a functor

$$H_n(-) : Ch_\bullet(\mathcal{A}) \to \mathcal{A}$$

from the category of chain complexes in $\mathcal{A}$ to $\mathcal{A}$ itself.

This establishes the basic objects that we are concerned with in the following. But as before, we are not so much interested in chain complexes up to chain map isomorphism, rather, we are interested in them up to a notion of homotopy equivalence. This we begin to study in the next section Homology exact sequences and homotopy fiber sequences. But in order to formulate that neatly, it is useful to have the tensor product of chain complexes. We close this section with introducing that notion.

Definition

For $X, Y \in Ch_\bullet(\mathcal{A})$ write $X \otimes Y \in Ch_\bullet(\mathcal{A})$ for the chain complex whose component in degree $n$ is given by the direct sum

$$(X \otimes Y)_n := \oplus_{i + j = n} X_i \otimes_R Y_j$$

over all tensor products of components whose degrees sum to $n$, and whose differential is given on elements $(x,y)$ of homogeneous degree by

$$\partial^{X \otimes Y} (x, y) = (\partial^X x, y) + (-1)^{deg(x)} (x, \partial^Y y) \,.$$

Example

(square as tensor product of interval with itself)

For $R$ some ring, let $I_\bullet \in Ch_\bullet(R Mod)$ be the chain complex given by

$$I_\bullet = \left[ \cdots \to 0 \to 0 \to R \stackrel{\partial^{I}_0}{\to} R \oplus R \right] \,,$$

where $\partial^I_0 = (-id, id)$.

This is the normalized chain complex of the simplicial chain complex of the standard simplicial interval, the 1-simplex $\Delta_1$, which means: we may think of

$$I_0 = R \oplus R \simeq R[ \{(0), (1)\} ]$$

as the $R$-linear span of two basis elements labelled “$(0)$” and “$(1)$”, to be thought of as the two 0-chains on the endpoints of the interval. Similarly we may think of

$$I_1 = R \simeq R[\{(0 \to 1)\}]$$

as the free $R$-module on the single basis element which is the unique non-degenerate 1-simplex $(0 \to 1)$ in $\Delta^1$.

Accordingly, the differential $\partial^I_0$ is the oriented boundary map of the interval, taking this basis element to

$$\partial^I_0 : (0 \to 1) \mapsto (1) - (0)$$

and hence a general element $r\cdot(0 \to 1)$ for some $r \in R$ to

$$\partial^I_0 : r\cdot(0 \to 1) \mapsto r\cdot (1) - r\cdot(0) \,.$$

We now write out in full details the tensor product of chain complexes of $I_\bullet$ with itself, according to def. :

$$S_\bullet \coloneqq I_\bullet \otimes I_\bullet \,.$$

By definition and using the above choice of basis element, this is in low degree given as follows:

$$\begin{aligned} S_0 &= I_0 \oplus I_0 \\ & = (R \oplus R) \otimes (R \oplus R) \\ & \simeq R \oplus R \oplus R \oplus R \\ & = \left\{ r_{00} \cdot ((0),(0)') + r_{01} \cdot ((0),(1)') + r_{10} \cdot ((1),(0)') + r_{11} \cdot ((1),(1)') | r_{\cdot, \cdot} \in R \right\} \end{aligned} \,,$$

where in the last line we express a general element as a linear combination of the canonical basis elements which are obtained as tensor products $(a,b) \in R\otimes R$ of the previous basis elements. Notice that by the definition of tensor product of modules we have relations like

$$r ( (0), (1)') = (r(0), (1)') = ((0), r(1)')$$

etc.

Similarly then, in degree-1 the tensor product chain complex is

$$\begin{aligned} (I \otimes I)_1 & = (I_0 \otimes I_1) \oplus (I_1 \otimes I_0) \\ & \simeq R \otimes (R \oplus R) \oplus (R \oplus R) \otimes R \\ & \simeq R \oplus R \oplus R \oplus R \\ & \simeq \left\{ r_{0} \cdot ((0),(0\to 1)') + r_{1} \cdot ((1), (0 \to 1)') + \bar r_0 \cdot ((0\to 1), (0)') + \bar r_1 \cdot ((0 \to 1), (1)') | r_{\cdot}, \bar r_{\cdot} \in R \right\} \end{aligned} \,.$$

And finally in degree 2 it is

$$\begin{aligned} (I \otimes I)_2 & \simeq I_1 \otimes I_1 \\ & \simeq R \otimes R \\ & \simeq R \\ & \simeq \left\{ r\cdot ((0 \to 1), (0 \to 1)') | r \in R \right\} \end{aligned} \,.$$

All other contributions that are potentially present in $(I \otimes I)_\bullet$ vanish (are the 0-module) because all higher terms in $I_\bullet$ are.

The tensor product basis elements appearing in the above expressions have a clear geometric interpretation: we can label a square with them as follows

$$\array{ ((0),(1)') &&\underset{((0\to 1),(0))}{\to}&& ((1),(1)') \\ \\ {}^{\mathllap{((0),(0\to 1)')}}\uparrow &&\righttoleftarrow^{((0 \to 1), (0\to 1)')}&& \uparrow^{\mathrlap{((1),(0 \to 1)')}} \\ \\ ((0),(0)') &&\underset{((0\to 1),(0)')}{\to}&& ((1),(0)') } \,.$$

This diagram indicates a cellular square and identifies its canonical singular chains with the elements of $(I \otimes I)_\bullet$. The arrows indicate the orientation. For instance the fact that

$$\begin{aligned} \partial^{I \otimes I} ((0 \to 1), (0)') & = (\partial^I (0 \to 1), (0)') + (-1)^1 ((0\to 1), \partial^I (0)) \\ & = ( (1) - (0), \;(0)' ) - 0 \\ & = ((1), (0)') - ((0), (0)') \end{aligned}$$

says that the oriented boundary of the bottom morphism is the bottom right element (its target) minus the bottom left element (its source), as indicated. Here we used that the differential of a degree-0 element in $I_\bullet$ is 0, and hence so is any tensor product with it.

Similarly the oriented boundary of the square itself is computed to

$$\begin{aligned} \partial^{I \otimes I} ((0 \to 1), (0 \to 1)') &= (\partial^I (0 \to 1), (0 \to 1)') - ((0 \to 1), \partial^I(0 \to 1)) \\ & = ((1)- (0), (0 \to 1)') - ((0 \to 1), (1)' - (0)') \\ & = ((1), (0 \to 1)') - ((0), (0 \to 1)') - ((0 \to 1), (1)') + ((0 \to 1), (0)') \end{aligned} \,,$$

which can be read as saying that the boundary is the evident boundary thought of as oriented by drawing it counterclockwise into the plane, so that the right arrow (which points up) contributes with a +1 prefactor, while the left arrow (which also points up) contributes with a -1 prefactor.

Proposition

Equipped with the standard tensor product of chain complexes $\otimes$, def. the category of chain complexes is a monoidal category $(Ch_\bullet(R Mod), \otimes)$. The unit object is the chain complex concentrated in degree 0 on the tensor unit $R$ of $R Mod$.

Definition

We write $Ch_\bullet^{ub}$ for the category of unbounded chain complexes.

Definition

For $X,Y \in Ch^{ub}_\bullet(\mathcal{A})$ any two objects, define a chain complex $[X,Y] \in Ch^{ub}_\bullet(\mathcal{A})$ to have components

$$[X,Y]_n := \prod_{i \in \mathbb{Z}} Hom_{R Mod}(X_i, Y_{i+n})$$

(the collection of degree-$n$ maps between the underlying graded modules) and whose differential is defined on homogeneously graded elements $f \in [X,Y]_n$ by

$$d f := d_Y \circ f - (-1)^{n} f \circ d_X \,.$$

This defines a functor

$$[-,-] : Ch^{ub}_\bullet(\mathcal{A})^{op} \times Ch^{ub}_\bullet(\mathcal{A}) \to Ch^{ub}_\bullet(\mathcal{A}) \,.$$

Proposition

This functor

$[-,-] : Ch^{ub}_\bullet \times Ch^{ub}_\bullet \to Ch^{ub}_\bullet$

is the internal hom of the category of chain complexes.

Proposition

The collection of cycles of the internal hom $[X,Y]_\bullet$ in degree 0 coincides with the external hom functor

$$Z_0([X,Y]) \simeq Hom_{Ch^{ub}_\bullet}(X,Y) \,.$$

The chain homology of the internal hom $[X,Y]$ in degree 0 coincides with the homotopy classes of chain maps.

Proof

By Definition the 0-cycles in $[X,Y]$ are collections of morphisms $\{f_k : X_k \to Y_k\}$ such that

$$f_{k+1} \circ d_X = d_Y \circ f_k \,.$$

This is precisely the condition for $f$ to be a chain map.

Similarly, the boundaries in degree 0 are precisely the collections of morphisms of the form

$$\lambda_{k+1} \circ d_X + d_Y \circ \lambda_k$$

for a collection of maps $\{\lambda_k : X_k \to Y_{k+1}\}$. This are precisely the null homotopies.

Proposition

The monoidal category $(Ch_\bullet, \otimes)$ is a closed monoidal category, the internal hom is the standard internal hom of chain complexes.

4) Homology exact sequences

With the basic definition of the category of chain complexes in hand, we now consider the first application, which is as simple as it is of ubiquituous use in mathematics: long exact sequences in homology. This is the “abelianization”, in the sense of the discussion in 2) above, of what in homotopy theory are long exact sequences of homotopy groups. But both concepts, in turn, are just the shadow on homology groups/homotopy groups, respectively of homotopy fiber sequences of the underlying chain complexes/topological spaces themselves. Since these are even more useful, in particular in chapter III) below, we discuss below in 5) how to construct these using chain homotopy and mapping cones.

First we need the fundamental notion of exact sequences. As before, we fix some commutative ring $R$ throughout and consider the category of modules over $R$, which we will abbreviate

$$\mathcal{A} \coloneqq R Mod \,.$$

Definition

An exact sequence in $\mathcal{A}$ is a chain complex $C_\bullet$ in $\mathcal{A}$ with vanishing chain homology in each degree:

$$\forall n \in \mathbb{N} . H_n(C) = 0 \,.$$

Definition

A short exact sequence is an exact sequence, def. of the form

$$\cdots \to 0 \to 0 \to A \to B \to C \to 0 \to 0 \to \cdots \,.$$

One usually writes this just “$0 \to A \to B \to C \to 0$” or even just “$A \to B \to C$”.

Remark

A general exact sequence is sometimes called a long exact sequence, to distinguish from the special case of a short exact sequence.

Beware that there is a difference between $A \to B \to C$ being exact (at $B$) and $A \to B \to C$ being a “short exact sequence” in that $0 \to A \to B \to C \to 0$ is exact at $A$, $B$ and $C$. This is illustrated by the following proposition.

Proposition

Explicitly, a sequence of morphisms

$$0 \to A \stackrel{i}\to B \stackrel{p}\to C \to 0$$

in $\mathcal{A}$ is short exact, def. , precisely if

1. $i$ is a monomorphism,

2. $p$ is an epimorphism,

3. and the image of $i$ equals the kernel of $p$ (equivalently, the coimage of $p$ equals the cokernel of $i$).

Proof

The third condition is the definition of exactness at $B$. So we need to show that the first two conditions are equivalent to exactness at $A$ and at $C$.

This is easy to see by looking at elements when $\mathcal{A} \simeq R$Mod, for some ring $R$ (and the general case can be reduced to this one using one of the embedding theorems):

The sequence being exact at

$$0 \to A \to B$$

means, since the image of $0 \to A$ is just the element $0 \in A$, that the kernel of $A \to B$ consists of just this element. But since $A \to B$ is a group homomorphism, this means equivalently that $A \to B$ is an injection.

Dually, the sequence being exact at

$$B \to C \to 0$$

means, since the kernel of $C \to 0$ is all of $C$, that also the image of $B \to C$ is all of $C$, hence equivalently that $B \to C$ is a surjection.

Example

Let $\mathcal{A} = \mathbb{Z}$Mod $\simeq$ Ab. For $n \in \mathbb{N}$ with $n \geq 1$ let $\mathbb{Z} \stackrel{\cdot n}{\to} \mathbb{Z}$ be the linear map/homomorphism of abelian groups which acts by the ordinary multiplication of integers by $n$. This is clearly an injection. The cokernel of this morphism is the projection to the quotient group, which is the cyclic group $\mathbb{Z}_n \coloneqq \mathbb{Z}/n\mathbb{Z}$. Hence we have a short exact sequence

$$0 \to \mathbb{Z} \stackrel{\cdot n}{\to} \mathbb{Z} \to \mathbb{Z}_n \,.$$

A typical use of a long exact sequence, notably of the homology long exact sequence to be discussed, is that it allows to determine some of its entries in terms of others.

The characterization of short exact sequences in prop. is one example for this. Another is this:

Proposition

If part of an exact sequence looks like

$$\cdots \to 0 \to C_{n+1} \stackrel{\partial_n}{\to} C_n \to 0 \to \cdots \,,$$

then $\partial_n$ is an isomorphism and hence

$$C_{n+1} \simeq C_n \,.$$

Often it is useful to make the following strengthening of short exactness explicit.

Definition

A short exact sequence $0\to A \stackrel{i}{\to} B \stackrel{p}{\to} C\to 0$ in $\mathcal{A}$ is called split if either of the following equivalent conditions hold

1. There exists a section of $p$, hence a homomorphism $s \colon C\to B$ such that $p \circ s = id_C$.

2. There exists a retract of $i$, hence a homomorphism $r \colon B\to A$ such that $r \circ i = id_A$.

3. There exists an isomorphism of sequences with the sequence

   $$0\to A\to A\oplus C\to C\to 0$$

   given by the direct sum and its canonical injection/projection morphisms.

Proposition

(splitting lemma)

The three conditions in def. are indeed equivalent.

Proof

It is clear that the third condition implies the first two: take the section/retract to be given by the canonical injection/projection maps that come with a direct sum.

Conversely, suppose we have a retract $r \colon B \to A$ of $i \colon A \to B$. Write $P \colon B \stackrel{r}{\to} A \stackrel{i}{\to} B$ for the composite. Notice that by $r\circ i = id$ this is an idempotent: $P \circ P = P$, hence a projector.

Then every element $b \in B$ can be decomposed as $b = (b - P(b)) + P(b)$ hence with $b - P(b) \in ker(r)$ and $P(b) \in im(i)$. Moreover this decomposition is unique since if $b = i(a)$ while at the same time $r(b) = 0$ then $0 = r(i(a)) = a$. This shows that $B \simeq im(i) \oplus ker(r)$ is a direct sum and that $i \colon A \to B$ is the canonical inclusion of $im(i)$. By exactness it then follows that $ker(r) \simeq ker(p)$ and hence that $B \simeq A \oplus C$ with the canonical inclusion and projection.

The implication that the second condition also implies the third is formally dual to this argument.

Moreover, of particular interest are exact sequences of chain complexes. We consider this concept in full beauty below in section 5). In order to motivate the discussion there we here content ourselves with the following quick definition, which already admits discussion of some of its rich consequences.

Definition

A sequence of chain maps of chain complexes

$$0 \to A_\bullet \to B_\bullet \to C_\bullet \to 0$$

is a short exact sequence of chain complexes in $\mathcal{A}$ if for each $n$ the component

$$0 \to A_n \to B_n \to C_n \to 0$$

is a short exact sequence in $\mathcal{A}$, according to def. .

Definition

Consider a short exact sequence of chain complexes as in def. . For $n \in \mathbb{Z}$, define a group homomorphism

$$\delta_n : H_n(C) \to H_{n-1}(A) \,,$$

called the $n$th connecting homomorphism of the short exact sequence, by sending

$$\delta_n : [c] \mapsto [\partial^B \hat c]_A \,,$$

where

1. $c \in Z_n(C)$ is a cycle representing the given homology group $[c]$;

2. $\hat c \in C_n(B)$ is any lift of that cycle to an element in $B_n$, which exists because $p$ is a surjection (but which no longer needs to be a cycle itself);

3. $[\partial^B \hat c]_A$ is the $A$-homology class of $\partial^B \hat c$ which is indeed in $A_{n-1} \hookrightarrow B_{n-1}$ by exactness (since $p(\partial^B \hat c) = \partial^C p(\hat c) = \partial^C c = 0$) and indeed in $Z_{n-1}(A) \hookrightarrow A_{n-1}$ since $\partial^A \partial^B \hat c = \partial^B \partial^B \hat c = 0$.

Proposition

Def. is indeed well defined in that the given map is independent of the choice of lift $\hat c$ involved and in that the group structure is respected.

Proof

To see that the construction is well-defined, let $\tilde c \in B_{n}$ be another lift. Then $p(\hat c - \tilde c) = 0$ and hence $\hat c - \tilde c \in A_n \hookrightarrow B_n$. This exhibits a homology-equivalence $[\partial^B\hat c]_A \simeq [\partial^B \tilde c]_A$ since $\partial^A(\hat c - \tilde c) = \partial^B \hat c - \partial^B \tilde c$.

To see that $\delta_n$ is a group homomorphism, let $[c] = [c_1] + [c_2]$ be a sum. Then $\hat c \coloneqq \hat c_1 + \hat c_2$ is a lift and by linearity of $\partial$ we have $[\partial^B \hat c]_A = [\partial^B \hat c_1] + [\partial^B \hat c_2]$.

Proposition

Under chain homology $H_\bullet(-)$ the morphisms in the short exact sequence together with the connecting homomorphisms yield the homology long exact sequence

$$\cdots \to H_n(A) \to H_n(B) \to H_n(C) \stackrel{\delta_n}{\to} H_{n-1}(A) \to H_{n-1}(B) \to H_{n-1}(C) \to \cdots \,.$$

Proof

Consider first the exactness of $H_n(A) \stackrel{H_n(i)}{\to} H_n(B) \stackrel{H_n(p)}{\to} H_n(C)$.

It is clear that if $a \in Z_n(A) \hookrightarrow Z_n(B)$ then the image of $[a] \in H_n(B)$ is $[p(a)] = 0 \in H_n(C)$. Conversely, an element $[b] \in H_n(B)$ is in the kernel of $H_n(p)$ if there is $c \in C_{n+1}$ with $\partial^C c = p(b)$. Since $p$ is surjective let $\hat c \in B_{n+1}$ be any lift, then $[b] = [b - \partial^B \hat c]$ but $p(b - \partial^B c) = 0$ hence by exactness $b - \partial^B \hat c \in Z_n(A) \hookrightarrow Z_n(B)$ and so $[b]$ is in the image of $H_n(A) \to H_n(B)$.

It remains to see that

1. the image of $H_n(B) \to H_n(C)$ is the kernel of $\delta_n$;

2. the kernel of $H_{n-1}(A) \to H_{n-1}(B)$ is the image of $\delta_n$.

This follows by inspection of the formula in def. . We spell out the first one:

If $[c]$ is in the image of $H_n(B) \to H_n(C)$ we have a lift $\hat c$ with $\partial^B \hat c = 0$ and so $\delta_n[c] = [\partial^B \hat c]_A = 0$. Conversely, if for a given lift $\hat c$ we have that $[\partial^B \hat c]_A = 0$ this means there is $a \in A_n$ such that $\partial^A a \coloneqq \partial^B a = \partial^B \hat c$. But then $\tilde c \coloneqq \hat c - a$ is another possible lift of $c$ for which $\partial^B \tilde c = 0$ and so $[c]$ is in the image of $H_n(B) \to H_n(C)$.

Example

The connecting homomorphism of the long exact sequence in homology induced from short exact sequences of the form in example is called a Bockstein homomorphism.

We now discuss a deeper, more conceptual way of understanding the origin of long exact sequences in homology and the nature of connecting homomorphisms. This will give first occasion to see some actual homotopy theory of chain complexes at work, and hence serves also as a motivating example for the discussions to follow in chapter III).

For this we need the notion of chain homotopy, which is the abelianized analog of the notion of homotopy of continuous maps above in def. . We now first introduce this concept by straightforwardly mimicking the construction in def. with topological spaces replaced by chain complexes. Then we use chain homotopies to construct mapping cones of chain maps. Finally we explain how these refine the above long exact sequences in homology groups to homotopy cofiber sequences of the chain complexes themselves.

A chain homotopy is a homotopy in $Ch_\bullet(\mathcal{A})$. We first give the explicit definition, the more abstract characterization is below in prop. .

Definition

A chain homotopy $\psi : f \Rightarrow g$ between two chain maps $f,g : C_\bullet \to D_\bullet$ in $Ch_\bullet(\mathcal{A})$ is a sequence of morphisms

$$\{ (\psi_n : C_n \to D_{n+1}) \in \mathcal{A} | n \in \mathbb{N} \}$$

in $\mathcal{A}$ such that

$$f_n - g_n = \partial^D \circ \psi_n + \psi_{n-1} \partial^C \,.$$

Remark

It may be useful to illustrate this with the following graphics, which however is not a commuting diagram:

$$\array{ \vdots && \vdots \\ \downarrow && \downarrow \\ C_{n+1} &\stackrel{f_{n+1} - g_{n+1}}{\to}& D_{n+1} \\ \downarrow^{\mathrlap{\partial^C_{n}}} &\nearrow_{\mathrlap{\psi_{n}}}& \downarrow^{\mathrlap{\partial^D_{n}}} \\ C_n &\stackrel{f_n - g_n}{\to}& D_n \\ \downarrow^{\mathrlap{\partial^C_{n-1}}} &\nearrow_{\mathrlap{\psi_{n-1}}}& \downarrow^{\mathrlap{\partial^D_{n-1}}} \\ C_{n-1} &\stackrel{f_{n-1} - g_{n-1}}{\to}& D_{n-1} \\ \downarrow && \downarrow \\ \vdots && \vdots } \,.$$

Instead, a way to encode chain homotopies by genuine diagrammatics is below in prop. , for which we introduce the interval object for chain complexes:

Definition

Let

$$I_\bullet \coloneqq N_\bullet(C(\Delta[1]))$$

be the normalized chain complex in $\mathcal{A}$ of the simplicial chains on the simplicial 1-simplex:

$$I_\bullet = [ \cdots \to 0 \to 0 \to R \stackrel{(-id,id)}{\to} R \oplus R ] \,.$$

Remark

This is the standard interval in chain complexes. Indeed it is manifestly the “abelianization” of the standard interval object $\Delta^1$ in sSet/Top: the 1-simplex.

Proposition

A chain homotopy $\psi : f \Rightarrow g$ is equivalently a commuting diagram

$$\array{ C_\bullet \\ \downarrow & \searrow^{\mathrlap{f}} \\ I_\bullet \otimes C_\bullet &\stackrel{(f,g,\psi)}{\to}& D_\bullet \\ \uparrow & \nearrow_{\mathrlap{g}} \\ C_\bullet }$$

in $Ch_\bullet(\mathcal{A})$, hence a genuine left homotopy with respect to the interval object in chain complexes.

Proof

For notational simplicity we discuss this in $\mathcal{A} =$ Ab.

Observe that $N_\bullet(\mathbb{Z}(\Delta[1]))$ is the chain complex

$$( \cdots \to 0 \to 0 \to \mathbb{Z} \stackrel{(-id,id)}{\to} \mathbb{Z} \oplus \mathbb{Z} \to 0 \to 0 \to \cdots)$$

where the term $\mathbb{Z} \oplus \mathbb{Z}$ is in degree 0: this is the free abelian group on the set $\{(0),(1)\}$ of 0-simplices in $\Delta[1]$. The other copy of $\mathbb{Z}$ is the free abelian group on the single non-degenerate edge $(0 \to 1)$ in $\Delta[1]$. (All other simplices of $\Delta[1]$ are degenerate and hence do not contribute to the normalized chain complex which we are discussing here.) The single nontrivial differential sends $1 \in \mathbb{Z}$ to $(-1,1) \in \mathbb{Z} \oplus \mathbb{Z}$, reflecting the fact that one of the vertices is the 0-boundary the other the 1-boundary of the single nontrivial edge.

It follows that the tensor product of chain complexes $I_\bullet \otimes C_\bullet$ is

$$\array{ && (I \otimes C)_2 &\to& (I \otimes C)_1 &\to& (I \otimes C)_0 &\to& \cdots \\ \cdots &\to& C_1 \oplus C_{2} \oplus C_2 &\to& C_0 \oplus C_{1} \oplus C_{1} &\to& C_{-1} \oplus C_0 \oplus C_0 &\to& \cdots } \,.$$

Therefore a chain map $(f,g,\psi) : I_\bullet \otimes C_\bullet \to D_\bullet$ that restricted to the two copies of $C_\bullet$ is $f$ and $g$, respectively, is characterized by a collection of commuting diagrams

$$\array{ C_{n+1}\oplus C_{n+1} \oplus C_{n} &\stackrel{(f_{n+1},g_{n+1}, \psi_n)}{\to}& D_n \\ {}^{\mathllap{\partial^{I \otimes C}}}\downarrow && \downarrow^{\mathrlap{\partial^D}} \\ C_{n} \oplus C_{n} \oplus C_{n-1} &\stackrel{(f_n,g_n,\psi_{n-1})}{\to} & D_{n-1} } \,.$$

On the elements $(1,0,0)$ and $(0,1,0)$ in the top left this reduces to the chain map condition for $f$ and $g$, respectively. On the element $(0,0,1)$ this is the equation for the chain homotopy

$$f_n - g_n - \psi_{n-1} d_C = d_D \psi_{n} \,.$$

Let $C_\bullet, D_\bullet \in Ch_\bullet(\mathcal{A})$ be two chain complexes.

Definition

Define the relation chain homotopic on $Hom(C_\bullet, D_\bullet)$ by

$$(f \sim g) \Leftrightarrow \exists (\psi : f \Rightarrow g) \,.$$

Proposition

Chain homotopy is an equivalence relation on $Hom(C_\bullet,D_\bullet)$.

Definition

Write $Hom(C_\bullet,D_\bullet)_{\sim}$ for the quotient of the hom set $Hom(C_\bullet,D_\bullet)$ by chain homotopy.

Proposition

This quotient is compatible with composition of chain maps.

Accordingly the following category exists:

Definition

Write $\mathcal{K}_\bullet(\mathcal{A})$ for the category whose objects are those of $Ch_\bullet(\mathcal{A})$, and whose morphisms are chain homotopy classes of chain maps:

$$Hom_{\mathcal{K}_\bullet(\mathcal{A})}(C_\bullet, D_\bullet) \coloneqq Hom_{Ch_\bullet(\mathcal{A})}(C_\bullet, D_\bullet)_\sim \,.$$

This is usually called the (strong) homotopy category of chain complexes in $\mathcal{A}$.

Remark

Beware, as we will discuss in detail below in 8), that another category that would deserve to carry this name instead is called the derived category of $\mathcal{A}$. In the derived category one also quotients out chain homotopy, but one allows that first the domain of the two chain maps $f$ and $g$ is refined along a quasi-isomorphism.

Definition

A chain map $f_\bullet : C_\bullet \to D_\bullet$ in $Ch_\bullet(\mathcal{A})$ is called a quasi-isomorphism if for each $n \in \mathbb{N}$ the induced morphisms on chain homology groups

$$H_n(f) \colon H_n(C) \to H_n(D)$$

is an isomorphism.

Remark

Quasi-isomorphisms are also called, more descriptively, homology isomorphisms or $H_\bullet$-isomorphisms. See at homology localization for more on this.

With the homotopy theoretic notions of chain homotopy and quasi-isomorphism in hand, we can now give a deeper explanation of long exact sequences in homology. We first give now a heuristic discussion that means to serve as a guide through the constructions to follow. The reader wishing to skip this may directly jump ahead to definition .

While the notion of a short exact sequence of chain complexes is very useful for computations, it does not have invariant meaning if one considers chain complexes as objects in (abelian) homotopy theory, where one takes into account chain homotopies between chain maps and takes equivalence of chain complexes not to be given by isomorphism, but by quasi-isomorphism.

For if a chain map $A_\bullet \to B_\bullet$ is the degreewise kernel of a chain map $B_\bullet \to C_\bullet$, then if $\hat A_\bullet \stackrel{\simeq}{\to} A_\bullet$ is a quasi-isomorphism (for instance a projective resolution of $A_\bullet$) then of course the composite chain map $\hat A_\bullet \to B_\bullet$ is in general far from being the degreewise kernel of $C_\bullet$. Hence the notion of degreewise kernels of chain maps and hence that of short exact sequences is not meaningful in the homotopy theory of chain complexes in $\mathcal{A}$ (for instance: not in the derived category of $\mathcal{A}$).

That short exact sequences of chain complexes nevertheless play an important role in homological algebra is due to what might be called a “technical coincidence”:

Proposition

If $A_\bullet \to B_\bullet \to C_\bullet$ is a short exact sequence of chain complexes, then the commuting square

$$\array{ A_\bullet &\to& 0 \\ \downarrow && \downarrow \\ B_\bullet &\to& C_\bullet }$$

is not only a pullback square in $Ch_\bullet(\mathcal{A})$, exhibiting $A_\bullet$ as the fiber of $B_\bullet \to C_\bullet$ over $0 \in C_\bullet$, it is in fact also a homotopy pullback.

This means it is universal not just among commuting such squares, but also among such squares which commute possibly only up to a chain homotopy $\phi$:

$$\array{ Q_\bullet &\to& 0 \\ \downarrow &\swArrow_{\phi}& \downarrow \\ B_\bullet &\to& C_\bullet }$$

and with morphisms between such squares being maps $A_\bullet \to A'_\bullet$ correspondingly with further chain homotopies filling all diagrams in sight.

Equivalently, we have the formally dual result

Proposition

If $A_\bullet \to B_\bullet \to C_\bullet$ is a short exact sequence of chain complexes, then the commuting square

$$\array{ A_\bullet &\to& 0 \\ \downarrow && \downarrow \\ B_\bullet &\to& C_\bullet }$$

is not only a pushout square in $Ch_\bullet(\mathcal{A})$, exhibiting $C_\bullet$ as the cofiber of $A_\bullet \to B_\bullet$ over $0 \in C_\bullet$, it is in fact also a homotopy pushout.

But a central difference between fibers/cofibers on the one hand and homotopy fibers/homotopy cofibers on the other is that while the (co)fiber of a (co)fiber is necessarily trivial, the homotopy (co)fiber of a homotopy (co)fiber is in general far from trivial: it is instead the looping $\Omega(-)$ or suspension $\Sigma(-)$ of the codomain/domain of the original morphism: by the pasting law for homotopy pullbacks the pasting composite of successive homotopy cofibers of a given morphism $f : A_\bullet \to B_\bullet$ looks like this:

$$\array{ A_\bullet &\stackrel{f}{\to}& B_\bullet &\to& 0 \\ \downarrow &\swArrow_{\mathrlap{\phi}}& \downarrow &\swArrow& \downarrow \\ 0 &\to& cone(f) &\to& A[1]_{\bullet} &\stackrel{}{\to}& 0 \\ && \downarrow &\swArrow& \downarrow^{\mathrlap{f[1]}} &\swArrow& \downarrow \\ && 0 &\to& B[1] &\to& cone(f)[1]_\bullet &\to& \cdots \\ && && \downarrow && \downarrow &\ddots& \\ && && \vdots && && }$$

here

• $cone(f)$ is a specific representative of the homotopy cofiber of $f$ called the mapping cone of $f$, whose construction comes with an explicit chain homotopy $\phi$ as indicated, hence $cone(f)$ is homology-equivalence to $C_\bullet$ above, but is in general a “bigger” model of the homotopy cofiber;

• $A[1]$ etc. is the suspension of a chain complex of $A$, hence the same chain complex but pushed up in degree by one.

In conclusion we get from every morphim of chain complexes a long homotopy cofiber sequence

$$\cdots \to A_\bullet \stackrel{f}{\to}B_\bullet \stackrel{}{\to} cone(f) \stackrel{}{\to} A[1]_\bullet \stackrel{f[1]}{\to} B[1]_\bullet \stackrel{}{\to} cone(f)[1]_\bullet \to \cdots \,.$$

And applying the chain homology functor to this yields the long exact sequence in chain homology which is traditionally said to be associated to the short exact sequence $A_\bullet \to B_\bullet \to C_\bullet$.

In conclusion this means that it is not really the passage to homology groups which “makes a short exact sequence become long”. It’s rather that passing to homology groups is a shadow of passing to chain complexes regarded up to quasi-isomorphism, and this is what makes every short exact sequence be realized as but a special presentation of a stage in a long homotopy fiber sequence.

We give a precise account of this story in the next section.

5) Homotopy fiber sequences and mapping cones

We have seen in 4) the long exact sequence in homology implied by a short exact sequence of chain complexes, constructed by an elementary if somewhat un-illuminating formula for the connecting homomorphism. We ended 4) by sketching how this formula arises as the shadow under the homology functor of a homotopy fiber sequence of chain complexes, constructed using mapping cones. This we now discuss in precise detail.

In the following we repeatedly mention that certain chain complexes are colimits of certain diagrams of chain complexes. The reader unfamiliar with colimits may simply ignore them and regard the given chain complex as arising by definition. However, even a vague intuitive understanding of the indicated colimits as formalizations of “gluing” of chain complexes along certain maps should help to motivate why these definitions are what they are. The reader unhappy even with this can jump ahead to prop. and take this and the following propositions up to and including prop. as definitions.

The notion of a mapping cone that we introduce now is something that makes sense whenever

1. there is a notion of cylinder object, such as the topological cylinder $[0,1] \times X$ over a topological space, or the chain complex cylinder $I_\bullet \otimes X_\bullet$ of a chain complex from def. .

2. there is a way to glue objects along maps between them, a notion of colimit.

Definition

For $f : X \to Y$ a morphism in a category with cylinder objects $cyl(-)$, the mapping cone or homotopy cofiber of $f$ is the colimit in the following diagram

$$\array{ && X &\stackrel{f}{\to}& Y \\ && \downarrow^{\mathrlap{i_1}} && \downarrow \\ X &\stackrel{i_0}{\to}& cyl(X) \\ \downarrow && &\searrow & \downarrow \\ {*} &\to& &\to& cone(f) }$$

in $C$ using any cylinder object $cyl(X)$ for $X$.

Remark

Heuristically this says that $cone(f)$ is the object obtained by

1. forming the cylinder over $X$;

2. gluing to one end of that the object $Y$ as specified by the map $f$.

3. shrinking the other end of the cylinder to the point.

Heuristically it is clear that this way every cycle in $Y$ that happens to be in the image of $X$ can be “continuously” translated in the cylinder-direction, keeping it constant in $Y$, to the other end of the cylinder, where it becomes the point. This means that every homotopy group of $Y$ in the image of $f$ vanishes in the mapping cone. Hence in the mapping cone the image of $X$ under $f$ in $Y$ is removed up to homotopy. This makes it clear how $cone(f)$ is a homotopy-version of the cokernel of $f$. And therefore the name “mapping cone”.

Another interpretation of the mapping cone is just as important:

Remark

A morphism $\eta : cyl(X) \to Y$ out of a cylinder object is a left homotopy $\eta : g \Rightarrow h$ between its restrictions $g\coloneqq \eta(0)$ and $h \coloneqq \eta(1)$ to the cylinder boundaries

$$\array{ X \\ \downarrow^{\mathrlap{i_0}} & \searrow^{\mathrlap{g}} \\ cyl(X) &\stackrel{\eta}{\to}& Y \\ \uparrow^{\mathrlap{i_1}} & \nearrow_{\mathrlap{h}} \\ X } \,.$$

Therefore prop. says that the mapping cone is the universal object with a morphism $i$ from $Y$ and a left homotopy from $i \circ f$ to the zero morphism.

$$\array{ X &\stackrel{f}{\to}& Y \\ \downarrow &\swArrow_{\eta}& \downarrow \\ * &\to& cone(f) }$$

The interested reader can find more on the conceptual background of this construction at factorization lemma and at homotopy pullback.

Proposition

This colimit, in turn, may be computed in two stages by two consecutive pushouts in $C$, and in two ways by the following pasting diagram:

$$\array{ && X &\stackrel{f}{\to}& Y \\ && \downarrow^{i_1} && \downarrow \\ X &\stackrel{i_0}{\to}& cyl(X) &\to & cyl(f) \\ \downarrow && \downarrow && \downarrow \\ {*} &\to& cone(X) &\to& cone(f) } \,.$$

Here every square is a pushout, (and so by the pasting law is every rectangular pasting composite).

This now is a basic fact in ordinary category theory. The pushouts appearing here go by the following names:

Definition

The pushout

$$\array{ X &\stackrel{i_0}{\to}& cyl(X) \\ \downarrow && \downarrow \\ {*} &\to& cone(X) }$$

defines the cone $cone(X)$ over $X$ (with respect to the chosen cylinder object): the result of taking the cylinder over $X$ and identifying one $X$-shaped end with the point.

The pushout

$$\array{ X &\stackrel{f}{\to}& Y \\ \downarrow && \downarrow \\ cyl(X) &\to& cyl(f) }$$

defines the mapping cylinder $cyl(f)$ of $f$, the result of identifying one end of the cylinder over $X$ with $Y$, using $f$ as the gluing map.

The pushout

$$\array{ cyl(x) &\to& cyl(f) \\ \downarrow && \downarrow \\ cone(X) &\to& cone(f) }$$

defines the mapping cone $cone(f)$ of $f$: the result of forming the cyclinder over $X$ and then identifying one end with the point and the other with $Y$, via $f$.

Remark

As in remark all these step have evident heuristic geometric interpretations:

1. $cone(X)$ is obtained from the cylinder over $X$ by contracting one end of the cylinder to the point;

2. $cyl(f)$ is obtained from the cylinder over $X$ by gluing $Y$ to one end of the cylinder, as specified by the map $f$;

We discuss now this general construction of the mapping cone $cone(f)$ for a chain map $f$ between chain complexes. The end result is prop. below, reproducing the classical formula for the mapping cone.

Definition

Write $*_\bullet \in Ch_\bullet(\mathcal{A})$ for the chain complex concentrated on $R$ in degree 0

$$*_\bullet 0 = [\cdots \to 0 \to 0 \to R] \,.$$

Remark

This may be understood as the normalized chain complex of chains of simplices on the terminal simplicial set $\Delta^0$, the 0-simplex.

Definition

Let $I_\bullet \in Ch_{\bullet}(\mathcal{A})$ be given by

$$I_\bullet = (\cdots 0 \to 0 \to R \stackrel{(-id,id)}{\to} R \oplus R) \,.$$

Denote by

$$i_0 : *_\bullet \to I_\bullet$$

the chain map which in degree 0 is the canonical inclusion into the second summand of a direct sum and by

$$i_1 : *_\bullet \to I_\bullet$$

correspondingly the canonical inclusion into the first summand.

Remark

This is the standard interval object in chain complexes.

It is in fact the normalized chain complex of chains on a simplicial set for the canonical simplicial interval, the 1-simplex:

$$I_\bullet = C_\bullet(\Delta[1]) \,.$$

The differential $\partial^I = (-id, id)$ here expresses the alternating face map complex boundary operator, which in terms of the three non-degenerate basis elements is given by

$$\partial ( 0 \to 1 ) = (1) - (0) \,.$$

We decompose the proof of this statement is a sequence of substatements.

Proposition

For $X_\bullet \in Ch_\bullet$ the tensor product of chain complexes

$$(I \otimes X)_\bullet \in Ch_\bullet$$

is a cylinder object of $X_\bullet$ for the structure of a category of cofibrant objects on $Ch_\bullet$ whose cofibrations are the monomorphisms and whose weak equivalences are the quasi-isomorphisms (the substructure of the standard injective model structure on chain complexes).

Example

In example above we saw the cyclinder over the interval itself: the square.

Proposition

The complex $(I \otimes X)_\bullet$ has components

$$(I \otimes X)_n = X_n \oplus X_n \oplus X_{n-1}$$

and the differential is given by

$$\array{ X_{n+1} \oplus X_{n+1} &\stackrel{\partial^X \oplus \partial^X}{\to}& X_n \oplus X_n \\ \oplus &\nearrow_{(-id,id)}& \oplus \\ X_{n} &\underset{-\partial^X}{\to}& X_{n-1} } \,,$$

hence in matrix calculus by

$$\partial^{I \otimes X} = \left( \array{ \partial^X \oplus \partial^X & (-id, id) \\ 0 & -\partial^X } \right) : (X_{n+1} \oplus X_{n+1}) \oplus X_{n} \to (X_{n} \oplus X_{n}) \oplus X_{n-1} \,.$$

Proof

By the formula discussed at tensor product of chain complexes the components arise as the direct sum

$$(I \otimes X )_n = (R_{(0)} \otimes X_n ) \oplus (R_{(1)} \otimes X_n ) \oplus (R_{(0 \to 1)} \otimes X_{(n-1)} )$$

and the differential picks up a sign when passed past the degree-1 term $R_{(0 \to 1)}$:

$$\begin{aligned} \partial^{I \otimes X} ( (0 \to 1), x ) &= ( (\partial^I (0 \to 1)), x ) - ( (0\to 1), \partial^X x ) \\ & = ( - (0) + (1), x ) - ( (0 \to 1), \partial^X x ) \\ & = -((0), x) + ((1), x) - ( (0 \to 1), \partial^X x ) \end{aligned} \,.$$

Remark

The two boundary inclusions of $X_\bullet$ into the cylinder are given in terms of def. by

$$i^X_0 : X_\bullet \simeq *_\bullet \otimes X_\bullet \stackrel{i_0 \otimes id_X}{\to} (I\otimes X)_\bullet$$

and

$$i^X_1 : X_\bullet \simeq *_\bullet \otimes X_\bullet \stackrel{i_1 \otimes id_X}{\to} (I\otimes X)_\bullet$$

which in components is the inclusion of the second or first direct summand, respectively

$$X_n \hookrightarrow X_n \oplus X_n \oplus X_{n-1} \,.$$

One part of definition now reads:

Definition

For $f_\bullet : X_\bullet \to Y_\bullet$ a chain map, the mapping cylinder $cyl(f)$ is the pushout

$$\array{ cyl(f)_\bullet &\leftarrow& Y_\bullet \\ \uparrow && \uparrow^{\mathrlap{f}} \\ I_\bullet \otimes X_\bullet &\stackrel{i_0}{\leftarrow}& X_\bullet } \,.$$

Proposition

The components of $cyl(f)$ are

$$cyl(f)_n = X_n \oplus Y_n \oplus X_{n-1}$$

and the differential is given by

$$\array{ X_{n+1} \oplus Y_{n+1} &\stackrel{\partial^X \oplus \partial^Y}{\to}& X_n \oplus Y_n \\ \oplus &\nearrow_{(-id,f)}& \oplus \\ X_{n} &\underset{-\partial^X}{\to}& X_{n-1} } \,,$$

hence in matrix calculus by

$$\partial^{cyl(f)} = \left( \array{ \partial^X \oplus \partial^Y & (-id, f_n) \\ 0 & -\partial^X } \right) : (X_{n+1} \oplus Y_{n+1}) \oplus X_{n} \to (X_{n} \oplus Y_{n}) \oplus X_{n-1} \,.$$

Proof

The colimits in a category of chain complexes $Ch_\bullet(\mathcal{A})$ are computed in the underlying presheaf category of towers in $\mathcal{A}$. There they are computed degreewise in $\mathcal{A}$ (since limits of presheaves are computed objectwise). Here the statement is evident:

the pushout identifies one direct summand $X_n$ with $Y_n$ along $f_n$ and so where previously a $id_{X_n}$ appeared on the diagonl, there is now $f_n$.

The last part of definition now reads:

Definition

For $f_\bullet : X_\bullet \to Y_\bullet$ a chain map, the mapping cone $cone(f)$ is the pushout

$$\array{ cone(f) &\leftarrow& cyl(f) \\ \uparrow && \uparrow \\ cone(X) &\leftarrow& X \otimes I \\ \uparrow && \uparrow^{\mathrlap{i_1}} \\ 0 &\leftarrow& X }$$

Proposition

The components of the mapping cone $cone(f)$ are

$$cone(f)_n = Y_n \oplus X_{n-1}$$

with differential given by

$$\array{ Y_{n+1} &\stackrel{\partial^Y}{\to}& Y_n \\ \oplus &\nearrow_{f_n}& \oplus \\ X_{n} &\underset{-\partial^X}{\to}& X_{n-1} } \,,$$

and hence in matrix calculus by

$$\partial^{cone(f)} = \left( \array{ \partial^Y_n & f_n \\ 0 & -\partial^X_n } \right) : Y_{n+1} \oplus X_{n} \to Y_{n} \oplus X_{n-1} \,.$$

Proof

As before the pushout is computed degreewise. This identifies the remaining unshifted copy of $X$ with 0.

Proposition

For $f : X_\bullet \to Y_\bullet$ a chain map, the canonical inclusion $i : Y_\bullet \to cone(f)_\bullet$ of $Y_\bullet$ into the mapping cone of $f$ is given in components

$$i_n : Y_n \to cone(f)_n = Y_n \oplus X_{n-1}$$

by the canonical inclusion of a summand into a direct sum.

Proof

This follows by starting with remark and then following these inclusions through the formation of the two colimits as discussed above.

Using these mapping cones of chain maps, we now explain how the long exact sequences of homology groups, prop. , are a shadow under homology of genuine homotopy cofiber sequences of the chain complexes themselves.

Let $f : X_\bullet \to Y_\bullet$ be a chain map and write $cone(f) \in Ch_\bullet(\mathcal{A})$ for its mapping cone as explicitly given in prop. .

Definition

Write $X[1]_\bullet \in Ch_\bullet(\mathcal{A})$ for the suspension of a chain complex of $X$. Write

$$p : cone(f) \to X[1]_\bullet$$

for the chain map which in components

$$p_n : cone(f)_n \to X[1]_n$$

is given, via prop. , by the canonical projection out of a direct sum

$$p_n : Y_\n \oplus X_{n-1} \to X_{n-1} \,.$$

This defines the mapping cone construction on chain complex. Its definition as a universal left homotopy should make the following proposition at least plausible, which we cannot prove yet at this point, but which we state nevertheless to highlight the meaning of the mapping cone construction. The tools for the proof of propositions like this are discussed further below in 7) Derived categories and derived functors.

Proposition

The chain map $p : cone(f)_\bullet \to X[1]_\bullet$ represents the homotopy cofiber of the canonical map $i : Y_\bullet \to cone(f)_\bullet$.

Proof

By prop. and def. the sequence

$$Y_\bullet \stackrel{i}{\to} cone(f)_\bullet \stackrel{p}{\to} X[1]_\bullet$$

is a short exact sequence of chain complexes (since it is so degreewise, in fact degreewise it is even a split exact sequence, def. ). In particular we have a cofiber pushout diagram

$$\array{ Y_\bullet &\stackrel{i}{\hookrightarrow}& cone(f)_\bullet \\ \downarrow && \downarrow \\ 0 &\to& X[1]_\bullet } \,.$$

Now, in the injective model structure on chain complexes all chain complexes are cofibrant objects and an inclusion such as $i : Y_\bullet \hookrightarrow cone(f)_\bullet$ is a cofibration. By the detailed discussion at homotopy limit this means that the ordinary colimit here is in fact a homotopy colimit, hence exhibits $p$ as the homotopy cofiber of $i$.

Accordingly one says:

Corollary

For $f_\bullet : X_\bullet \to Y_\bullet$ a chain map, there is a homotopy cofiber sequence of the form

$$X_\bullet \stackrel{f_\bullet}{\to} Y_\bullet \stackrel{i_\bullet}{\to} cone(f)_\bullet \stackrel{p_\bullet}{\to} X[1]_\bullet \stackrel{f[1]_\bullet}{\to} Y_\bullet \stackrel{i[1]_\bullet}{\to} cone(f)_\bullet \stackrel{p[1]_\bullet}{\to} X[2]_\bullet \to \cdots$$

In order to compare this to the discussion of connecting homomorphisms, we now turn attention to the case that $f_\bullet$ happens to be a monomorphism. Notice that this we can always assume, up to quasi-isomorphism, for instance by prolonging $f$ by the map into its mapping cylinder

$$X_\bullet \to Y_\bullet \stackrel{\simeq}{\to} cyl(f) \,.$$

By the axioms on an abelian category in this case we have a short exact sequence

$$0 \to X_\bullet \stackrel{f_\bullet}{\to} Y_\bullet \stackrel{p_\bullet}{\to} Z_\bullet \to 0$$

of chain complexes. The following discussion revolves around the fact that now $cone(f)_\bullet$ as well as $Z_\bullet$ are both models for the homotopy cofiber of $f$.

Lemma

Let

$$X_\bullet \stackrel{f_\bullet}{\to} Y_\bullet \stackrel{p_\bullet}{\to} Z_\bullet$$

be a short exact sequence of chain complexes.

The collection of linear maps

$$h_n : Y_n \oplus X_{n-1} \to Y_n \stackrel{}{\to} Z_n$$

constitutes a chain map

$$h_\bullet : cone(f)_\bullet \to Z_\bullet \,.$$

This is a quasi-isomorphism. The inverse of $H_n(h_\bullet)$ is given by sending a representing cycle $z \in Z_n$ to

$$(\hat z_n, \partial^Y \hat z_n) \in Y_n \oplus X_{n+1} \,,$$

where $\hat z_n$ is any choice of lift through $p_n$ and where $\partial^Y \hat z_n$ is the formula expressing the connecting homomorphism in terms of elements, as discussed at Connecting homomorphism – In terms of elements.

Finally, the morphism $i_\bullet : Y_\bullet \to cone(f)_\bullet$ is eqivalent in the homotopy category (the derived category) to the zigzag

$$\array{ && cone(f)_\bullet \\ && \downarrow^{\mathrlap{h}}_{\mathrlap{\simeq}} \\ Y_\bullet &\to& Z_\bullet } \,.$$

Proof

To see that $h_\bullet$ defines a chain map recall the differential $\partial^{cone(f)}$ from prop. , which acts by

$$\partial^{cone(f)} (x_{n-1}, \hat z_n) = ( -\partial^X x_{n-1} , \partial^Y \hat z_n + x_{n-1} )$$

and use that $x_{n-1}$ is in the kernel of $p_n$ by exactness, hence

$$\begin{aligned} h_{n-1}\partial^{cone(f)}(x_{n-1}, \hat z_n) &= h_{n-1}( -\partial^X x_{n-1}, \partial^Y \hat z_n + x_{n-1} ) \\ & = p_{n-1}( \partial^Y \hat z_n + x_{n-1}) \\ & = p_{n-1}( \partial^Y \hat z_n ) \\ & = \partial^Z p_n \hat z_n \\ & = \partial^Z h_n(x_{n-1}, \hat z_n) \end{aligned} \,.$$

It is immediate to see that we have a commuting diagram of the form

$$\array{ && cone(f)_\bullet \\ & {}^{\mathllap{i_\bullet}}\nearrow& \downarrow^{\mathrlap{h}}_{\mathrlap{\simeq}} \\ Y_\bullet &\to& Z_\bullet }$$

since the composite morphism is the inclusion of $Y$ followed by the bottom morphism on $Y$.

Abstractly, this already implies that $cone(f)_\bullet \to Z_\bullet$ is a quasi-isomorphism, for this diagram gives a morphism of cocones under the diagram defining $cone(f)$ in prop. and by the above both of these cocones are homotopy-colimiting.

But in checking the claimed inverse of the induced map on homology groups, we verify this also explicity:

We first determine those cycles $(x_{n-1}, y_n) \in cone(f)_n$ which lift a cycle $z_n$. By lemma a lift of chains is any pair of the form $(x_{n-1}, \hat z_n)$ where $\hat z_n$ is a lift of $z_n$ through $Y_n \to X_n$. So $x_{n-1}$ has to be found such that this pair is a cycle. By prop. the differential acts on it by

$$\partial^{cone(f)} (x_{n-1}, \hat z_n) = ( -\partial^X x_{n-1} , \partial^Y \hat z_n + x_{n-1} )$$

and so the condition is that

$x_{n-1} \coloneqq -\partial^Y \hat z_n$ (which implies $\partial^X x_{n-1} = -\partial^X \partial^Y \hat z_n = -\partial^Y \partial^Y \hat z_n = 0$ due to the fact that $f_n$ is assumed to be an inclusion, hence that $\partial^X$ is the restriction of $\partial^Y$ to elements in $X_n$).

This condition clearly has a unique solution for every lift $\hat z_n$ and a lift $\hat z_n$ always exists since $p_n : Y_n \to Z_n$ is surjective, by assumption that we have a short exact sequence of chain complexes. This shows that $H_n(h_\bullet)$ is surjective.

To see that it is also injective we need to show that if a cycle $(-\partial^Y \hat z_n, \hat z_n) \in cone(f)_n$ maps to a cycle $z_n = p_n(\hat z_n)$ that is trivial in $H_n(Z)$ in that there is $c_{n+1}$ with $\partial^Z c_{n+1} = z_n$, then also the original cycle was trivial in homology, in that there is $(x_n, y_{n+1})$ with

$$\partial^{cone(f)}(x_n, y_{n+1}) \coloneqq (-\partial^X x_n, \partial^Y y_{n+1} + x_n) = (-\partial^Y \hat z_n, \hat z_n) \,.$$

For that let $\hat c_{n+1} \in Y_{n+1}$ be a lift of $c_{n+1}$ through $p_n$, which exists again by surjectivity of $p_{n+1}$. Observe that

$$p_{n}( \hat z_n - \partial^Y \hat c_{n+1}) = z_n -\partial^Z ( p_n \hat c_{n+1} ) = z_n - \partial^Z ( c_{n+1} ) = 0$$

by assumption on $z_n$ and $c_{n+1}$, and hence that $\hat z_n - \partial^Y \hat c_{n+1}$ is in $X_n$ by exactness.

Hence $(z_n - \partial^Y \hat c_{n+1}, \hat c_{n+1}) \in cone(f)_n$ trivializes the given cocycle:

$$\begin{aligned} \partial^{cone(f)}( \hat z_n - \partial^Y \hat c_{n+1} , \hat c_{n+1}) & = (-\partial^X(\hat z_n - \partial^Y \hat c_{n+1} ), \partial^Y \hat c_{n+1} + (\hat z_n - \partial^Y \hat c_{n+1} ) ) \\ & = (-\partial^Y(\hat z_n - \partial^Y \hat c_{n+1}), \hat z_n ) \\ & = ( -\partial^Y \hat z_n, \hat z_n ) \end{aligned} \,.$$

Theorem

Let

$$X_\bullet \stackrel{f_\bullet}{\to} Y_\bullet \to Z_\bullet$$

be a short exact sequence of chain complexes.

Then the chain homology functor

$$H_n(-) : Ch_\bullet(\mathcal{A}) \to \mathcal{A}$$

sends the homotopy cofiber sequence of $f$, cor. , to the long exact sequence in homology induced by the given short exact sequence, hence to

$$H_n(X_\bullet) \to H_n(Y_\bullet) \to H_n(Z_\bullet) \stackrel{\delta}{\to} H_{n-1}(X_\bullet) \to H_{n-1}(Y_\bullet) \to H_{n-1}(Z_\bullet) \stackrel{\delta}{\to} H_{n-2}(X_\bullet) \to \cdots \,,$$

where $\delta_n$ is the $n$th connecting homomorphism.

Proof

By lemma the homotopy cofiber sequence is equivalen to the zigzag

$$\array{ && && && && && cone(f)[1]_\bullet &\to& \cdots \\ && && && && && \downarrow^{\mathrlap{h[1]_\bullet}}_{\mathrlap{\simeq}} \\ && && cone(f)_\bullet &\to& X[1]_\bullet &\stackrel{f[1]_\bullet}{\to}& Y[1]_\bullet &\to& Z[1]_\bullet \\ && && \downarrow^{\mathrlap{h_\bullet}}_{\mathrlap{\simeq}} \\ X_\bullet &\stackrel{f_\bullet}{\to}& Y_\bullet &\stackrel{}{\to}& Z_\bullet } \,.$$

Observe that

$$H_n( X[k]_\bullet) \simeq H_{n-k}(X_\bullet) \,.$$

It is therefore sufficient to check that

$$H_n \left( \array{ cone(f)_\bullet &\to& X[1]_\bullet \\ \downarrow^{\mathrlap{\simeq}} \\ Z_\bullet } \right) \;\; : \;\; H_n(Z_\bullet) \to H_n(cone(f)_\bullet) \to H_{n-1}(X_\bullet)$$

equals the connecting homomorphism $\delta_n$ induced by the short exact sequence.

By prop. the inverse of the vertical map is given by choosing lifts and forming the corresponding element given by the connecting homomorphism. By prop. the horizontal map is just the projection, and hence the assignment is of the form

$$[z_n] \mapsto [x_{n-1}, y_n] \mapsto [x_{n-1}] \,.$$

So in total the image of the zig-zag under homology sends

$$[z_n]_Z \mapsto -[\partial^Y \hat z_n]_X \,.$$

By the discussion there, this is indeed the action of the connecting homomorphism.

In summary, the above says that for every chain map $f_\bullet : X_\bullet \to Y_\bullet$ we obtain maps

$$X_\bullet \stackrel{f}{\to} Y_\bullet \stackrel{ \left( \array{ 0 \\ id_{Y_\bullet} } \right) }{\to} cone(f)_\bullet \stackrel{ \left( \array{ id_{X[1]_\bullet} & 0 } \right) }{\to} X[1]_\bullet$$

which form a homotopy fiber sequence and such that this sequence continues by forming suspensions, hence for all $n \in \mathbb{Z}$ we have

$$X[n]_\bullet \stackrel{f}{\to} Y[n]_\bullet \stackrel{ \left( \array{ 0 \\ id_{Y[n]_\bullet} } \right) }{\to} cone(f)[n]_\bullet \stackrel{ \left( \array{ id_{X[n+11]_\bullet} & 0 } \right) }{\to} X[n+1]_\bullet$$

To amplify this quasi-cyclic behaviour one sometimes depicts the situation as follows:

$$\array{ X_\bullet &&\stackrel{f}{\to}&& Y_\bullet \\ & {}_{\mathllap{[1]}}\nwarrow && \swarrow \\ && cone(f)_\bullet }$$

and hence speaks of a “triangle”, or distinguished triangle or mapping cone triangle of $f$.

• distinguished triangle = period of homotopy fiber sequence .

Due to these “triangles” one calls the homotopy category of chain complexes localized at the quasi-isomorphisms, hence the derived category which we discuss below in 8), a triangulated category.

6) Double complexes and the diagram chasing lemmas

We have seen in the discussion of the connecting homomorphism in the homology long exact sequence in 4) above that given an exact sequence of chain complexes – hence in particular a chain complex of chain complexes – there are interesting ways to relate elements on the far right to elements on the far left in lower degree. In 5) we had given the conceptual explanation of this phenomenon in terms of long homotopy fiber sequences. But often it is just computationally useful to be able to efficiently establish and compute these “long diagram chase”-relations, independently of a homotopy-theoretic interpretation. Such computational tools we discuss here.

A chain complex of chain complex is called a double complex and so we first introduce this elementary notion and the corresponding notion notion of total complex. (Total complexes are similarly elementary to define but will turn out to play a deeper role as models for homotopy colimits, this we indicate further below in chapter V)).

There is a host of classical diagram-chasing lemmas that relate far-away entries in double complexes that enjoy suitable exactness properties. These go by names such as the snake lemma or the 3x3 lemma. The underlying mechanism of all these lemmas is made most transparent in the salamander lemma. This is fairly trivial to establish, and the notions it induces allow quick transparent proofs of all the other diagram-chasing lemmas.

The discussion to go here is kept at salamander lemma. See there.

Abelian homotopy theory

We have seen in section II) that the most interesting properties of the category of chain complexes is all secretly controled by the phenomenon of chain homotopy and quasi-isomorphism. Strictly speaking these two phenomena point beyond plain category theory to the richer context of general abstract homotopy theory. Here we discuss properties of the category of chain complexes from this genuine homotopy-theoretic point of view. The result of passing the category of chain complexes to genuine homotopy theory is called the derived category (of the underlying abelian category $\mathcal{A}$, say of modules) and we start in 7) with a motivation of the phenomenon of this “homotopy derivation” and the discussion of the necessary resolutions of chain complexes. This naturally gives rise to the general notion of derived functors which we discuss in 8). Examples of these are ubiquituous in homological algebra, but as in ordinary enriched category theory two stand out as being of more fundamental importance, the derived functor “Ext” of the hom-functor and the derived functor “Tor” of the tensor product functor. Their properties and uses we discuss in 9).

7) Chain homotopy and resolutions

We now come back to the category $\mathcal{K}(\mathcal{A})$ of def. , the “homotopy category of chain complexes” in which chain-homotopic chain maps are identified. This would seem to be the right context to study the homotopy theory of chain complexes, but one finds that there are still chain maps which ought to be identified in homotopy theory, but which are still not identified in $\mathcal{K}(\mathcal{A})$. This is our motivating example below.

We discuss then how this problem is fixed by allowing to first “resolve” chain complexes quasi-isomorphically by “good representatives” called projective resolutions or injective resolutions. Many of the computations in the following sections – and in homological algebra in general – come down to operating on such resolutions. We end this section by prop. below, which shows that the above problem indeed goes away when allowing chain complexes to be resolved.

In the next section, 8), we discuss how this process of forming resolutions functorially extends to the whole category of modules.

So we start here with this simple example that shows the problem with bare chain homotopies and indicates how these have to be resolved:

Example

In $Ch_\bullet(\mathcal{A})$ for $\mathcal{A} =$ Ab consider the chain map

$$\array{ \cdots &\to& 0 &\to& 0 &\to& 0 &\to& \mathbb{Z}_2 \\ && \downarrow && \downarrow && \downarrow && \downarrow^{\mathrlap{id}} \\ \cdots &\to& 0 &\to& \mathbb{Z} &\stackrel{\cdot 2}{\to}& \mathbb{Z} &\stackrel{mod\,2}{\to}& \mathbb{Z}_2 } \,.$$

The codomain of this map is an exact sequence, hence is quasi-isomorphic to the 0-chain complex. Thereofore in homotopy theory it should behave entirely as the 0-complex itself. In particular, every chain map to it should be chain homotopic to the zero morphism (have a null homotopy).

But the above chain map is chain homotopic precisely only to itself. This is because the degree-0 component of any chain homotopy out of this has to be a homomorphism of abelian groups $\mathbb{Z}_2 \to \mathbb{Z}$, and this must be the 0-morphism, because $\mathbb{Z}$ is a free group, but $\mathbb{Z}_2$ is not.

This points to the problem: the components of the domain chain complex are not free enough to admit sufficiently many maps out of it.

Consider therefore a free resolution of the above domain complex by the quasi-isomorphism

$$\array{ \cdots &\to& 0 &\to& 0 &\to& \mathbb{Z} &\stackrel{\cdot 2}{\to}& \mathbb{Z} \\ && \downarrow && \downarrow && \downarrow && \downarrow^{\mathrlap{mod\,2}} \\ \cdots &\to& 0 &\to& 0 &\to& 0 &\to& \mathbb{Z}_2 } \,,$$

where now the domain complex consists entirely of free groups. The composite of this with the original chain map is now

$$\array{ \cdots &\to& 0 &\to& 0 &\to& \mathbb{Z} &\stackrel{\cdot 2}{\to}& \mathbb{Z} \\ && \downarrow && \downarrow && \downarrow^{0} && \downarrow^{\mathrlap{mod\,2}} \\ \cdots &\to& 0 &\to& \mathbb{Z} &\stackrel{\cdot 2}{\to}& \mathbb{Z} &\stackrel{mod\,2}{\to}& \mathbb{Z}_2 } \,.$$

This is the corresponding resolution of the original chain map. And this indeed has a null homotopy:

$$\array{ \cdots &\to& 0 &\to& 0 &\to& \mathbb{Z} &\stackrel{\cdot 2}{\to}& \mathbb{Z} \\ && \downarrow &\swarrow& \downarrow &\swarrow_{-id}& \downarrow^{0} &\swarrow_{\mathrlap{id}}& \downarrow^{\mathrlap{mod\,2}} \\ \cdots &\to& 0 &\to& \mathbb{Z} &\stackrel{\cdot 2}{\to}& \mathbb{Z} &\stackrel{mod\,2}{\to}& \mathbb{Z}_2 } \,.$$

So resolving the domain by a sufficiently free complex makes otherwise missing chain homotopies exist. Below in lemma we discuss the general theory behind the kind of situation of this example. But to get there we first need some basic notions and facts.

Notably, in general it is awkward to insist on actual free resolutions. But it is easy to see, and this we discuss now, that essentially just as well is a resolution by modules which are direct summands of free modules.

Definition

An object $P$ of a category $C$ is a projective object if it has the left lifting property against epimorphisms.

This means that $P$ is projective if for any morphism $f:P \to B$ and any epimorphism $q:A \to B$, $f$ factors through $q$ by some morphism $P\to A$.

$$\array{ && A \\ &{}^{\mathllap{\exists}}\nearrow& \downarrow^{\mathrlap{q}} \\ P &\stackrel{f}{\to}& B } \,.$$

An equivalent way to say this is that:

Definition

An object $P$ is projective precisely if the hom-functor $Hom(P,-)$ preserves epimorphisms.

Remark

The point of this lifting property will become clear when we discuss the construction of projective resolutions a bit further below: they are built by applying this property degreewise to obtain suitable chain maps.

We will be interested in projective objects in the category $R$Mod: projective modules. Before we come to that, notice the following example (which the reader may on first sight feel is pedantic and irrelevant, but for the following it is actually good to make this explicit).

Example

In the category Set of sets the following are equivalent

• every object is projective;

• the axiom of choice holds.

Remark

We will assume here throughout the axiom of choice in Set, as usual. The point of the above example, however, is that one could just as well replace Set by another “base topos” which will behave essentially precisely like Set, but in general will not validate the axiom of choice. Homological algebra in such a more general context is the theory of complexes of abelian sheaves/sheaves of abelian groups and ultimately the theory of abelian sheaf cohomology.

This is a major aspect of homological algebra. While we will not discuss this further here in this introduction, the reader might enjoy keeping in mind that all of the following discussion of resolutions of $R$-modules goes through in this wider context of sheaves of modules except for subtleties related to the (partial) failure of example for the category of sheaves.

We now characterize projective modules.

Lemma

Assuming the axiom of choice, a free module $N \simeq R^{(S)}$ is projective.

Proof

Explicitly: if $S \in Set$ and $F(S) = R^{(S)}$ is the free module on $S$, then a module homomorphism $F(S) \to N$ is specified equivalently by a function $f : S \to U(N)$ from $S$ to the underlying set of $N$, which can be thought of as specifying the images of the unit elements in $R^{(S)} \simeq \oplus_{s \in S} R$ of the ${\vert S\vert}$ copies of $R$.

Accordingly then for $\tilde N \to N$ an epimorphism, the underlying function $U(\tilde N) \to U(N)$ is an epimorphism, and the axiom of choice in Set says that we have all lifts $\tilde f$ in

$$\array{ && U(\tilde N) \\ & {}^{\tilde f} \nearrow & \downarrow \\ S &\stackrel{f}{\to}& U(N) } \,.$$

By adjunction these are equivalently lifts of module homomorphisms

$$\array{ && \tilde N \\ & \nearrow & \downarrow \\ R^{(S)} &\stackrel{}{\to}& N } \,.$$

Lemma

If $N \in R Mod$ is a direct summand of a free module, hence if there is $N' \in R Mod$ and $S \in Set$ such that

$$R^{(S)} \simeq N \oplus N' \,,$$

then $N$ is a projective module.

Proof

Let $\tilde K \to K$ be a surjective homomorphism of modules and $f : N \to K$ a homomorphism. We need to show that there is a lift $\tilde f$ in

$$\array{ && \tilde K \\ & {}^{\mathllap{\tilde f}}\nearrow & \downarrow \\ N &\stackrel{f}{\to}& K } \,.$$

By definition of direct sum we can factor the identity on $N$ as

$$id_N : N \to N \oplus N' \to N \,.$$

Since $N \oplus N'$ is free by assumption, and hence projective by lemma , there is a lift $\hat f$ in

$$\array{ && && \tilde K \\ && & {}^{\mathllap{\hat f}}\nearrow & \downarrow \\ N &\to& N \oplus N' &\to& K } \,.$$

Hence $\tilde f : N \to N \oplus N' \stackrel{\hat f}{\to} \tilde K$ is a lift of $f$.

Proposition

An $R$-module $N$ is projective precisely if it is the direct summand of a free module.

Proof

By lemma if $N$ is a direct summand then it is projective. So we need to show the converse.

Let $F(U(N))$ be the free module on the set $U(N)$ underlying $N$, hence the direct sum

$$F(U(n)) = \oplus_{n \in U(n)} R \,.$$

There is a canonical module homomorphism

$$\oplus_{n \in U(N)} R \to N$$

given by sending the unit $1 \in R_n$ of the copy of $R$ in the direct sum labeled by $n \in U(n)$ to $n \in N$.

(Abstractly this is the counit $\epsilon : F(U(N)) \to N$ of the free/forgetful-adjunction $(F \dashv U)$.)

This is clearly an epimorphism. Thefore if $N$ is projective, there is a section $s$ of $\epsilon$. This exhibits $N$ as a direct summand of $F(U(N))$.

We discuss next how to build resolutions of chain complexes by projective modules. But before we come to that it is useful to also introduce the dual notion. So far we have concentrated on chain complexes with degrees in the natural numbers: non-negative degrees. For a discussion of resolutions we need a more degree-symmetric perspective, which of course is straightforward to obtain.

Definition

A cochain complex $C^\bullet$ in $\mathcal{A} = R Mod$ is a sequence of morphism

$$C^0 \stackrel{d^0}{\to} C^1 \stackrel{d^1}{\to} C^2 \stackrel{d^2}{\to} \cdots$$

in $\mathcal{A}$ such that $d\circ d = 0$. A homomorphism of cochain complexes $f^\bullet : C^\bullet \to D^\bullet$ is a collection of morphisms $\{f^n : C^n \to D^n\}$ such that $d^n_D \circ f^n = f^n \circ d^n_C$ for all $n \in \mathbb{N}$.

We write $Ch^\bullet(\mathcal{A})$ for the category of cochain complexes.

Example

Let $N \in \mathcal{A}$ be a fixed module and $C_\bullet \in Ch_\bullet(\mathcal{A})$ a chain complex. Then applying degreewise the hom-functor out of the components of $C_\bullet$ into $N$ yields a cochain complex in $\mathbb{Z} Mod \simeq$ Ab:

$$Hom_{\mathcal{A}}(C_\bullet, N) = \left[ Hom_{\mathcal{A}}(C_0, N) \stackrel{Hom_{\mathcal{A}}(\partial_0, N)}{\to} Hom_{\mathcal{A}}(C_1, N) \stackrel{Hom_{\mathcal{A}}(\partial_1, N)}{\to} Hom_{\mathcal{A}}(C_2, N) \stackrel{Hom_{\mathcal{A}}(\partial_2, N)}{\to} \cdots \right] \,.$$

Example

In example let $\mathcal{A} = \mathbb{Z}$Mod $=$ Ab, let $N = \mathbb{Z}$ and let $C_\bullet = \mathbb{Z}[Sing(X)]$ be the singular simplicial complex of a topological space $X$. Write

$$C^\bullet(X) \coloneqq Hom_{\mathbb{Z}[Sing X], \mathbb{Z}} \,.$$

Then $H^\bullet(C(X))$ is called the singular cohomology of $X$.

Remark

Example is just a special case of the internal hom of def. : we may regard cochain complexes in non-negative degree equivalently as chain complexes in positive degree.

Accordingly we say for $C^\bullet$ a cochain complex that

• an element in $C^n$ is an $n$-cochain

• an element in $im(d^{n-1})$ is an $n$-coboundary

• al element in $ker(d^n)$ is an $n$-cocycle.

But equivalently we may regard a cochain in degree $n$ as a chain in degree $(-n)$ and so forth. And this is the perspective used in all of the following.

The role of projective objects, def. , for chain complexes is played, dually, by injective objects for cochain complexes:

Definition

An object $I$ a category is injective if all diagrams of the form

$$\array{ X &\to& I \\ {}^{}\downarrow \\ Z }$$

with $X \to Z$ a monomorphism admit an extension

$$\array{ X &\to& I \\ {}^{}\downarrow & \nearrow_{\mathrlap{\exists}} \\ Z } \,.$$

Since we are interested in refining modules by projective or injective modules, we have the following terminology.

Definition

A category

• has enough projectives if for every object $X$ there is a projective object $Q$ equipped with an epimorphism $Q \to X$;

• has enough injectives if for every object $X$ there is an injective object $P$ equipped with a monomorphism $X \to P$.

We have essentially already seen the following statement.

Proposition

Assuming the axiom of choice, the category $R$Mod has enough projectives.

Proof

Let $F(U(N))$ be the free module on the set $U(N)$ underlying $N$. By lemma this is a projective module.

The canonical morphism

$$F(U(n)) = \oplus_{n \in U(n)} R \to N$$

is clearly a surjection, hence an epimorphism in $R$Mod.

We now show that similarly $R Mod$ has enough injectives. This is a little bit more work and hence we proceed with a few preparatory statements.

The following basic statement of algebra we cite here without proof (but see at injective object for details).

Proposition

Assuming the axiom of choice, an abelian group $A$ is injective as a $\mathbb{Z}$-module precisely if it is a divisible group, in that for all integers $n \in \mathbb{N}$ we have $n G = G$.

Example

By prop. the following abelian groups are injective in Ab.

The group of rational numbers $\mathbb{Q}$ is injective in Ab, as is the additive group of real numbers $\mathbb{R}$ and generally that underlying any field. The additive group underlying any vector space is injective. The quotient of any injective group by any other group is injective.

Example

Not injective in Ab are the cyclic groups $\mathbb{Z}/n\mathbb{Z}$.

Proposition

Assuming the axiom of choice, the category $\mathbb{Z}$Mod $\simeq$ Ab has enough injectives.

Proof

By prop. an abelian group is an injective $\mathbb{Z}$-module precisely if it is a divisible group. So we need to show that every abelian group is a subgroup of a divisible group.

To start with, notice that the group $\mathbb{Q}$ of rational numbers is divisible and hence the canonical embedding $\mathbb{Z} \hookrightarrow \mathbb{Q}$ shows that the additive group of integers embeds into an injective $\mathbb{Z}$-module.

Now by the discussion at projective module every abelian group $A$ receives an epimorphism $(\oplus_{s \in S} \mathbb{Z}) \to A$ from a free abelian group, hence is the quotient group of a direct sum of copies of $\mathbb{Z}$. Accordingly it embeds into a quotient $\tilde A$ of a direct sum of copies of $\mathbb{Q}$.

$$\array{ ker &\stackrel{=}{\to}& ker \\ \downarrow && \downarrow \\ (\oplus_{s \in S} \mathbb{Z}) &\hookrightarrow& (\oplus_{s \in S} \mathbb{Q}) \\ \downarrow && \downarrow \\ A &\hookrightarrow& \tilde A }$$

Here $\tilde A$ is divisible because the direct sum of divisible groups is again divisible, and also the quotient group of a divisible groups is again divisble. So this exhibits an embedding of any $A$ into a divisible abelian group, hence into an injective $\mathbb{Z}$-module.

Proposition

Assuming the axiom of choice, for $R$ a ring, the category $R$Mod has enough injectives.

The proof uses the following lemma.

Write $U\colon R Mod \to Ab$ for the forgetful functor that forgets the $R$-module structure on a module $N$ and just remembers the underlying abelian group $U(N)$.

Lemma

The functor $U\colon R Mod \to Ab$ has a right adjoint

$$R_* : Ab \to R Mod$$

given by sending an abelian group $A$ to the abelian group

$$U(R_*(A)) \coloneqq Ab(U(R),A)$$

equipped with the $R$-module struture by which for $r \in R$ an element $(U(R) \stackrel{f}{\to} A) \in U(R_*(A))$ is sent to the element $r f$ given by

$$r f : r' \mapsto f(r' \cdot r) \,.$$

This is called the coextension of scalars along the ring homomorphism $\mathbb{Z} \to R$.

The unit of the $(U \dashv R_*)$ adjunction

$$\epsilon_N : N \to R_*(U(N))$$

is the $R$-module homomorphism

$$\epsilon_N : N \to Hom_{Ab}(U(R), U(N))$$

given on $n \in N$ by

$$j(n) : r \mapsto r n \,.$$

Proof

of prop.

Let $N \in R Mod$. We need to find a monomorphism $N \to \tilde N$ such that $\tilde N$ is an injective $R$-module.

By prop. there exists a monomorphism

$$i \colon U(N) \hookrightarrow D$$

of the underlying abelian group into an injective abelian group $D$.

Now consider the $(U \dashv R_*)$-adjunct

$$N \to R_*(D)$$

of $i$, hence the composite

$$N \stackrel{\eta_N}{\to} R_*(U(N)) \stackrel{R_*(i)}{\to} R_*(D)$$

with $R_*$ and $\eta_N$ from lemma . On the underlying abelian groups this is

$$U(N) \stackrel{U(\eta_N)}{\to} Hom_{Ab}(U(R), U(N)) \stackrel{Hom_{Ab}(U(R),i)}{\to} Hom_{Ab}(U(R),U(D)) \,.$$

Hence this is monomorphism. Therefore it is now sufficient to see that $Hom_{Ab}(U(R), U(D))$ is an injective $R$-module.

This follows from the existence of the adjunction isomorphism given by lemma

$$Hom_{Ab}(U(K),U(D)) \simeq Hom_{R Mod}(K, Hom_{Ab}(U(R), U(D)))$$

natural in $K \in R Mod$ and from the injectivity of $D \in Ab$.

$$\array{ U(K) &\to& D \\ \downarrow & \nearrow \\ U(L) } \;\;\;\;\; \leftrightarrow \;\;\;\;\; \array{ K &\to& R_*D \\ \downarrow & \nearrow \\ L } \,.$$

Now we can state the main definition of this section and discuss its central properties.

Definition

For $X \in \mathcal{A}$ an object, an injective resolution of $X$ is a cochain complex $J^\bullet \in Ch^\bullet(\mathcal{A})$ (in non-negative degree) equipped with a quasi-isomorphism

$$i : X \stackrel{\sim}{\to} J^\bullet$$

such that $J^n \in \mathcal{A}$ is an injective object for all $n \in \mathbb{N}$.

Remark

In components the quasi-isomorphism of def. is a chain map of the form

$$\array{ X &\to& 0 &\to& \cdots &\to& 0 &\to& \cdots \\ \downarrow^{\mathrlap{i^0}} && \downarrow && && \downarrow \\ J^0 &\stackrel{d^0}{\to}& J^1 &\stackrel{d^1}{\to}& \cdots &\to& J^n &\stackrel{d^n}{\to}&\cdots } \,.$$

Since the top complex is concentrated in degree 0, this being a quasi-isomorphism happens to be equivalent to the sequence

$$0 \to X \stackrel{i^0}{\to} J^0 \stackrel{d^0}{\to} J^1 \stackrel{d^1}{\to} J^2 \stackrel{d^2}{\to} \cdots$$

being an exact sequence. In this form one often finds the definition of injective resolution in the literature.

Definition

For $X \in \mathcal{A}$ an object, a projective resolution of $X$ is a chain complex $J_\bullet \in Ch_\bullet(\mathcal{A})$ (in non-negative degree) equipped with a quasi-isomorphism

$$p : J_\bullet \stackrel{\sim}{\to} X$$

such that $J_n \in \mathcal{A}$ is a projective object for all $n \in \mathbb{N}$.

Remark

In components the quasi-isomorphism of def. is a chain map of the form

$$\array{ \cdots &\stackrel{\partial_n}{\to}& J_n &\stackrel{\partial_{n-1}}{\to}& \cdots &\to& J_1 &\stackrel{\partial_0}{\to}& J_0 \\ && \downarrow && && \downarrow && \downarrow^{\mathrlap{p_0}} \\ \cdots &\to& 0 &\to& \cdots &\to& 0 &\to& X } \,.$$

Since the bottom complex is concentrated in degree 0, this being a quasi-isomorphism happens to be equivalent to the sequence

$$\cdots J_2 \stackrel{\partial_1}{\to} J_1 \stackrel{\partial_0}{\to} J_0 \stackrel{p_0}{\to} X \to 0$$

being an exact sequence. In this form one often finds the definition of projective resolution in the literature.

We first discuss the existence of injective/projective resolutions, and then the functoriality of their constructions.

Proposition

Let $\mathcal{A}$ be an abelian category with enough injectives, such as our $R$Mod for some ring $R$.

Then every object $X \in \mathcal{A}$ has an injective resolution, def. .

Proof

Let $X \in \mathcal{A}$ be the given object. By remark we need to construct an exact sequence of the form

$$0 \to X \to J^0 \stackrel{d^0}{\to} J^1 \stackrel{d^1}{\to} J^2 \stackrel{d^2}{\to} \cdots \to J^n \to \cdots$$

such that all the $J^\cdot$ are injective objects.

This we now construct by induction on the degree $n \in \mathbb{N}$.

In the first step, by the assumption of enough injectives we find an injective object $J^0$ and a monomorphism

$$X \hookrightarrow J^0$$

hence an exact sequence

$$0 \to X \to J^0 \,.$$

Assume then by induction hypothesis that for $n \in \mathbb{N}$ an exact sequence

$$X \to J^0 \stackrel{d^0}{\to} \cdots \to J^{n-1} \stackrel{d^{n-1}}{\to} J^n$$

has been constructed, where all the $J^\cdot$ are injective objects. Forming the cokernel of $d^{n-1}$ yields the short exact sequence

$$0 \to J^{n-1} \stackrel{d^{n-1}}{\to} J^n \stackrel{p}{\to} J^n/J^{n-1} \to 0 \,.$$

By the assumption that there are enough injectives in $\mathcal{A}$ we may now again find a monomorphism $J^n/J^{n-1} \stackrel{i}{\hookrightarrow} J^{n+1}$ into an injective object $J^{n+1}$. This being a monomorphism means that

$$J^{n-1} \stackrel{d^{n-1}}{\to} J^n \stackrel{d^n \coloneqq i \circ p}{\longrightarrow} J^{n+1}$$

is exact in the middle term. Therefore we now have an exact sequence

$$0 \to X \to J^0 \to \cdots \to J^{n-1} \stackrel{d^{n-1}}{\to} J^n \stackrel{d^{n}}{\to} J^{n+1}$$

which completes the induction step.

The following proposition is formally dual to prop. .

Proposition

Let $\mathcal{A}$ be an abelian category with enough projectives (such as $R$Mod for some ring $R$).

Then every object $X \in \mathcal{A}$ has a projective resolution, def. .

Proof

Let $X \in \mathcal{A}$ be the given object. By remark we need to construct an exact sequence of the form

$$\cdots \stackrel{\partial_2}{\to} J_2 \stackrel{\partial_1}{\to} J_1 \stackrel{\partial_0}{\to} J_0 \to X \to 0$$

such that all the $J_\cdot$ are projective objects.

This we we now construct by induction on the degree $n \in \mathbb{N}$.

In the first step, by the assumption of enough projectives we find a projective object $J_0$ and an epimorphism

$$J_0 \to X$$

hence an exact sequence

$$J_0 \to X \to 0 \,.$$

Assume then by induction hypothesis that for $n \in \mathbb{N}$ an exact sequence

$$J_n \stackrel{\partial_{n-1}}{\to} J_{n-1} \to \cdots \stackrel{\partial_0}{\to} J_0 \to X \to 0$$

has been constructed, where all the $J_\cdot$ are projective objects. Forming the kernel of $\partial_{n-1}$ yields the short exact sequence

$$0 \to ker(\partial_{n-1}) \stackrel{i}{\to} J_n \stackrel{\partial_{n-1}}{\to} J_{n-1} \to 0 \,.$$

By the assumption that there are enough projectives in $\mathcal{A}$ we may now again find an epimorphism $p : J_{n+1} \to ker(\partial_{n-1})$ out of a projective object $J_{n+1}$. This being an epimorphism means that

$$J_{n+1} \stackrel{\partial_{n} \coloneqq i\circ p}{\to} J_n \stackrel{\partial_{n-1}}{\to}$$

is exact in the middle term. Therefore we now have an exact sequence

$$J_{n+1} \stackrel{\partial_n}{\to} J_n \stackrel{\partial_{n-1}}{\to} \cdots \stackrel{\partial_0}{\to} J_0 \to X \to 0 \,,$$

which completes the induction step.

To conclude this section we now show that all this work indeed serves to solve the problem indicated above in example .

Proposition

Let $f^\bullet : X^\bullet \to J^\bullet$ be a chain map of cochain complexes in non-negative degree, out of an exact complex $0 \simeq_{qi} X^\bullet$ to a degreewise injective complex $J^\bullet$. Then there is a null homotopy

$$\eta : 0 \Rightarrow f^\bullet$$

Proof

By definition of chain homotopy we need to construct a sequence of morphisms $(\eta^{n+1} : X^{n+1} \to J^{n})_{n \in \mathbb{N}}$ such that

$$f^n = \eta^{n+1} \circ d^n_X + d^{n-1}_J \circ \eta^n \,.$$

for all $n$. We now construct this by induction over $n$.

It is convenient to start at $n = -1$, take $\eta^{\leq 0} \coloneqq 0$ and $f^{\lt 0} \coloneqq 0$. Then the above condition holds for $n = -1$.

Then in the induction step assume that for given $n \in \mathbb{N}$ we have constructed $\eta^{\bullet \leq n}$ satisfying the above condition for $f^{\lt n}$

First define now

$$g^n \coloneqq f^n - d_J^{n-1} \circ \eta^n$$

and observe that by induction hypothesis

$$\begin{aligned} g^n \circ d_X^{n-1} & = f^n \circ d^{n-1}_X - d^{n-1}_J \circ \eta^n \circ d^{n-1}_X \\ & = f^n \circ d^{n-1}_X - d^{n-1}_J \circ f^{n-1} + d^{n-1}_J \circ d^{n-2}_J \circ \eta^{n-1} \\ & = 0 + 0 \\ & = 0 \end{aligned} \,.$$

This means that $g^n$ factors as

$$X^n \to X^n / im(d^{n-1}_X) \stackrel{g^n}{\to} J^n \,,$$

where the first map is the projection to the quotient.

Observe then that by exactness of $X^\bullet$ the morphism $X^n / im(d^{n-1}_X) \stackrel{d^n_X}{\to} X^{n+1}$ is a monomorphism. Together this gives us a diagram of the form

$$\array{ X^n / im(d^{n-1}_X) &\stackrel{d^n_X}{\to}& X^{n+1} \\ \downarrow^{\mathrlap{g^n}} & \swarrow_{\mathrlap{\eta^{n+1}}} \\ J^n } \,,$$

where the morphism $\eta^{n+1}$ may be found due to the defining right lifting property of the injective object $J^n$ against the top monomorphism.

Observing that the commutativity of this diagram is the chain homotopy condition involving $\eta^n$ and $\eta^{n+1}$, this completes the induction step.

The formally dual statement of prop is the following.

Lemma

Let $f_\bullet : P_\bullet \to Y_\bullet$ be a chain map of chain complexes in non-negative degree, into an exact complex $0 \simeq_{qi} Y_\bullet$ from a degreewise projective complex $X^\bullet$. Then there is a null homotopy

$$\eta : 0 \Rightarrow f_\bullet$$

Proof

This is formally dual to the proof of prop .

Hence we have seen now that injective and projective resolutions of chain complexes serve to make chain homotopy interact well with quasi-isomorphism. In the next section we show that this construction lifts from single chain complexes to chain maps between chain complexes and in fact to the whole category of chain complexes. The resulting “resolved” category of chain complexes is the derived category, the true home of the abelian homotopy theory of chain complexes.

8) The derived category

In the previous section we have seen that every object $A \in \mathcal{A}$ admits an injective resolution and a projective resolution. Here we lift this construction to morphisms and then to the whole category of chain complexes, up to chain homotopy.

The following proposition says that, when injectively resolving objects, the morphisms between these objects lift to the resolutions, and the following one, prop. , says that this lift is unique up to chain homotopy.

Proposition

Let $f : X \to Y$ be a morphism in $\mathcal{A}$. Let

$$i_Y : Y \stackrel{\sim}{\to} Y^\bullet$$

be an injective resolution of $Y$ and

$$i_X : X \stackrel{\sim}{\to} X^\bullet$$

any monomorphism that is a quasi-isomorphism (possibly but not necessarily an injective resolution). Then there is a chain map $f^\bullet : X^\bullet \to Y^\bullet$ giving a commuting diagram

$$\array{ X &\stackrel{\sim}{\to}& X^\bullet \\ \downarrow^{\mathrlap{f}} && \downarrow^{\mathrlap{f^\bullet}} \\ Y &\stackrel{\sim}{\to}& Y^\bullet } \,.$$

Proof

By definition of chain map we need to construct morphisms $(f^n : X^n \to Y^n)_{n \in \mathbb{N}}$ such that for all $n \in \mathbb{N}$ the diagrams

$$\array{ X^{n} &\stackrel{d^n_X}{\to}& X^{n+1} \\ \downarrow^{\mathrlap{f^n}} && \downarrow^{\mathrlap{f^{n+1}}} \\ Y^{n} &\stackrel{d^n_Y}{\to}& Y^{n+1} }$$

commute (the defining condition on a chain map) and such that the diagram

$$\array{ X &\stackrel{i_X}{\to}& X^0 \\ \downarrow^{f} && \downarrow^{\mathrlap{f^0}} \\ Y &\stackrel{i_Y}{\to}& Y^0 }$$

commutes in $\mathcal{A}$ (which makes the full diagram in $Ch^\bullet(\mathcal{A})$ commute).

We construct these $f^\bullet = (f^n)_{n \in \mathbb{N}}$ by induction.

To start the induction, the morphism $f^0$ in the last diagram above can be found by the defining right lifting property of the injective object $Y^0$ against the monomorphism $i_X$.

Assume then that for some $n \in \mathbb{N}$ component maps $f^{\bullet \leq n}$ have been obtained such that $d^k_Y\circ f^k = f^{k+1}\circ d^k_X$ for all $0 \leq k \lt n$ . In order to construct $f^{n+1}$ consider the following diagram, which we will describe/construct stepwise from left to right:

$$\array{ X^n &\stackrel{}{\to}& X^n/im(d^{n-1}_X) &\stackrel{d^n_X}{\hookrightarrow}& X^{n+1} \\ {}^{\mathllap{f^n}}\downarrow & \searrow^{\mathrlap{g^n}} & \downarrow^{\mathrlap{h^n}} & \swarrow_{\mathrlap{f^{n+1}}} \\ Y^n &\underset{d^n_Y}{\to}& Y^{n+1} } \,.$$

Here the morphism $f^n$ on the left is given by induction assumption and we define the diagonal morphism to be the composite

$$g^n \coloneqq d^n_Y \circ f^n \,.$$

Observe then that by the chain map property of the $f^{\bullet \leq n}$ we have

$$d^n_Y \circ f^n \circ d^{n-1}_X = d^n_Y \circ d^{n-1}_Y \circ f^{n-1} = 0$$

and therefore $g^n$ factors through $X^n/im(d^{n-1}_X)$ via some $h^n$ as indicated in the middle of the above diagram. Finally the morphism on the top right is a monomorphism by the fact that $X^{\bullet}$ is exact in positive degrees (being quasi-isomorphic to a complex concentrated in degree 0) and so a lift $f^{n+1}$ as shown on the far right of the diagram exists by the defining lifting property of the injective object $Y^{n+1}$.

The total outer diagram now commutes, being built from commuting sub-diagrams, and this is the required chain map property of $f^{\bullet \leq n+1}$ This completes the induction step.

Proposition

The morphism $f_\bullet$ in prop. is the unique one up to chain homotopy making the given diagram commute.

Proof

Given two cochain maps $g_1^\bullet, g_2^\bullet$ making the diagram commute, a chain homotopy $g_1^\bullet \Rightarrow g_2^\bullet$ is equivalently a null homotopy $0 \Rightarrow g_2^\bullet - g_1^\bullet$ of the difference, which sits in a square of the form

$$\array{ X &\underoverset{h^\bullet}{\sim}{\to}& X^\bullet \\ \downarrow^{\mathrlap{0}} && \downarrow^{\mathrlap{f^\bullet \coloneqq g_2^\bullet - g_1^\bullet}} \\ Y &\stackrel{\sim}{\to}& Y^\bullet }$$

with the left vertical morphism being the zero morphism (and the bottom an injective resolution). Hence we have to show that in such a diagram $f^\bullet$ is null-homotopic.

This we may reduce to the statement of prop. by considering instead of $f^\bullet$ the induced chain map of augmented complexes

$$\array{ 0 &\stackrel{}{\to}& X &\stackrel{h^0}{\to}& X^0 &\stackrel{d^0_X}{\to}& X^1 &\to& \cdots \\ \downarrow^{\mathrlap{f^{-2} = 0}} && \downarrow^{\mathrlap{f^{-1} = 0}} && \downarrow^{f^0} && \downarrow^{f^1} \\ 0 &\to& Y &\to& Y^0 &\stackrel{d^0_J}{\to}& Y^1 &\to& \cdots } \,,$$

where the second square from the left commutes due to the commutativity of the original square of chain complexes in degree 0.

Since $h^\bullet$ is a quasi-isomorphism, the top chain complex is exact, by remark . Moreover the bottom complex consists of injective objects from the second degree on (the former degree 0). Hence the induction in the proof of prop. implies the existence of a null homotopy

$$\array{ 0 &\stackrel{}{\to}& X &\stackrel{}{\to}& X^0 &\stackrel{d^0_X}{\to}& X^1 &\to& \cdots \\ \downarrow^{\mathrlap{f^{-2} = 0}} &\swarrow_{\mathrlap{\eta^{-1} = 0}}& \downarrow^{\mathrlap{f^{-1} = 0}} &\swarrow_{\mathrlap{\eta^0 = 0} }& \downarrow^{f^0} &\swarrow_{\mathrlap{\eta^1}}& \downarrow^{f^1} \\ 0 &\to& Y &\to& Y^0 &\stackrel{d^0_Y}{\to}& Y^1 &\to& \cdots }$$

starting with $\eta^{-1} = 0$ and $\eta^{0 } = 0$ (notice that the proof prop. was formulated exactly this way), which works because $f^{-1} = 0$. The de-augmentation $\{f^{\bullet \geq 0}\}$ of this is the desired null homotopy of $f^\bullet$.

We now discuss how the injective/projective resolutions constructed above are functorial if regarded in the homotopy category of chain complexes, def. . For definiteness, to be able to distinguish chain complexes from cochain complexes, introduce the following notation.

Definition

(the derived category)

Write as before

$$\mathcal{K}_\bullet(\mathcal{A}) \in Cat$$

for the strong chain homotopy category of chain complexes, from def. .

Write similarly now

$$\mathcal{K}^\bullet(\mathcal{A}) \in Cat$$

for the strong chain homotopy category of co-chain complexes.

Write furthermore

$$\mathcal{D}_\bullet(\mathcal{A}) \coloneqq \mathcal{K}_\bullet(\mathcal{P}_{\mathcal{A}}) \hookrightarrow \mathcal{K}_\bullet(\mathcal{A})$$

for the full subcategory on the degreewise projective chain complexes, and

$$\mathcal{D}^\bullet(\mathcal{A}) \coloneqq \mathcal{K}^\bullet(\mathcal{I}_{\mathcal{A}}) \hookrightarrow \mathcal{K}^\bullet(\mathcal{A})$$

for the full subcategory on the degreewise injective cochain complexes.

These subcategories – or any category equivalent to them – are called the (strictly bounded above/below) derived category of $\mathcal{A}$.

Remark

Often one defines the derived category by more general abstract means than we have introduced here, namely as the localization of the category of chain complexes at the quasi-isomorphims. If one does this, then the simple definition def. is instead a theorem. The interested reader can find more details and further pointers here.

Theorem

If $\mathcal{A}$ has enough injectives, def. , then there exists a functor

$$P : \mathcal{A} \to \mathcal{D}^\bullet(\mathcal{A}) = \mathcal{K}^\bullet(\mathcal{I}_{\mathcal{A}})$$

together with natural isomorphisms

$$H^0(-) \circ P \simeq id_{\mathcal{A}}$$

and

$$H^{n \geq 1}(-) \circ P \simeq 0 \,.$$

Proof

By prop. every object $X^\bullet \in Ch^\bullet(\mathcal{A})$ has an injective resolution. Proposition says that for $X \to X^\bullet$ and $X \to \tilde X^\bullet$ two resolutions there is a morphism $X^\bullet \to \tilde X^\bullet$ in $\mathcal{K}^\bullet(\mathcal{A})$ and prop. says that this morphism is unique in $\mathcal{K}^\bullet(\mathcal{A})$. In particular it is therefore an isomorphism in $\mathcal{K}^\bullet(\mathcal{A})$ (since the composite with the reverse lifted morphism, also being unique, has to be the identity).

So choose one such injective resolution $P(X)^\bullet$ for each $X^\bullet$.

Then for $f : X \to Y$ any morphism in $\mathcal{A}$, proposition again says that it can be lifted to a morphism between $P(X)^\bullet$ and $P(Y)^\bullet$ and proposition says that there is an image in $\mathcal{K}^\bullet(\mathcal{A})$, unique for morphism making the given diagram commute.

This implies that this assignment of morphisms is functorial, since then also the composites are unique.

Dually we have:

Theorem

If $\mathcal{A}$ has enough projectives, def. , then there exists a functor

$$Q : \mathcal{A} \to \mathcal{D}_\bullet(\mathcal{A}) =\mathcal{K}_\bullet(\mathcal{P}_{\mathcal{A}})$$

together with natural isomorphisms

$$H_0(-) \circ P \simeq id_{\mathcal{A}}$$

and

$$H_{n \geq 1}(-) \circ P \simeq 0 \,.$$

For actually working with the derived category, the following statement is of central importance, which we record here without proof (which requires a bit of localization theory). It says that for computing hom-sets in the derived category, it is in fact sufficient to just resolve the domain or the codomain.

Proposition

Let $X_\bullet, Y_\bullet \in Ch_\bullet(\mathcal{A})$. We have natural isomorphisms

$$\mathcal{D}_\bullet(Q(X)_\bullet, Q(Y)_\bullet) \simeq \mathcal{K}_\bullet(Q(X)_\bullet, Y_\bullet) \,.$$

Dually, for $X^\bullet, Y^\bullet \in Ch^\bullet(\mathcal{A})$, we have a natural isomorphism

$$\mathcal{D}^\bullet(P(X)^\bullet, P(Y)^\bullet) \simeq \mathcal{K}^\bullet(X^\bullet, P(Y)^\bullet) \,.$$

In conclusion we have found that there are resolution functors that embed $\mathcal{A}$ in the homotopically correct context of resolved chain complexes with chain maps up to chain homotopy between them.

In the next section we discuss the general properties of this “homotopically correct context”: the derived category.

9) Derived functors

In the previous section we have seen how the entire category $\mathcal{A}$ (= $R$Mod) embeds into its derived category, the category of degreewise injective cochain complexes

$$P : \mathcal{A} \to \mathcal{D}^\bullet(\mathcal{A}) = \mathcal{K}^\bullet(\mathcal{I}_{\mathcal{A}})$$

or degreewise projective chain complexes

$$Q : \mathcal{A} \to \mathcal{D}_\bullet(\mathcal{A}) = \mathcal{K}_\bullet(\mathcal{P}_{\mathcal{A}})$$

modulo chain homotopy. This construction of the derived category naturally gives rise to the following notion of derived functors.

Definition

For $\mathcal{A}, \mathcal{B}$ two abelian categories (e.g. $R$Mod and $R'$Mod), a functor

$$F \colon \mathcal{A} \to \mathcal{B}$$

is called an additive functor if

1. $F$ maps the zero object to the zero object, $F(0) \simeq 0 \in \mathcal{B}$;

2. given any two objects $x, y \in \mathcal{A}$, there is an isomorphism $F(x \oplus y) \cong F(x) \oplus F(y)$, and this respects the inclusion and projection maps of the direct sum:

$$\array { x & & & & y \\ & {}_{\mathllap{i_X}}\searrow & & \swarrow_{\mathrlap{i_y}} \\ & & x \oplus y \\ & {}^{\mathllap{p_x}}\swarrow & & \searrow^{\mathrlap{p_y}} \\ x & & & & y } \quad\quad\stackrel{F}{\mapsto}\quad\quad \array { F(x) & & & & F(y) \\ & {}_{\mathllap{i_{F(x)}}}\searrow & & \swarrow_{\mathrlap{i_{F(y)}}} \\ & & F(x \oplus y) \cong F(x) \oplus F(y) \\ & {}^{\mathllap{p_{F(X)}}}\swarrow & & \searrow^{\mathrlap{p_{F(y)}}} \\ F(x) & & & & F(y) }$$

Definition

Given an additive functor $F : \mathcal{A} \to \mathcal{A}'$, it canonically induces a functor

$$Ch_\bullet(F) \colon Ch_\bullet(\mathcal{A}) \to Ch_\bullet(\mathcal{A}')$$

between categories of chain complexes (its “prolongation”) by applying it to each chain complex and to all the diagrams in the definition of a chain map. Similarly it preserves chain homotopies and hence it passes to the quotient given by the strong homotopy category of chain complexes

$$\mathcal{K}(F) \colon \mathcal{K}(\mathcal{A}) \to \mathcal{K}(\mathcal{A}') \,.$$

Remark

If $\mathcal{A}$ and $\mathcal{A}'$ have enough projectives, then their derived categories are

$$\mathcal{D}_\bullet(\mathcal{A}) \simeq \mathcal{K}_\bullet(\mathcal{P}_{\mathcal{A}})$$

and

$$\mathcal{D}^\bullet(\mathcal{A}) \simeq \mathcal{K}^\bullet(\mathcal{I}_{\mathcal{A}})$$

etc. One wants to accordingly derive from $F$ a functor $\mathcal{D}_\bullet(\mathcal{A}) \to \mathcal{D}_\bullet(\mathcal{A})$ between these derived categories. It is immediate to achieve this on the domain category, there we can simply precompose and form

$$\mathcal{A} \to \mathcal{D}_\bullet(\mathcal{A}) \simeq \mathcal{K}_\bullet(\mathcal{P}_{\mathcal{A}}) \hookrightarrow \mathcal{K}_\bullet(\mathcal{A}) \stackrel{\mathcal{K}_\bullet(F)}{\to} \mathcal{K}_\bullet(\mathcal{A}') \,.$$

But the resulting composite lands in $\mathcal{K}_\bullet(\mathcal{A}')$ and in general does not factor through the inclusion $\mathcal{D}_\bullet(\mathcal{A}') = \mathcal{K}_\bullet(\mathcal{P}_{\mathcal{A}'}) \hookrightarrow \mathcal{K}_\bullet(\mathcal{A}')$.

In a more general abstract discussion than we present here, one finds that by applying a projective resolution functor on chain complexes, one can enforce this factorization. However, by definition of resolution, the resulting chain complex is quasi-isomorphic to the one obtained by the above composite.

This means that if one is only interested in the “weak chain homology type” of the chain complex in the image of a derived functor, then forming chain homology groups of the chain complexes in the images of the above composite gives the desired information. This is what def. and def. below do.

Definition

Let $\mathcal{A}, \mathcal{A}'$ be two abelian categories, for instance $\mathcal{A} = R$Mod and $\mathcal{A}' = R'$Mod. Then a functor $F \colon \mathcal{A} \to \mathcal{A}'$ which preserves direct sums (and hence in particular the zero object) is called

• a left exact functor if it preserves kernels;

• a right exact functor if it preserves cokernels;

• an exact functor if it is both left and right exact.

Here to “preserve kernels” means that for every morphism $X \stackrel{f}{\to} Y$ in $\mathcal{A}$ we have an isomorphism on the left of the following commuting diagram

$$\array{ F(ker(f)) &\to& F(X) & \stackrel{F(f)}{\to} & F(Y) \\ \downarrow^{\mathrlap{\simeq}} && \downarrow^{\mathrlap{=}} && \downarrow^{\mathrlap{=}} \\ ker(F(f)) &\to& F(X) &\stackrel{F(f)}{\to}& F(Y) } \,,$$

hence that both rows are exact. And dually for right exact functors.

We record the following immediate consequence of this definition (which in the literature is often taken to be the definition).

Proposition

If $F$ is a left exact functor, then for every exact sequence of the form

$$0 \to A \to B \to C$$

also

$$0 \to F(A) \to F(B) \to F(C)$$

is an exact sequence. Dually, if $F$ is a right exact functor, then for every exact sequence of the form

$$A \to B \to C \to 0$$

also

$$F(A) \to F(B) \to F(C) \to 0$$

is an exact sequence.

Proof

If $0 \to A \to B \to C$ is exact then $A \hookrightarrow B$ is a monomorphism by prop. . But then the statement that $A \to B \to C$ is exact at $B$ says precisely that $A$ is the kernel of $B \to C$. So if $F$ is left exact then by definition also $F(A) \to F(B)$ is the kernel of $F(B) \to F(C)$ and so is in particular also a monomorphism. Dually for right exact functors.

Remark

Proposition is clearly the motivation for the terminology in def. : a functor is left exact if it preserves short exact sequences to the left, and right exact if it preserves them to the right.

Now we can state the main two definitions of this section.

Definition

Let

$$F : \mathcal{A} \to \mathcal{A}'$$

be a left exact functor between abelian categories such that $\mathcal{A}$ has enough injectives. For $n \in \mathbb{N}$ the $n$th right derived functor of $F$ is the composite

$$R^n F : \mathcal{A} \stackrel{P}{\to} \mathcal{K}^\bullet(\mathcal{I}_{\mathcal{A}}) \stackrel{\mathcal{K}^\bullet(F)}{\to} \mathcal{K}^\bullet(\mathcal{A}') \stackrel{H^n(-)}{\to} \mathcal{A}' \,,$$

where

• $P$ is the injective resolution functor of theorem ;

• $\mathcal{K}(F)$ is the prolongation of $F$ according to def. ;

• $H^n(-)$ is the $n$-chain homology functor. Hence

$$(R^n F)(X^\bullet) \coloneqq H^n(F(P(X)^\bullet)) \,.$$

Dually:

Definition

Let

$$F : \mathcal{A} \to \mathcal{A}'$$

be a right exact functor between abelian categories such that $\mathcal{A}$ has enough projectives. For $n \in \mathbb{N}$ the $n$th left derived functor of $F$ is the composite

$$L_n F : \mathcal{A} \stackrel{Q}{\to} K_\bullet(\mathcal{P}_{\mathcal{A}}) \stackrel{\mathcal{K}_\bullet(F)}{\to} \mathcal{K}_\bullet(\mathcal{A}') \stackrel{H_n(-)}{\to} \mathcal{A}' \,,$$

where

• $Q$ is the projective resolution functor of theorem ;

• $\mathcal{K}(F)$ is the prolongation of $F$ according to def. ;

• $H_n(-)$ is the $n$-chain homology functor. Hence

$$(L_n F)(X_\bullet) \coloneqq H_n(F(Q(X)_\bullet)) \,.$$

The following proposition says that in degree 0 these derived functors coincide with the original functors.

Proposition

Let $F \colon \mathcal{A} \to \mathcal{B}$ a left exact functor, def. in the presence of enough injectives. Then for all $X \in \mathcal{A}$ there is a natural isomorphism

$$R^0F(X) \simeq F(X) \,.$$

Dually, if $F$ is a right exact functor in the presence of enough projectives, then

$$L_0 F(X) \simeq F(X) \,.$$

Proof

We discuss the first statement, the second is formally dual.

By remark an injective resolution $X \stackrel{\simeq_{qi}}{\to} X^\bullet$ is equivalently an exact sequence of the form

$$0 \to X \hookrightarrow X^0 \to X^1 \to \cdots \,.$$

If $F$ is left exact then it preserves this excact sequence by definition of left exactness, and hence

$$0 \to F(X) \hookrightarrow F(X^0) \to F(X^1) \to \cdots$$

is an exact sequence. But this means that

$$R^0 F(X) \coloneqq ker(F(X^0) \to F(X^1)) \simeq F(X) \,.$$

The following immediate consequence of the definition is worth recording:

Proposition

Let $F$ be an additive functor.

• If $F$ is right exact and $N \in \mathcal{A}$ is a projective object, then

  $$L_n F(N) = 0 \;\;\;\; \forall n \geq 1 \,.$$

• If $F$ is left exact and $N \in \mathcal{A}$ is a injective object, then

  $$R^n F(N) = 0 \;\;\;\; \forall n \geq 1 \,.$$

Proof

If $N$ is projective then the chain complex $[\cdots \to 0 \to 0 \to N]$ is already a projective resolution and hence by definition $L_n F(N) \simeq H_n(0)$ for $n \geq 1$. Dually if $N$ is an injective object.

For proving the basic property of derived functors below in prop. which continues these basis statements to higher degree, in a certain way, we need the following technical lemma.

Lemma

For $0 \to A \stackrel{i}{\to} B \stackrel{p}{\to} C \to 0$ a short exact sequence in an abelian category with enough projectives, there exists a commuting diagram of chain complexes

$$\array{ 0 &\to& A_\bullet &\to& B_\bullet &\to& C_\bullet &\to& 0 \\ && \downarrow^{\mathrlap{f_\bullet}} && \downarrow^{\mathrlap{g_\bullet}} && \downarrow^{\mathrlap{h_\bullet}} \\ 0 &\to& A &\stackrel{i}{\to}& B &\stackrel{p}{\to}& C &\to& 0 }$$

where

• each vertical morphism is a projective resolution;

and in addition

• the top row is again a short exact sequence of chain complexes.

Proof

By prop. we can choose $f_\bullet$ and $h_\bullet$. The task is now to construct the third resolution $g_\bullet$ such as to obtain a short exact sequence of chain complexes, hence degreewise a short exact sequence, in the two row.

To construct this, let for each $n \in \mathbb{N}$

$$B_n \coloneqq A_n \oplus C_n$$

be the direct sum and let the top horizontal morphisms be the canonical inclusion and projection maps of the direct sum.

Let then furthermore (in matrix calculus notation)

$$g_0 = \left( \array{ (j_0)_A & (j_0)_B } \right) : A_0 \oplus C_0 \to B$$

be given in the first component by the given composite

$$(g_0)_A : A_0 \oplus C_0 \stackrel{}{\to} A_0 \stackrel{f_0}{\to} A \stackrel{i}{\hookrightarrow} B$$

and in the second component we take

$$(j_0)_C : A_0 \oplus C_0 \to C_0 \stackrel{\zeta}{\to} B$$

to be given by a lift in

$$\array{ && B \\ & {}^{\mathllap{\zeta}}\nearrow & \downarrow^{\mathrlap{p}} \\ C_0 &\stackrel{h_0}{\to}& C } \,,$$

which exists by the left lifting property of the projective object $C_0$ (since $C_\bullet$ is a projective resolution) against the epimorphism $p : B \to C$ of the short exact sequence.

In total this gives in degree 0

$$\array{ A_0 &\hookrightarrow& A_0 \oplus C_0 &\to& C_0 \\ {}^{\mathllap{f_0}}\downarrow && {}^{\mathllap{((g_0)_A, (g_0)_C)}}\downarrow &\swarrow_{\zeta}& \downarrow^{\mathrlap{h_0}} \\ A &\stackrel{i}{\hookrightarrow}& B &\stackrel{p}{\to}& C } \,.$$

Let then the differentials of $B_\bullet$ be given by

$$d_k^{B_\bullet} = \left( \array{ d_k^{A_\bullet} & (-1)^k e_k \\ 0 & d_k^{C_\bullet} } \right) : A_{k+1} \oplus C_{k+1} \to A_k \oplus C_k \,,$$

where the $\{e_k\}$ are constructed by induction as follows. Let $e_0$ be a lift in

$$\array{ & && A_0 \\ & & {}^{\mathllap{e_0}}\nearrow & \downarrow^{\mathrlap{f_0}} \\ \zeta \circ d^{C_\bullet}_0 \colon & C_1 &\stackrel{}{\to}& A &\hookrightarrow B& }$$

which exists since $C_1$ is a projective object and $A_0 \to A$ is an epimorphism by $A_\bullet$ being a projective resolution. Here we are using that by exactness the bottom morphism indeed factors through $A$ as indicated, because the definition of $\zeta$ and the chain complex property of $C_\bullet$ gives

$$\begin{aligned} p \circ \zeta \circ d^{C_\bullet}_0 &= h_0 \circ d^{C_\bullet}_0 \\ & = 0 \circ h_1 \\ & = 0 \end{aligned} \,.$$

Now in the induction step, assuming that $e_{n-1}$ has been been found satisfying the chain complex property, let $e_n$ be a lift in

$$\array{ & && A_n \\ & & {}^{\mathllap{e_{n}}}\nearrow & \downarrow^{\mathrlap{d^{A_\bullet}_{n-1}}} \\ e_{n-1}\circ d_n^{C_\bullet} \colon & C_{n+1} &\stackrel{}{\hookrightarrow}& ker(d^{A_\bullet}_{n-1}) = im(d^{A_\bullet}_{n-1})) &\to& A_{n-1} } \,,$$

which again exists since $C_{n+1}$ is projective. That the bottom morphism factors as indicated is the chain complex property of $e_{n-1}$ inside $d^{B_\bullet}_{n-1}$.

To see that the $d^{B_\bullet}$ defines this way indeed squares to 0 notice that

$$d^{B_\bullet}_{n} \circ d^{B_\bullet}_{n+1} = \left( \array{ 0 & (-1)^{n}\left(e_{n} \circ d^{C_\bullet}_{n+1} - d^{A_\bullet}_n \circ e_{n+1} \right) \\ 0 & 0 } \right) \,.$$

This vanishes by the very commutativity of the above diagram.

This establishes $g_\bullet$ such that the above diagram commutes and the bottom row is degreewise a short exact sequence, in fact a split exact sequence, by construction.

To see that $g_\bullet$ is indeed a quasi-isomorphism, consider the homology long exact sequence associated to the short exact sequence of cochain complexes $0 \to A_\bullet \to B_\bullet \to C_\bullet \to 0$. In positive degrees it implies that the chain homology of $B_\bullet$ indeed vanishes. In degree 0 it gives the short sequence $0 \to A \to H_0(B_\bullet) \to B\to 0$ sitting in a commuting diagram

$$\array{ 0 &\to& A &\hookrightarrow& H_0(B_\bullet) &\to& C &\to& 0 \\ \downarrow && \downarrow^{\mathrlap{=}} && \downarrow && \downarrow^{\mathrlap{=}} && \downarrow \\ 0 &\to& A &\hookrightarrow& B &\to& C &\to& 0 \,, }$$

where both rows are exact. That the middle vertical morphism is an isomorphism then follows by the five lemma.

The formally dual statement to lemma is the following.

Lemma

For $0 \to A \to B \to C \to 0$ a short exact sequence in an abelian category with enough injectives, there exists a commuting diagram of cochain complexes

$$\array{ 0 &\to& A &\to& B &\to& C &\to& 0 \\ && \downarrow^{\mathrlap{}} && \downarrow^{\mathrlap{}} && \downarrow^{\mathrlap{}} \\ 0 &\to& A^\bullet &\to& B^\bullet &\to& C^\bullet &\to& 0 }$$

where

• each vertical morphism is an injective resolution;

and in addition

• the bottom row is again a short exact sequence of cochain complexes.

The central general fact about derived functors to be discussed here is now the following.

Proposition

Let $\mathcal{A}, \mathcal{B}$ be abelian categories and assume that $\mathcal{A}$ has enough injectives.

Let $F : \mathcal{A} \to \mathcal{B}$ be a left exact functor and let

$$0 \to A \to B \to C \to 0$$

be a short exact sequence in $\mathcal{A}$.

Then there is a long exact sequence of images of these objects under the right derived functors $R^\bullet F(-)$ of def.

$$\array{ 0 &\to& R^0F (A) &\to& R^0 F(B) &\to& R^0 F(C) &\stackrel{\delta_0}{\to}& R^1 F(A) &\to& R^1 F(B) &\to& R^1F(C) &\stackrel{\delta_1}{\to}& R^2 F(A) &\to& \cdots \\ && \downarrow^{\mathrlap{\simeq}} && \downarrow^{\mathrlap{\simeq}} && \downarrow^{\mathrlap{\simeq}} \\ 0 &\to& F(A) &\to& F(B) &\to& F(C) }$$

in $\mathcal{B}$.

Proof

By lemma we can find an injective resolution

$$0 \to A^\bullet \to B^\bullet \to C^\bullet \to 0$$

of the given exact sequence which is itself again an exact sequence of cochain complexes.

Since $A^n$ is an injective object for all $n$, its component sequences $0 \to A^n \to B^n \to C^n \to 0$ are indeed split exact sequences (see the discussion there). Splitness is preserved by any functor $F$ (and also since $F$ is additive it even preserves the direct sum structure that is chosen in the proof of lemma ) and so it follows that

$$0 \to F(\tilde A^\bullet) \to F(\tilde B^\bullet) \to F(\tilde C^\bullet) \to 0$$

is a again short exact sequence of cochain complexes, now in $\mathcal{B}$. Hence we have the corresponding homology long exact sequence from prop. :

$$\cdots \to H^{n-1}(F(A^\bullet)) \to H^{n-1}(F(B^\bullet)) \to H^{n-1}(F(C^\bullet)) \stackrel{\delta}{\to} H^n(F(A^\bullet)) \to H^n(F(B^\bullet)) \to H^n(F(C^\bullet)) \stackrel{\delta}{\to} H^{n+1}(F(A^\bullet)) \to H^{n+1}(F(B^\bullet)) \to H^{n+1}(F(C^\bullet)) \to \cdots \,.$$

By construction of the resolutions and by def. , this is equal to

$$\cdots \to R^{n-1}F(A) \to R^{n-1}F(B) \to R^{n-1}F(C) \stackrel{\delta}{\to} R^{n}F(A) \to R^{n}F(B) \to R^{n}F(C) \stackrel{\delta}{\to} R^{n+1}F(A) \to R^{n+1}F(B) \to R^{n+1}F(C) \to \cdots \,.$$

Finally the equivalence of the first three terms with $F(A) \to F(B) \to F(C)$ is given by prop. .

Remark

Prop. implies that one way to interpret $R^1 F(A)$ is as a “measure for how a left exact functor $F$ fails to be an exact functor”. For, with $A \to B \to C$ any short exact sequence, this proposition gives the exact sequence

$$0 \to F(A) \to F(B) \to F(C) \to R^1 F(A)$$

and hence $0 \to F(A) \to F(B) \to F(C) \to 0$ is a short exact sequence itself precisely if $R^1 F(A) \simeq 0$.

Dually, if $F$ is right exact functor, then $L_1 F (C)$ “measures how $F$ fails to be exact” for then

$$L_1F (C) \to F(A) \to F(B) \to F(C) \to 0$$

is an exact sequence and hence is a short exact sequence precisely if $L_1F(C) \simeq 0$.

Notice that in fact we even have the following statement (following directly from the definition).

Proposition

Let $F$ be an additive functor which is an exact functor. Then

$$R^{\geq 1} F = 0$$

and

$$L_{\geq 1} F = 0 \,.$$

Proof

Because an exact functor preserves all exact sequences. If $Y_\bullet \to A$ is a projective resolution then also $F(Y)_\bullet$ is exact in all positive degrees, and hence $L_{n\geq 1} F(A) ) H_{n \geq}(F(Y)) = 0$. Dually for $R^n F$.

Conversely:

Definition

Let $F \colon \mathcal{A} \to \mathcal{B}$ be a left or right exact additive functor. An object $A \in \mathcal{A}$ is called an $F$-acyclic object if all positive-degree right/left derived functors of $F$ are zero on $A$.

Acyclic objects are useful for computing derived functors on non-acyclic objects. More generally, we now discuss how the derived functor of an additive functor $F$ may also be computed not necessarily with genuine injective/projective resolutions, but with (just) “$F$-injective”/“$F$-projective resolutions”.

While projective resolutions in $\mathcal{A}$ are sufficient for computing every left derived functor on $Ch_\bullet(\mathcal{A})$ and injective resolutions are sufficient for computing every right derived functor on $Ch^\bullet(\mathcal{A})$, if one is interested just in a single functor $F$ then such resolutions may be more than necessary. A weaker kind of resolution which is still sufficient is then often more convenient for applications. These $F$-projective resolutions and $F$-injective resolutions, respectively, we discuss now. A special case of both are $F$-acyclic resolutions.

Let $\mathcal{A}, \mathcal{B}$ be abelian categories and let $F \colon \mathcal{A} \to \mathcal{B}$ be an additive functor.

Definition

Assume that $F$ is left exact. An additive full subcategory $\mathcal{I} \subset \mathcal{A}$ is called $F$-injective (or: consisting of $F$-injective objects) if

1. for every object $A \in \mathcal{A}$ there is a monomorphism $A \to \tilde A$ into an object $\tilde A \in \mathcal{I} \subset \mathcal{A}$;

2. for every short exact sequence $0 \to A \to B \to C \to 0$ in $\mathcal{A}$ with $A, B \in \mathcal{I} \subset \mathcal{A}$ also $C \in \mathcal{I} \subset \mathcal{A}$;

3. for every short exact sequence $0 \to A \to B \to C \to 0$ in $\mathcal{A}$ with $A\in \mathcal{I} \subset \mathcal{A}$ also $0 \to F(A) \to F(B) \to F(C) \to 0$ is a short exact sequence in $\mathcal{B}$.

And dually:

Definition

Assume that $F$ is right exact. An additive full subcategory $\mathcal{P} \subset \mathcal{A}$ is called $F$-projective (or: consisting of $F$-projective objects) if

1. for every object $A \in \mathcal{A}$ there is an epimorphism $\tilde A \to A$ from an object $\tilde A \in \mathcal{P} \subset \mathcal{A}$;

2. for every short exact sequence $0 \to A \to B \to C \to 0$ in $\mathcal{A}$ with $B, C \in \mathcal{P} \subset \mathcal{A}$ also $A \in \mathcal{P} \subset \mathcal{A}$;

3. for every short exact sequence $0 \to A \to B \to C \to 0$ in $\mathcal{A}$ with $C\in \mathcal{I} \subset \mathcal{A}$ also $0 \to F(A) \to F(B) \to F(C) \to 0$ is a short exact sequence in $\mathcal{B}$.

With the $\mathcal{I},\mathcal{P}\subset \mathcal{A}$ as above, we say:

Definition

For $A \in \mathcal{A}$,

• an $F$-injective resolution of $A$ is a cochain complex $I^\bullet \in Ch^\bullet(\mathcal{I}) \subset Ch^\bullet(\mathcal{A})$ and a quasi-isomorphism

  $$A \stackrel{\simeq_{qi}}{\to} I^\bullet$$

• an $F$-projective resolution of $A$ is a chain complex $Q_\bullet \in Ch_\bullet(\mathcal{P}) \subset Ch^\bullet(\mathcal{A})$ and a quasi-isomorphism

  $$Q_\bullet \stackrel{\simeq_{qi}}{\to} A \,.$$

Let now $\mathcal{A}$ have enough projectives / enough injectives, respectively, def. .

Example

For $F \colon \mathcal{A} \to \mathcal{B}$ an additive functor, let $Ac \subset \mathcal{A}$ be the full subcategory on the $F$-acyclic objects, def. . Then

• if $F$ is left exact, then $\mathcal{I} \coloneqq Ac$ is a subcategory of $F$-injective objects;

• if $F$ is right exact, then $\mathcal{P} \coloneqq Ac$ is a subcategory of $F$-projective objects.

Proof

Consider the case that $F$ is right exact. The other case works dually. Then the first condition of def. is satisfied because every injective object is an $F$-acyclic object and by assumption there are enough of these.

For the second and third condition of def. use that there is the long exact sequence of derived functors prop.

$$0 \to A \to B \to C \to R^1 F(A) \to R^1 F(B) \to R^1 F(C) \to R^2 F(A) \to R^2 F(B) \to R^2 F(C) \to \cdot \,.$$

For the second condition, by assumption on $A$ and $B$ and definition of $F$-acyclic object we have $R^n F(A) \simeq 0$ and $R^n F(B) \simeq 0$ for $n \geq 1$ and hence short exact sequences

$$0 \to 0 \to R^n F(C) \to 0$$

which imply that $R^n F(C)\simeq 0$ for all $n \geq 1$, hence that $C$ is acyclic.

Similarly, the third condition is equivalent to $R^1 F(A) \simeq 0$.

Example

The $F$-projective/injective resolutions by acyclic objects as in example are called $F$-acyclic resolutions.

Let $\mathcal{A}$ be an abelian category with enough injectives. Let $F \colon \mathcal{A} \to \mathcal{B}$ be an additive left exact functor with right derived functor $R_\bullet F$, def. . Finally let $\mathcal{I} \subset \mathcal{A}$ be a subcategory of $F$-injective objects, def. .

Lemma

If a cochain complex $A^\bullet \in Ch^\bullet(\mathcal{I}) \subset Ch^\bullet(\mathcal{A})$ is quasi-isomorphic to 0,

$$X^\bullet \stackrel{\simeq_{qi}}{\to} 0$$

then also $F(X^\bullet) \in Ch^\bullet(\mathcal{B})$ is quasi-isomorphic to 0

$$F(X^\bullet) \stackrel{\simeq_{qi}}{\to} 0 \,.$$

Proof

Consider the following collection of short exact sequences obtained from the long exact sequence $X^\bullet$:

$$0 \to X^0 \stackrel{d^0}{\to} X^1 \stackrel{d^1}{\to} im(d^1) \to 0$$

$$0 \to im(d^1) \to X^2 \stackrel{d^2}{\to} im(d^2) \to 0$$

$$0 \to im(d^2) \to X^3 \stackrel{d^3}{\to} im(d^3) \to 0$$

and so on. Going by induction through this list and using the second condition in def. we have that all the $im(d^n)$ are in $\mathcal{I}$. Then the third condition in def. says that all the sequences

$$0 \to F(im(d^n)) \to F(X^{n+1}) \to F(im(d^{n+1})) \to 0$$

are exact. But this means that

$$0 \to F(X^0)\to F(X^1) \to F(X^2) \to \cdots$$

is exact, hence that $F(X^\bullet)$ is quasi-isomorphic to 0.

Theorem

For $A \in \mathcal{A}$ an object with $F$-injective resolution $A \stackrel{\simeq_{qi}}{\to} I_F^\bullet$, def. , we have for each $n \in \mathbb{N}$ an isomorphism

$$R^n F(A) \simeq H^n(F(I_F^\bullet))$$

between the $n$th right derived functor, def. of $F$ evaluated on $A$ and the cochain cohomology of $F$ applied to the $F$-injective resolution $I_F^\bullet$.

Proof

By prop. we can also find an injective resolution $A \stackrel{\simeq_{qi}}{\to} I^\bullet$. By prop. there is a lift of the identity on $A$ to a chain map $I^\bullet_F \to I^\bullet$ such that the diagram

$$\array{ A &\stackrel{\simeq_{qi}}{\to}& I_F^\bullet \\ \downarrow^{\mathrlap{id}} && \downarrow^{\mathrlap{f}} \\ A &\stackrel{\simeq_{qi}}{\to}& I^\bullet }$$

commutes in $Ch^\bullet(\mathcal{A})$. Therefore by the 2-out-of-3 property of quasi-isomorphisms it follows that $f$ is a quasi-isomorphism

Let $Cone(f) \in Ch^\bullet(\mathcal{A})$ be the mapping cone of $f$ and let $I^\bullet \to Cone(f)$ be the canonical chain map into it. By the explicit formulas for mapping cones, we have that

1. there is an isomorphism $F(Cone(f)) \simeq Cone(F(f))$;

2. $Cone(f) \in Ch^\bullet(\mathcal{I})\subset Ch^\bullet(\mathcal{A})$ (because $F$-injective objects are closed under direct sum). %I is injective but not F-injective, so this does not seem to follow%

The first implies that we have a homology exact sequence

$$\cdots \to H^n(I^\bullet) \to H^n(I_F^\bullet) \to H^n(Cone(f)^\bullet) \to H^{n+1}(I^\bullet) \to H^{n+1}(I_F^\bullet) \to H^{n+1}(Cone(f)^\bullet) \to \cdots \,.$$

Observe that with $f^\bullet$ a quasi-isomorphism $Cone(f^\bullet)$ is quasi-isomorphic to 0. Therefore the second item above implies with lemma that also $F(Cone(f))$ is quasi-isomorphic to 0. This finally means that the above homology exact sequences consists of exact pieces of the form

$$0 \to (R^n F(A)\coloneqq) H^n(I^\bullet) \stackrel{\simeq}{\to} H^n(I_F^\bullet) \to 0 \,.$$

This concludes the discussion of the general definition and the general properties of derived functors that we will consider here. In the next section we discuss the two archetypical examples.

10) Fundamental examples of derived functors

We introduce here the two archetypical examples of derived functors and discuss their basic properties. In the next chapter IV) The fundamental theorems we discuss how to use these derived functors for obtaining deeper statements.

Above we have seen the definition and basic general properties of derived functors obtained from left/right exact functors between abelian categories.

Of all functors, a most fundamental one is the hom-functor of a given category. For categories such as $R$Mod considered here, it comes with its left adjoint, the tensor product functor, which is hence equally fundamentally important. Here we discuss the derived functors of these two basic functors in detail.

For simplicity – this here being an introduction – we will discuss various statements only over $R = \mathbb{Z}$, hence for abelian groups. The main simplification that this leads to is the following.

Proposition

Every subgroup of a free abelian group is itself a free group.

This is a classical fact going back to Dedekind, now known (in its generalization to not-necessarily abelian groups) as the Nielsen-Schreier theorem. For us it is interesting due to the following consequence

Proposition

Assuming the axiom of choice, every abelian group $A$ admits a projective resolution, def. , concentrated in degree 0 and degree 1, hence a resolution which under remark corresponds to a short exact sequence

$$0 \to F_1 \to F_0 \to A \to 0$$

where $F_0$ and $F_1$ are projective, indeed free.

Proof

By the proof of prop. there is an epimorphism $F_0 \to A$ out of a free abelian group (take for instance $F_0 = F(U(A))$, the free abelian group in the underlying set of $A$). By prop. the kernel of this epimorphism is itself a free group, and hence by prop. is itself projective. Take this kernel to be $F_1 \hookrightarrow F_0$.

This fact drastically constrains the complexity of right derived functors on abelian groups:

Proposition

Let $F \colon Ab \to Ab$ be an additive functor which is left exact functor. Then its right derived functors $R^n F$ vanish for all $n \geq 2$.

Proof

By prop. there is a projective resolution of any $A \in Ab$ of the form $F_\bullet = [\cdots \to 0 \to 0 \to F_1 \to F_0]$. This implies the claim by def. .

Remark

The conclusion of prop. holds more generally over every ring which is a principal ideal domain. This includes in particular $R = k$ a field, in which case $R Mod \simeq k$Vect. On the other hand, every $k$-vector space is already projective itself, so that in this case the whole theory of right derived functors trivializes.

a) The derived Hom functor and group extensions

For $\mathcal{A}$ an abelian category, such as $R$Mod, the hom-sets naturally have the structure of an abelian group themselves. This means that the hom-functor of $\mathcal{A}$ is

$$Hom_{\mathcal{A}}(-,-) \colon \mathcal{A}^{op}\times \mathcal{A} \to Ab \,,$$

where $\mathcal{A}^{op}$ is the opposite category of $\mathcal{A}$. This functor sends a morphism

$$\array{ (X_1 , A_1) \\ (\uparrow , \downarrow) \\ (X_2, A_2) } \;\;\; \in \mathcal{A}^{op} \times \mathcal{A}$$

to the linear map which sends a homomorphism $(X_1 \stackrel{f}{\to} A_1) \in Hom(X_1,A_1)$ to the composite homomorphism

$$\array{ X_1 &\stackrel{f}{\to}& A_2 \\ \uparrow^{\mathrlap{}} && \downarrow \\ X_2 && A_2 } \;\;\;\; \in Hom(X_2, A_2) \,.$$

In particular if we hold the first argument fixed on an object $X \in \mathcal{A}$, then this yields a functor

$$Hom(X,-) \colon \mathcal{A} \to Ab$$

and if we keep the second argument fixed on an object $A \in \mathcal{A}$, then this yields a functor

$$Hom(-,A) \colon \mathcal{A}^{op} \to Ab \,.$$

This functor we have already seen above in example .

A very basic fact is the following.

Proposition

The functor $Hom(-,-)\colon \mathcal{A}^{op} \times \mathcal{A} \to Ab$ is a left exact functor, def. . In particular for every $X \in \mathcal{A}$ the functor $Hom(X,-)\colon \mathcal{A} \to Ab$ is left exact, and for every $A \in \mathcal{A}$ the functor $Hom(-,A) \colon \mathcal{A}^{op} \to Ab$ is left exact.

Remark

A kernel in the opposite category $\mathcal{A}^{op}$ is equivalently a cokernel in $\mathcal{A}$. Hence if we regard $Hom(-,A)$ instead as a contravariant functor from $\mathcal{A}$ to Ab, then the statement that it is left exact means that (on top of preserving direct sums) it sends cokernels in $\mathcal{A}$ to kernels in Ab.

We therefore have the corresponding right derived functor:

Definition

For given $A \in \mathcal{A}$, write

$$Ext^\bullet(-,A) \coloneqq R^\bullet Hom(-,A) \colon \mathcal{A} \to Ab$$

for the right derived functor, def. , of the hom-functor in the first argument, according to prop. .

This is called the Ext-functor.

The basic property of the derived Hom-functor/Ext-functor is that it classifies group extensions by (suspensions of) $A$. This we now discuss in detail, starting from a basic discussion of group extensions themselves.

The following definition essentially just repeats that of a short exact sequence above in def. , but now we consider it for $G$ a possibly nonabelian group and think of it slightly differently regarding these sequences up to homomorphisms as in def. below. Equivalently we may think of the following as a discussion of the classification of short exact sequences when the leftmost and rightmost component are held fixed.

Definition

Two consecutive homomorphisms of groups

(2)$$A \overset{i}\hookrightarrow \hat G\overset{p}\to G$$

are a short exact sequence if

1. $i$ is monomorphism,

2. $p$ an epimorphism

3. the image of $i$ is all of the kernel of $p$: $ker(p)\simeq im(i)$.

We say that such a short exact sequence exhibits $\hat G$ as a group extension of $G$ by $A$.

If $A \hookrightarrow \hat G$ factors through the center of $\hat G$ we say that this is a central extension.

Remark

Sometimes in the literature one sees $\hat G$ called an extension “of $A$ by $G$”. This is however in conflict with terms such as central extension, extension of principal bundles, etc, where the extension is always regarded of the base, by the fiber.

Definition

A homomorphism of extensions $f : \hat G_1 \to \hat G_2$ of a given $G$ by a given $A$ is a group homomorphism of this form which fits into a commuting diagram

$$\array{ && \hat G_1 \\ & \nearrow && \searrow \\ A &&\downarrow^{\mathrlap{f}}&& G \\ & \searrow && \nearrow \\ && \hat G_2 } \,.$$

Proposition

A morphism of extensions as in def. is necessarily an isomorphism.

(3)$$\array{ 1\to &A&\stackrel{i}\to &\hat G_1&\stackrel{p}\to &G&\to 1 \\ &\downarrow\mathrlap{=}&&\downarrow\mathrlap\epsilon&&\downarrow\mathrlap=& \\ 1\to &A&\stackrel{i'}\to &\hat G_2&\stackrel{p'}\to& G&\to 1 } \,.$$

Proof

By the short five lemma.

Definition

For $G$ and $A$ groups, write $Ext(G,A)$ for the set of equivalence classes of extensions of $G$ by $A$, as above and $CentrExt(G,A) \hookrightarrow Ext(G,A)$ for for the central extensions. If $G$ and $A$ are both abelian, write

$$AbExt(G,A) \hookrightarrow CentrExt(G,A)$$

for the subset of abelian groups $\hat G$ that are (necessarily central) extensions of $G$ by $A$.

We discuss now the following two ways that the $Ext^1$ knows about such group extensions.

1. Central extensions of a possibly non-abelian group $G$ are classified by the degree-2 group cohomology $H^2_{Grp}(G,A)$ of $G$ with coefficients in $A$, and this in turn is equivalently computed by $Ext^1_{\mathbb{Z}[G] Mod}(\mathbb{Z}, A)$, where $\mathbb{Z}[G]$ is the group ring of $G$.

   This is theorem below.

2. Abelian extensions of an abelian group $G$ are classified by $Ext^1_{Ab}(G,A)$. In fact, generally, in an abelian category $\mathcal{A}$ extensions of $G \in \mathcal{A}$ by $A \in \mathcal{A}$ (in the sense of short exact sequences $A \to \hat G \to G$) are classified by $Ext^1_{\mathcal{A}}(G,A)$.

This is prop below.

We first discuss now group cohomology:

Definition

Let $G$ be group and $A$ an abelian group (regarded as being equipped with the trivial $G$-action).

Then a group 2-cocycle on $G$ with coefficients in $A$ is a function

$$c \colon G \times G \to A$$

such that for all $(g_1, g_2) \in G \times G$ it satisfies the equation

(4)$$c(g_1, g_2) - c(g_1, g_2 \cdot g_3) + c(g_1 \cdot g_2, g_3) - c(g_2, g_3) = 0 \;\;\;\; \in A$$

(called the 2-cocycle condition).

For $c, \tilde c$ two such cocycles, a coboundary $h \colon c \to \tilde c$ between them is a function

$$h \colon G \to A$$

such that for all $(g_1,g_2) \in G \times G$ the equation

(5)$$\tilde c(g_1, g_2) = (c + d h)(g_1, g_2) \,,$$

holds in $A$, where

$$(d h)(g_1, g_2) \coloneqq h(g_1 g_2) - h(g_1) - h(g_2)$$

is a 2-coboundary.

The degree-2 group cohomology is the set

$$H^2_{Grp}(G,A) = 2Cocycles(G,A) / Coboundaries(G,A)$$

of equivalence classes of group 2-cocycles modulo group coboundaries. This is itself naturally an abelian group under pointwise addition of cocycles in $A$

$$[c_1] + [c_2] = [c_1 + c_2]$$

where

$$c_1 + c_2 \colon (g_1, g_2) \mapsto c_1(g_1,g_2) + c_2(g_1, g_2) \,.$$

The following says that in the computation of $H^2_{Grp}(G,A)$ one may concentrate on nice representatives that are called normalized cocycles:

Definition

A group 2-cocycle $c \colon G \times G \to A$, def. is called normalized if

$$\forall_{g_0,g_1 \in G} \;\; \left(g_0 = e \;or\; g_1 = e \right) \Rightarrow \left( c(g_0,g_1) = 0 \right) \,.$$

Lemma

For $c \colon G \times G \to A$ a group 2-cocycle, we have for all $g \in G$ that

$$c(e,g) = c(e,e) = c(g,e) \,.$$

Proof

The cocycle condition (4) evaluated on

$$(g^{-1}, g, e) \in G^3$$

says that

$$c(g^{-1}, g) + c(e, e) = c(g, e) + c(g^{-1}, g )$$

hence that

$$c(e,e) = c(g, e) \,.$$

Similarly the 2-cocycle c