nLab projective resolution

Idea

In the context of homological algebra a projective/injective resolution of an object or chain complex in an abelian category is a resolution by a quasi-isomorphic chain complex that consists of projective objects or injective objects, respectively.

Under suitable conditions these are precisely the cofibrant resolution or fibrant resolution with respect to a standard model structure on chain complexes.

For instance for non-negatively graded chain complexes of abelian groups there is a model structure with weak equivalences are the quasi-isomorphisms and the fibrations are the positive-degreewise surjections. Here every object is a fibrant object and hence no fibrant resolution is necessary; while the cofibrant resolutions are precisely the projective resolutions.

Dually, for non-negatively graded chain complexes of abelian groups there is a model structure with weak equivalences are the quasi-isomorphisms and the cofibrations the positive-degreewise injections. Here every object is a cofibrant object and hence no cofibrant resolution is necessary; while the fibrant resolutions are precisely the projective resolutions.

Definition

We first discuss, as is traditional, projective/injective resolutions of single objects, and then the general cases of projective/injective resolutions of chain complexes. This subsumes the previous case by regarding an object as a chain complex concentrated in degree 0.

Resolution of an object

Let $\mathcal{A}$ be an abelian category.

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.

$F$-Resolutions of an object

Projective and injective resolutions are typically used for computing the derived functor of some additive functor $F \colon \mathcal{A} \to \mathcal{B}$; see at derived functor in homological algebra. 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 here. 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 $C\in \mathcal{P} \subset \mathcal{A}$ also $0 \to F(A) \to F(B) \to F(C) \to 0$ is a short exact sequence in $\mathcal{B}$.

For instance (Schapira, def. 4.6.5).

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.

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. 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 left exact. The other case works dually.

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.

Resolution of a chain complex

The above definition of a projective resolution of an object has an immediate generalization to resolutions of chain complexes.

Definition

For $C_\bullet \in Ch_\bullet(\mathcal{A})$ a chain complex, a projective resolution of $C$ is an exact sequence of chain complexes

$$\cdots \to Q_{\bullet,2} \to Q_{\bullet,1} \to Q_{\bullet,0} \to C_\bullet \to 0$$

such that for each $n \in \mathbb{N}$ the component $Q_{n,\bullet} \to C_n$ is a projective resolution of the object $C_n$, according to def. .

Remark

A projective resolution as above may in particular also be regarded as a double complex $Q_{\bullet, \bullet}$ equipped with a morphism of double complex to $C_\bullet$ regarded as a vertically constant double complex.

For purposes of computations one is often interested in the following stronger notion.

For any chain complex $C_\bullet$, write $Z_\bullet$, $B_\bullet$, and $H_\bullet$ for the graded objects of cycles, boundaries and homology groups, respectively, regarded as chain complexes with vanishing differentials.

Definition

A projective resolution $Q_{\bullet,\bullet} \to C_\bullet$ of a chain complex $C_\bullet$, def. , is called fully projective (or proper) if furthermore for all $n \in \mathbb{N}$ the induced sequence of (horizontal) cycles

$$\cdots \to Z_{\bullet,2} \to Z_{\bullet,1} \to Z_{\bullet,0} \to Z(C)_\bullet \to 0$$

and (horizontal) boundaries

$$\cdots \to B_{\bullet,2} \to B_{\bullet,1} \to B_{\bullet,0} \to B(C)_\bullet \to 0$$

and (horizontal) homology groups

$$\cdots \to H_{\bullet,2} \to H_{\bullet,1} \to H_{\bullet,0} \to H(C)_\bullet \to 0$$

are each projective resolutions, def. , themselves.

Properties

Existence and construction of resolutions for objects

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 $R$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 enjectives 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.

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 $n$, where we take $\eta^0 \coloneqq 0$.

Then in the induction step assume that for given $n \in \mathbb{N}$ we have constructed $\eta^{\bullet \leq n}$ satisfying the above conditions.

First define now

$$g^n \coloneqq f^n - d_J^{n-1} \circ \eta^n$$

and observe that

$$\begin{aligned} g^n \circ d_X^{n-1} & = f^n \circ d^{n-1}_X - d^{n-1}_J \circ \eta^n \circ d^{n-1}_X \\ & = f^n \circ d^{n-1}_X - d^{n-1}_J \circ f^{n-1} + d^{n-1}_J \circ d^{n-2}_J \circ \eta^{n-1} \\ & = 0 + 0 \\ & 0 \end{aligned} \,.$$

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.

Remark

Without the assumption above that $J^\bullet$ is injective, such a null-homotopy indeed need not exist. Basic counterexamples are discussed in the section Chain homotopies that ought to exist but do not at homotopy category of chain complexes.

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 $P^\bullet$. Then there is a null homotopy

$$\eta : 0 \Rightarrow f_\bullet$$

Proof

This is formally dual to the proof of prop. .

The following proposition says that, when injectively resolving objects, the morphisms between these objects lift to the resolutions, uniquely 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 first 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 chain 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 . Morover 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 $f^\bullet$.

Sometimes one needs to construct resolutions of sequences of morphisms in a more controled way, for instance such that some degreewise exactness is preserved:

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.

This appears for instance in (May, lemma 3.4) or (Murfet, cor. 33).

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

$$\array{ A_0 &\hookrightarrow& A_0 \oplus C_0 &\to& C_0 \\ \downarrow^{\mathrlap{f_0}} && {}^{\mathllap{((g_0)_A, (g_0)_C)}}\downarrow &\swarrow_{\zeta}& \downarrow^{\mathrlap{h_0}} \\ A &\stackrel{i}{\hookrightarrow}& B &\stackrel{p}{\to}& C } \,.$$

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.

Proof

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 bottom horizontal morphisms be the canonical inclusion and projection maps of the direct sum.

Let then furthermore (in matrix calculus notation)

$$j^0 = \left( \array{ j^0_A \\ j^0_B } \right) : B \to A^0 \oplus C^0$$

be given in the second component by the given composite

$$j^0_B : B \to C \to C^0$$

and in the first component we take

$$j^0_A : B \to A^0$$

to be given by a lift in

$$\array{ A &\to& A^0 \\ \downarrow & \nearrow_{\mathrlap{j^0_A}} \\ B } \,,$$

which exists by the right lifting property of the injective object $A^0$ (since $A^\bullet$ is an injective resolution) against the monomorphism $A \to B$ of the short exact sequence.

Let the differentials be given by (…).

This establishes $j^\bullet$ 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 $j^\bullet$ is indeed a quasi-isomorphism, consider the homology long exact sequence associated to the short exact sequence of cochain complexes $0 \to A^\bullet \to B^\bullet \to C^\bullet \to 0$ (…).

Existence and construction of resolutions of complexes

Definition

If $\mathcal{A}$ has enough projectives, then every chain complex $C_\bullet \in Ch_\bullet(\mathcal{A})$ has a fully projective (proper) resolution, def. .

Proof

Notice that for each $n \in \mathbb{N}$ we have short exact sequences of chains, cycles, boundaries and homology groups as

$$0 \to B_n(C) \to Z_n(C) \to H_n(C) \to 0$$

$$0 \to Z_n(C) \to C_n \to B_{n-1}(C) \to 0 \,.$$

Now by prop. we find for each $n \in \mathbb{N}$ projective resolutions of the objects $H_n(C)$ and $B_n(C)$:

$$H_{n,\bullet} \stackrel{\simeq_{qi}}{\to} H_n(C)$$

$$B_{n,\bullet} \stackrel{\simeq_{qi}}{\to} B_n(C) \,.$$

Moreover, by prop. we find for each $n \in \mathbb{N}$ a projective resolution $Z_{p,\bullet}(C) \stackrel{\simeq_{qi}}{\to} Z_n(C)$ of the object $Z_p(C)$ such that its fits into a short exact sequence of chain complexes with the previous two chosen resolutions:

$$0 \to B_{n,\bullet}(C) \to Z_{n,\bullet}(C) \to H_{n,0}(C) \to 0 \,.$$

Analogously, we find for each $n$ a projective resolution $C_{n,\bullet} \to C_n$ that sits in a short exact sequence

$$0 \to Z_{n,\bullet} \to C_{n,\bullet} \to B_{n+1,\bullet} \to 0 \,.$$

Using the exactness of these sequences one checks now that

1. The $\{C_{n,\bullet}\}_{n \in \mathbb{N}}$ arrange into a double complex by taking the horizontal differential to be the composite

   $$C_{n,k} \to B_{n+1,k} \hookrightarrow Z_{n+1,k} \to C_{n+1,k} \,;$$

2. this double complex $C_{\bullet,\bullet}$ is indeed a fully projective resolution of $C_\bullet$.

Functorial resolutions and derived functors

We discuss how the injective/projective resolutions constructed in Existence and construction are functorial if regarded in the homotopy category of chain complexes and how this yields the construction of derived functors in homological algebra.

Write

$$\mathcal{K}^{+}(\mathcal{A}) \hookrightarrow \mathcal{K}(\mathcal{A})$$

for the full subcategory of the homotopy category of chain complexes on the one bounded above or bounded below, respectively. Write

$$\mathcal{K}^+(\mathcal{I}_{\mathcal{A}}) \hookrightarrow \mathcal{K}^+(\mathcal{A})$$

for the full subcategory on the degreewise injective complexes, and

$$\mathcal{K}^-(\mathcal{P}_{\mathcal{A}}) \hookrightarrow \mathcal{K}^-(\mathcal{A})$$

for the full subcategory on the degreewise projective objects.

Theorem

If $\mathcal{A}$ has enough injectives then there exists a functor

$$P : \mathcal{A} \to \mathcal{K}^+(\mathcal{I}_{\mathcal{A}})$$

together with a 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 the there is a morphism $X^\bullet \to \tilde X^\bullet$ in $\mathcal{K}^+()$ and prop. says that this morphism is unique in $\mathcal{K}^+(\mathcal{A})$. In particular it is therefore an isomorphism in $\mathcal{K}^+(\mathcal{A})$ (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 a unique such image in $\mathcal{K}^+(\mathcal{A})$ 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 then there exists a functor

$$Q : \mathcal{A} \to \mathcal{K}^-(\mathcal{P}_{\mathcal{A}})$$

together with a natural isomorphisms

$$H_0(-) \circ P \simeq id_{\mathcal{A}}$$

and

$$H_{n \geq 1}(-) \circ P \simeq 0 \,.$$

This is sufficient for the definition and construction of (non-total) derived functors in the next definition . But since that definition is but a model and just for a special case of derived functors, the reader might want to keep the following definition and remark in mind, for conceptual orientation.

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 achive this on the domain category, there we can simply precompose and form

$$\mathcal{A} \to \mathcal{D}_\bullet(\mathcal{A}) \simeq \mathcal{K}(\mathcal{P}_{\mathcal{A}}) \hookrightarrow \mathcal{K}(\mathcal{A}) \stackrel{\mathcal{K}(F)}{\to} \mathcal{K}(\mathcal{A}') \,.$$

But the resulting composite lands in $\mathcal{K}(\mathcal{A}')$ and in general does not factor through the inclusion $\mathcal{D}_\bullet(\mathcal{A}') = \mathcal{K}(\mathcal{P}_{\mathcal{A}'}) \hookrightarrow \mathcal{K}(\mathcal{A}')$.

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

$$F : \mathcal{A} \to \mathcal{A}'$$

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

$$R^n F : \mathcal{A} \stackrel{P}{\to} K^+(\mathcal{I}_{\mathcal{A}}) \stackrel{\mathcal{K}(F)}{\to} \mathcal{K}^+(\mathcal{A}') \stackrel{H^n(-)}{\to} \mathcal{A}' \,,$$

where

• $P$ is the injective resolution functor of theorem ;

• $\mathcal{K}(F)$ is the evident prolongation of $F$ to $\mathcal{K}^+(\mathcal{A})$;

• $H^n(-)$ 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

$$L_n F : \mathcal{A} \stackrel{Q}{\to} K^-(\mathcal{P}_{\mathcal{A}}) \stackrel{\mathcal{K}(F)}{\to} \mathcal{K}^-(\mathcal{A}') \stackrel{H_n(-)}{\to} \mathcal{A}' \,,$$

where

• $Q$ is the projective resolution functor of theorem ;

• $\mathcal{K}(F)$ is the evident prolongation of $F$ to $\mathcal{K}^+(\mathcal{A})$;

• $H_n(-)$ is the $n$-chain homology functor. Hence

$$(L_n F)(X_\bullet) \coloneqq H_n(F(Q(X)_\bullet)) \,.$$

We discuss now the basic general properties of such derived functors.

Proposition

Let $F \colon \mathcal{A} \to \mathcal{B}$ a left exact functor 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, of $F$ 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

$$R^0 F(X) \coloneqq ker(F(X^0) \to F(X^1)) \simeq F(X) \,.$$

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 a functor $F$ 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

$$\cdots \to H^{n-1}(F(A^\bullet)) \to H^{n-1}(F(B^\bullet)) \to H^{n-1}(F(C^\bullet)) \stackrel{\delta}{\to} H^n(F(A^\bullet)) \to H^n(F(B^\bullet)) \to H^n(F(C^\bullet)) \stackrel{\delta}{\to} H^{n+1}(F(A^\bullet)) \to H^{n+1}(F(B^\bullet)) \to H^{n+1}(F(C^\bullet)) \to \cdots \,.$$

But 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 $0 \to F(A) \to F(B) \to F(C) \to$ is a short exact sequence itself precisely if $R^1 F(A) \simeq 0$.

In fact we even have the following.

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 $R^n F$.

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, such as $F$-acyclic resolutions, as defined above.

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).

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

$$0 \to (R^n F(A)\coloneqq H^n(I^\bullet) \stackrel{\simeq}{\to} H^n(I_F^\bullet) \to 0 \,.$$

Derived Hom-functor/$Ext$-functor and extensions

Consider the derived functor of the hom functor.

Definition

For $A \in \mathcal{A}$, write

$$Ext^n(-,A) \coloneqq R^n Hom(-,A)$$

for the right derived functor, def. .

We discuss the use of projective resolutions in the computation of Ext-functors and group extensions.

Definition

Given $A, G \in \mathcal{A}$, an extension of $G$ by $A$ is a short exact sequence of the form

$$0 \to A \to \hat G \to G \to 0 \,.$$

Two extensions $\hat G_1$ and $\hat G_2$ are called equivalent if there is a morphism $f : \hat G_1 \to \hat G_2$ in $\mathcal{A}$ such that we have a commuting diagram

$$\array{ && \hat G_1 \\ & \nearrow && \searrow \\ A &&\downarrow^{\mathrlap{f}}&& G \\ & \searrow && \nearrow \\ && \hat G_2 } \,.$$

Write $Ext(G,A)$ for the set of equivalence classes of extensions of $G$ by $A$.

Remark

By the short five lemma a morphism $f$ as above is necessarily an isomorphism and hence we indeed have an equivalence relation.

Definition

If $\mathcal{A}$ has enough projectives, define a function

$$Extr \colon Ext(G,A) \to Ext^1(G,A)$$

from the group of extensions, def. , to the first Ext functor group as follows. Choose any projective resolution $Y_\bullet \stackrel{\simeq_{qi}}{\to} G$, which exists by prop. . Regard then $A \to \hat G \to G\to 0$ as a resolution

$$\array{ \cdots &\to& 0 &\to& 0 &\to& A &\to& \hat G \\ && \downarrow && \downarrow && \downarrow && \downarrow \\ \cdots &\to& 0 &\to& 0 &\to& 0 &\to& G }$$

of $G$, by remark . By prop. there exists then a commuting diagram of the form

$$\array{ Y_2 &\to& 0 \\ \downarrow^{\mathrlap{\partial_1^{Y}}} && \downarrow \\ Y_1 &\stackrel{c}{\to}& A \\ \downarrow^{\mathrlap{\partial_0^Y}} && \downarrow \\ Y_0 &\to& \hat G \\ \downarrow && \downarrow \\ G &\stackrel{id}{\to}& G }$$

lifting the identity map on $G$ to a chain map between the two resolutions.

By the commutativity of the top square, the morphism $c$ is 1-cocycle in $Hom(Y_\bullet,N)$, hence defines an element in $Ext^1(G,A) \coloneqq H^1(Hom(Y_\bullet,N))$.

Proposition

The construction of def. is indeed well defined in that it is independent of the choice of projective resolution as well as of the choice of chain map between the projective resolutions.

Proof

First consider the same projective resolution but another lift $\tilde c$ of the identity. By prop. any other choice $\tilde c$ fitting into a commuting diagram as above is related by a chain homotopy to $c$.

$$\array{ Y_2 &\to& 0 \\ \downarrow^{\mathrlap{\partial_1^{Y}}} &\nearrow_{\eta_1 = 0}& \downarrow \\ Y_1 &\stackrel{c - \tilde c}{\to}& A \\ \downarrow^{\mathrlap{\partial_0^Y}} &\nearrow_{\eta_0}& \downarrow \\ Y_0 &\to& \hat G \\ \downarrow &\nearrow_{}& \downarrow \\ G &\to& G } \,.$$

The chain homotopy condition here says that

$$c - \tilde c = \eta_0 \circ \partial^{Y}_0$$

and hence that in $Hom(Y_\bullet,N)$ we have that $d \eta_0 = c - \tilde c$ is a coboundary. Therefore for the given choice of resolution $Y_\bullet$ we have obtained a well-defined map

$$Ext(G,A) \to Ext^1(G,A) \,.$$

If moreover $Y'_\bullet \stackrel{\simeq_{qi}}{\to} G$ is another projective resolution, with respect to which we define such a map as above, then lifting the identity map on $G$ to a chain map between these resolutions in both directions, by prop. , establishes an isomorphism between the resulting maps, and hence the construction is independent also of the choice of resolution.

Definition

Define a function

$$Rec \colon Ext^1(G,A) \to Ext(G,A)$$

as follows. For $Y_\bullet \to G$ a projective resolution of $G$ and $[c] \in Ext^1(G,A) \simeq H^1(Hom_{\mathcal{A}}(F_\bullet,A))$ an element of the $Ext$-group, let

$$\array{ Y_2 &\to& 0 \\ \downarrow && \downarrow \\ Y_1 &\stackrel{c}{\to}& A \\ \downarrow \\ Y_0 \\ \downarrow \\ G }$$

be a representative. By the commutativity of the top square this restricts to a morphism

$$\array{ Y_1/Y_2 &\stackrel{c}{\to}& A \\ \downarrow \\ Y_0 \\ \downarrow \\ G } \,,$$

where now the left column is itself an extension of $G$ by the cokernel $Y_1/Y_2$ (because by exactness the kernel of $Y_1 \to Y_0$ is the image of $Y_2$ so that the kernel of $Y_1/Y_2 \to Y_0$ is zero). Form then the pushout of the horizontal map along the two vertical maps. This yields

$$\array{ Y_1/Y_2 &\stackrel{c}{\to}& A \\ \downarrow && \downarrow \\ Y_0 &\to& Y_0 \coprod_{Y_1/Y_2} A \\ \downarrow && \downarrow \\ G &\stackrel{id}{\to}& G } \,.$$

Here the bottom right is indeed $G$, by the pasting law for pushouts and using that the left vertical composite is the zero morphism. Moreover, the top right morphism is indeed a monomorphism as it is the pushout of a map of modules along an injection. Similarly the bottom right morphism is an epimorphism.

Hence $A \to Y_0 \coprod_{Y_1/Y_2} Y_0 \to G$ is an element in $Ext(G,A)$ which we assign to $c$.

Proposition

The construction of def. is indeed well defined in that it is independent of the choice of projective resolution as well as of the choice of representative of the $Ext$-element.

Proof

The coproduct $Y_0 \coprod_{Y_1/Y_2} A$ is equivalently

$$coker(Y_1/Y_2 \stackrel{(incl,-c)}{\to} Y_0 \oplus A) \,.$$

For a different representative $\tilde c$ of $[c]$ there is by construction a

$$\array{ Y_1 &\stackrel{\tilde c - c}{\to}& A \\ {}^{\mathllap{\partial_0}}\downarrow & \nearrow_{\lambda} \\ Y_0 } \,.$$

Define from this a map between the two cokernels induced by the commuting diagram

$$\array{ Y_1/Y_2 &\stackrel{id}{\to}& Y_1/Y_2 \\ \downarrow^{\mathrlap{(id,-c)}} && \downarrow^{\mathrlap{(id,-\tilde c)}} \\ Y_0 \oplus A &\stackrel{\left(\array{ id & 0 \\ \lambda & id }\right)}{\to}& Y_0 \oplus A } \,.$$

By construction this respects the inclusion of $A \stackrel{(0,id)}{\hookrightarrow} Y_0 \oplus A \to Y_0 \coprod_{Y_1/Y_2} A$. It also manifestly respects the projection to $G$. Therefore this defines a morphism and hence by remark even an isomorphism of extensions.

Proposition

The functions

$$Extr \colon Ext(G,A) \leftrightarrow Ext^1(G,A) \colon Rec$$

from def. to def. are inverses of each other and hence exhibit a bijection between extensions of $G$ by $A$ and $Ext^1(G,A)$.

Proof

By straightforward unwinding of the definitions.

In one direction, starting with a $c \in Ext^1(G,A)$ and constructing the extension by pushout, the resulting pushout diagram

$$\array{ Y_1 &\stackrel{c}{\to}& A \\ \downarrow && \downarrow \\ Y_0 &\to& Y_0 \coprod^c_{Y_1/Y_2} A \\ \downarrow && \downarrow \\ G &\stackrel{id}{\to}& G }$$

at the same time exhibits $c$ as the cocycle extracted from the extension $A \to Y_0 \coprod^c_{Y_1/Y_2} A \to G$.

Conversely, when starting with an extension $A \to \hat G \to G$ then extracting a $c$ by a choice of projective resolution and constructing from that another extension by pushout, the universal property of the pushout yields a morphism of exensions, which by remark is an isomorphism of extensions, hence an equality in $Ext(G,A)$.

Relation to syzygies

(…) syzygy (…)

Examples

Length-1 resolutions

Proposition

Assuming the axiom of choice, over $R = \mathbb{Z}$ hence in $R Mod =$ Ab every object $A$ has a projective resolution, even a free resolution, of length 1, hence a short exact sequence

$$0 \to F_1 \to F_0 \to A \to 0$$

with $F_1$ and $F_0$ being free abelian groups.

Proof

By the discussion at free modules - submodules of free modules a subgroup of a free abelian group is again free. Therefore for $p \colon F_0 \to A$ the surjection out of the free group $F_0 \coloneqq F(U(A))$ on the underlying set of $A$, setting $F_1 \coloneqq ker(p)$ yields the desired short exact sequence.

The same argument holds true for $R$ any principal ideal domain.

Projective resolutions adapted to abelian group cocycles

(…)

Projective resolutions adapted to general group cohomology

Let $G$ be a discrete group. Write $\mathbb{Z}[G]$ for the group ring over $G$. Notice from module – Abelian groups with G-action as modules over the group ring that linear $G$-actions on abelian groups $A$ are equivalently $\mathbb{Z}[G]$-module structures in $A$.

We discuss how cocycles in the group cohomology of $G$ with coefficients in such a module $A$ are naturally encoded in morphisms out of projective resolutions of the trivial $\mathbb{Z}[G]$-module.

Definition

Write

$$\epsilon \colon \mathbb{Z}[G] \to \mathbb{Z}$$

for the homomorphism of abelian groups which forms the sum of $R$-coefficients of the formal linear combinations that constitute the group ring

$$\epsilon \colon r \mapsto \sum_{g \in G} r_g \,.$$

This is called the augmentation map.

Definition

For $n \in \mathbb{N}$ let

$$Q^u_n \coloneqq F(U(G)^{\times^{n}})$$

be the free module over the group ring $\mathbb{Z}[G]$ on $n$-tuples of elements of $G$ (hence $Q^u_0 \simeq \mathbb{Z}[G]$ is the free module on a single generator).

For $n \geq 1$ let $\partial_{n-1} \colon Q^u_n \to Q^u_{n-1}$ be given on basis elements by

$$\partial_{n-1} (g_1, \cdots, g_n) \coloneqq g_1 [g_2, \cdots, g_n] + \sum_{i = 1}^{n-1} (-1)^i [g_1, \cdots, g_i g_{i+1}, g_{i+2}, \cdots, g_n] + (-1)^n [g_1, \cdots, g_{n-1}] \,,$$

where in the first summand we have the coefficient $g_1 \in G \hookrightarrow \mathbb{Z}[G]$ times the basis element $[g_2, \cdots, g_n]$ in $F(U(G)^{n-1})$.

In particular

$$\partial_0 \colon [g] \mapsto g[*] - [*] = g-e \in \mathbb{Z}[G] \,. \,.$$

Write furthermore $Q_n$ for the quotient module $Q^u_n \to Q^n$ which is the cokernel of the inclusion of those elements for which one of the $g_i$ is the unit element.

Proposition

The construction in def. defines chain complexes $Q^u_\bullet$ and $Q_\bullet$ of $\mathbb{Z}[G]$-modules. Moreover, with the augmentation map of def. these are projective resolutions

$$\epsilon \colon Q^u_\bullet \stackrel{\simeq_{qi}}{\to} \mathbb{Z}$$

$$\epsilon \colon Q_\bullet \stackrel{\simeq_{qi}}{\to} \mathbb{Z}$$

of $\mathbb{Z}$ equipped with the trivial $\mathbb{Z}[G]$-module structure in $\mathbb{Z}[G]$Mod.

Proof

The proof that we have indeed a chain complex is much like the proof of the existence of the alternating face map complex of a simplicial group, because writing

$$\partial^0_n [g_1, \cdots, g_n] \coloneqq g_1 [g_2, \cdots, g_n]$$

$$\partial^i_n [g_1, \cdots, g_n] \coloneqq [g_1, \cdots, g_i g_{i+1}, g_{i+2}, \cdots, g_n] \;\; for 1 \leq i \leq n-1$$

$$\partial_n [g_1, \cdots, g_n] \coloneqq [g_1, \cdots, g_{n-1}]$$

one finds that these satisfy the simplicial identities and that $\partial_n = \sum_{i = 0}^n (-1)^i \partial^i_n$.

That the augmentation map is a quasi-isomorphism is equivalent, by remark , to the augmentation

$$\cdots \stackrel{\partial_2}{\to} \mathbb{Z}[G]^2 \stackrel{\partial_1}{\to} \mathbb{Z}[G] \stackrel{\epsilon}{\to} \mathbb{Z} \to 0$$

being an exact sequence. In fact we show that it is a split exact sequence by constructing for the canonical chain map to the 0-complex a null homotopy $s_\bullet$. To that end, let

$$s_{-1} \colon \mathbb{Z} \to Q^u_0$$

be given by sending $1 \in \mathbb{Z}$ to the single basis element in $Q^u_0 \coloneqq \mathbb{Z}[G][*] \simeq \mathbb{Z}[G]$, and let for $n \in \mathbb{N}$

$$s_n \colon Q^u_n \to Q^u_{n+1}$$

be given on basis elements by

$$s_n(g[g_1, \cdots, g_n]) \coloneqq [g, g_1, \cdots, g_n] \,.$$

In the lowest degrees we have

$$\epsilon \circ s_{-1} = id_{\mathbb{Z}}$$

because

$$\epsilon(s_{-1}(1)) = \epsilon([*]) = \epsilon(e) = 1$$

and

$$\partial_0 \circ s_0 + s_{-1}\circ \epsilon = id_{Q^u_0}$$

because for all $g \in G$ we have

$$\begin{aligned} \partial_0 (s_0(g[*])) + s_{-1}(\epsilon(g[*])) & = \partial_0( [g] ) + s_{-1}(1) \\ & = g[*] - [*] + [*] \\ & = g[*] \end{aligned} \,.$$

For all remaining $n \geq 1$ we find

$$\partial_n \circ s_n + s_{n-1} \circ \partial_{n-1} = id_{Q^u_n}$$

by a lengthy but straightforward computation. This shows that every cycle is a boundary, hence that we have a resolution.

Finally, since the chain complex $Q^u_\bullet$ consists by construction degreewise of free modules hence in particular of a projective module, it is a projective resolution.

Propoition

For $A$ an abelian group equipped with a linear $G$-action and for $n \in \mathbb{N}$, the degree-$n$ group cohomology $H^n_{grp}(G, A)$ of $G$ with coefficients in $A$ is equivalently given by

$$\begin{aligned} H^n_{Grp}(G,A) & \simeq Ext^n_{\mathbb{Z}[G]}(\mathbb{Z}, A) \\ & \simeq H^n(Hom_{\mathbb{Z}[G]}(Q^u_n, A)) \\ & \simeq H^n(Hom_{\mathbb{Z}[G]}(Q_n, A)) \,. \end{aligned} \,,$$

where on the right we canonically regard $A \in \mathbb{Z}[G]$Mod.

Proof

By the free functor adjunction we have that

$$Hom_{\mathbb{Z}[G]}(F^u_n, A) \simeq Hom_{Set}(U(G)^{\times n}, U(A))$$

is the set of functions from $n$-tuples of elements of $G$ to elements of $A$. It is immediate to check that these are in the kernel of $Hom_{\mathbb{Z}[G]}(\partial_{n}, A)$ precisely if they are cocycles in the group cohomology (by comparison with the explicit formulas there) and that they are group cohomology coboundaries precisely if they are in the image of $Hom_{\mathbb{Z}[G]}(\partial_{n-1}, A)$. This establishes the first equivalences.

Similarly one finds that $H^n(Hom(F_n, A)))$ is the sub-group of normalized cocycles. By the discussion at group cohomology these already support the entire group cohomology (every cocycle is comologous to a normalized one).

Cohomology of cyclic groups

Let $G = C_k$ be a cyclic group of finite order $k$, with generator $g$. We discuss the group cohomology of $G$, as discussed at group cohomology - In terms of homological algebra.

Define special elements in the group algebra $\mathbb{Z}G$:

$$N \coloneqq 1 + g + g^2 + \ldots + g^{k-1}$$

$$\,$$

$$D \coloneqq g - 1,$$

and denote the corresponding multiplications by these elements by the same letters $N, D \colon \mathbb{Z}G \to \mathbb{Z}G$.

Then a very simple and useful projective resolution of the trivial $\mathbb{Z}G$-module $\mathbb{Z}$ is based on an exact sequence of $\mathbb{Z}G$-modules

$$\ldots \stackrel{N}{\to} \mathbb{Z}G \stackrel{D}{\to} \mathbb{Z}G \stackrel{N}{\to} \mathbb{Z}G \stackrel{D}{\to} \mathbb{Z}G \to \mathbb{Z} \to 0$$

where the last map $\mathbb{Z}G \to \mathbb{Z}$ is induced from the trivial group homomorphism $G \to 1$, hence is the map that forms the sum of all coefficients of all group elements.

It follows from this resolution that the cohomology groups $H^n(C_k, A)$ for a $C_k$-module $A$ are periodic of order 2:

$$H^{n+2}(C_k, A) \cong H^n(C_k, A)$$

for $n \geq 1$. More precisely,

Proposition

For $G = C_k$, we have

• $H^0(G, A) = A^G = \ker(D) \colon A \to A$,

• $H^{2 j + 1}(G, A) = \ker(N)/im(D)$ for $j \geq 0$,

• $H^{2 j}(G, A) = \ker(D)/im(N)$ for $j \geq 1$.

A well-known calculation in the cohomology of cyclic groups is Hilbert's Theorem 90.

Theorem

Suppose $K$ be a finite Galois extension of a field $k$, with a cyclic Galois group $G = \langle g \rangle$ of order $n$. Regard the multiplicative group $K^\ast$ as a $G$-module. Then $H^1(G, K^\ast) = 0$.

Proof

Let $\sigma \in \mathbb{Z}G$, and denote the action of $\sigma$ on an element $\beta \in K$ by exponential notation $\beta^\sigma$. The action of the element $N \in \mathbb{Z}G$ is

$$\beta^N = \beta^{1 + g + \ldots + g^{n-1}} = \beta \cdot \beta^g \cdot \ldots \beta^{g^{n-1}}$$

which is precisely the norm $N(\beta)$. We are to show that if $N(\beta) = 1$, then there exists $\alpha \in K$ such that $\beta = \alpha/g(\alpha)$. By the lemma that follows, the homomorphisms $1, g, \ldots, g^{n-1}: K^\ast \to K^\ast$ are, when considered as elements in a vector space of $K$-valued functions, $K$-linearly independent. It follows in particular that

$$1 + \beta g + \beta^{1+g}g^2 + \ldots + \beta^{1 + g + \ldots + g^{n-2}}g^{n-1}$$

is not identically zero, and therefore there exists $\theta \in K^\ast$ such that the element

$$\alpha = \theta + \beta \theta^g + \beta^{1+g}\theta^{g^2} + \ldots + \beta^{1 + g + \ldots + g^{n-2}}\theta^{g^{n-1}}$$

is non-zero. Using the fact that $N(\beta) = 1$, one may easily calculate that $\beta \alpha^g = \alpha$, as was to be shown.

The next result may be thought of as establishing “independence of characters” (where “characters” are valued in the multiplicative group of a field):

Lemma

Let $K$ be a field, let $G$ be a monoid, and let $\chi_1, \ldots, \chi_n \colon G \to K^\ast$ be distinct monoid homomorphisms. Then the functions $\chi_i$, considered as functions valued in $K$, are $K$-linearly independent.

Proof

A single $\chi \colon G \to K^\ast$ obviously forms a linearly independent set. Now suppose we have an equation

$$a_1 \chi_1 + \ldots + a_n \chi_n = 0$$

where $a_i \in K$, and assume $n$ is as small as possible. In particular, no $a_i$ is equal to $0$, and $n \geq 2$. Choose $g \in G$ such that $\chi_1(g) \neq \chi_2(g)$. Then for all $h \in G$ we have

$$a_1 \chi_1(g h) + \ldots + a_n \chi_n(g h) = 0$$

so that

$$a_1 \chi_1(g) \chi_1 + \ldots + a_n \chi_n(g)\chi_n = 0.$$

Dividing equation 2 by $\chi_1(g)$ and subtracting from it equation 1, the first term cancels, and we are left with a shorter relation

$$(a_2\frac{\chi_2(g)}{\chi_1(g)} - a_2)\chi_2 + \ldots = 0$$

which is a contradiction.

Related concepts

• projective object, projective presentation, projective cover, projective resolution

  • projective module

• injective object, injective presentation, injective envelope, injective resolution

  • injective module

• free object, free resolution

  • free module

• flat object, flat resolution

  • flat module

References

• Peter Hilton, Urs Stammbach, p. 129 of: A course in homological algebra, Springer-Verlag, New York, 1971, Graduate Texts in Mathematics, Vol. 4 (doi:10.1007/978-1-4419-8566-8, pdf)

• Pierre Schapira, section 4.5 of: Categories and homological algebra (2011) (pdf)

• Peter May, sections 3.1 and 4.2 in: Notes on Tor and Ext (pdf)

• Daniel Murfet, section 4 of: Derived functors (pdf)