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derived functor in homological algebra

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This entry discusses the general notion of derived functor specified to the special context of homological algebra, hence to functors between categories of chain complexes.

In the literature this is often understood to be the default case of derived functors. For more discussion of how the following relates to the more general concepts of derived functors see at derived functor – In homological algebra.


Contents

Idea

The general concept of derived functor is in homological algebra usually called the total hyper-derived functor, with just “derived functor” being reserved for a more restrictive case. In this tradition, we consider the special case first and then generalize it in stages. The relation between all these notions is discussed below.

Definition

Let 𝒜\mathcal{A} be an abelian category. Without real restruction of generality, we may assume that 𝒜=RMod\mathcal{A} = R Mod is the category of modules over some ring, an we will often speak in terms of this case.

The category 𝒜\mathcal{A} embeds into its derived category, the category of degreewise injective cochain complexes

P:𝒜𝒟 (𝒜)=𝒦 ( 𝒜) P : \mathcal{A} \to \mathcal{D}^\bullet(\mathcal{A}) = \mathcal{K}^\bullet(\mathcal{I}_{\mathcal{A}})

or degreewise projective chain complexes

Q:𝒜𝒟 (𝒜)=𝒦 (𝒫 𝒜) 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, which we now discuss.

Definition

For 𝒜,\mathcal{A}, \mathcal{B} two abelian categories (e.g. RRMod and RR'Mod), a functor

F:𝒜 F \colon \mathcal{A} \to \mathcal{B}

is called an additive functor if

  1. FF maps the zero object to the zero object, F(0)0F(0) \simeq 0 \in \mathcal{B};

  2. given any two objects x,y𝒜x, y \in \mathcal{A}, there is an isomorphism F(xy)F(x)F(y)F(x \oplus y) \cong F(x) \oplus F(y), and this respects the inclusion and projection maps of the direct sum:

x y i X i y xy p x p y x yFF(x) F(y) i F(x) i F(y) F(xy)F(x)F(y) p F(X) p F(y) F(x) F(y) \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:𝒜𝒜F : \mathcal{A} \to \mathcal{A}', it canonically induces a functor

Ch (F):Ch (𝒜)Ch (𝒜) 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

𝒦(F):𝒦(𝒜)𝒦(𝒜). \mathcal{K}(F) \colon \mathcal{K}(\mathcal{A}) \to \mathcal{K}(\mathcal{A}') \,.
Remark

If 𝒜\mathcal{A} and 𝒜\mathcal{A}' have enough projectives, then their derived categories are

𝒟 (𝒜)𝒦 (𝒫 𝒜) \mathcal{D}_\bullet(\mathcal{A}) \simeq \mathcal{K}_\bullet(\mathcal{P}_{\mathcal{A}})

and

𝒟 (𝒜)𝒦 ( 𝒜) \mathcal{D}^\bullet(\mathcal{A}) \simeq \mathcal{K}^\bullet(\mathcal{I}_{\mathcal{A}})

etc. One wants to accordingly derive from FF a functor 𝒟 (𝒜)𝒟 (𝒜)\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

𝒜𝒟 (𝒜)𝒦 (𝒫 𝒜)𝒦 (𝒜)𝒦 (F)𝒦 (𝒜). \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 𝒦 (𝒜)\mathcal{K}_\bullet(\mathcal{A}') and in general does not factor through the inclusion 𝒟 (𝒜)=𝒦 (𝒫 𝒜)𝒦 (𝒜)\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. 4 and def. 5 below do.

Definition

Let 𝒜,𝒜\mathcal{A}, \mathcal{A}' be two abelian categories, for instance 𝒜=R\mathcal{A} = R Mod and 𝒜=R\mathcal{A}' = R'Mod. Then a functor F:𝒜𝒜F \colon \mathcal{A} \to \mathcal{A}' which preserves direct sums (and hence in particular the zero object) is called

Here to “preserve kernels” means that for every morphism XfYX \stackrel{f}{\to} Y in 𝒜\mathcal{A} we have an isomorphism on the left of the following commuting diagram

F(ker(f)) F(X) F(f) F(Y) = = ker(F(f)) F(X) F(f) F(Y), \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 FF is a left exact functor, then for every exact sequence of the form

0ABC 0 \to A \to B \to C

also

0F(A)F(B)F(C) 0 \to F(A) \to F(B) \to F(C)

is an exact sequence. Dually, if FF is a right exact functor, then for every exact sequence of the form

ABC0 A \to B \to C \to 0

also

F(A)F(B)F(C)0 F(A) \to F(B) \to F(C) \to 0

is an exact sequence.

Proof

If 0ABC0 \to A \to B \to C is exact then ABA \hookrightarrow B is a monomorphism. But then the statement that ABCA \to B \to C is exact at BB says precisely that AA is the kernel of BCB \to C. So if FF is left exact then by definition also F(A)F(B)F(A) \to F(B) is the kernel of F(B)F(C)F(B) \to F(C) and so is in particular also a monomorphism. Dually for right exact functors.

Remark

Proposition 1 is clearly the motivation for the terminology in def. 3: 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:𝒜𝒜 F : \mathcal{A} \to \mathcal{A}'

be a left exact functor between abelian categories such that 𝒜\mathcal{A} has enough injectives. For nn \in \mathbb{N} the nnth right derived functor of FF is the composite

R nF:𝒜P𝒦 ( 𝒜)𝒦 (F)𝒦 (𝒜)H n()𝒜, R^n F \;\colon\; \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

(R nF)(X )H n(F(P(X) )). (R^n F)(X^\bullet) \coloneqq H^n(F(P(X)^\bullet)) \,.

Dually:

Definition

Let

F:𝒜𝒜 F \colon \mathcal{A} \to \mathcal{A}'

be a right exact functor between abelian categories such that 𝒜\mathcal{A} has enough projectives. For nn \in \mathbb{N} the nnth left derived functor of FF is the composite

L nF:𝒜QK (𝒫 𝒜)𝒦 (F)𝒦 (𝒜)H n()𝒜, L_n F \;\colon\; \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

(L nF)(X )H n(F(Q(X) )). (L_n F)(X_\bullet) \coloneqq H_n(F(Q(X)_\bullet)) \,.

Properties

Basic properties

The following proposition says that in degree 0 these derived functors coincide with the original functors.

Proposition

Let F:𝒜F \colon \mathcal{A} \to \mathcal{B} a left exact functor, def. 3 in the presence of enough injectives. Then for all X𝒜X \in \mathcal{A} there is a natural isomorphism

R 0F(X)F(X). R^0F(X) \simeq F(X) \,.

Dually, if FF is a right exact functor in the presence of enough projectives, then

L 0F(X)F(X). L_0 F(X) \simeq F(X) \,.
Proof

We discuss the first statement, the second is formally dual.

By remark \ref{InjectiveResolutionInComponents} an injective resolution X qiX X \stackrel{\simeq_{qi}}{\to} X^\bullet is equivalently an exact sequence of the form

0XX 0X 1. 0 \to X \hookrightarrow X^0 \to X^1 \to \cdots \,.

If FF is left exact then it preserves this excact sequence by definition of left exactness, and hence

0F(X)F(X 0)F(X 1) 0 \to F(X) \hookrightarrow F(X^0) \to F(X^1) \to \cdots

is an exact sequence. But this means that

R 0F(X)ker(F(X 0)F(X 1))F(X). 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 FF be an additive functor.

  • If FF is right exact and N𝒜N \in \mathcal{A} is a projective object, then

    L nF(N)=0n1. L_n F(N) = 0 \;\;\;\; \forall n \geq 1 \,.
  • If FF is left exact and N𝒜N \in \mathcal{A} is a injective object, then

    R nF(N)=0n1. R^n F(N) = 0 \;\;\;\; \forall n \geq 1 \,.
Proof

If NN is projective then the chain complex [00N][\cdots \to 0 \to 0 \to N] is already a projective resolution and hence by definition L nF(N)H n(0)L_n F(N) \simeq H_n(0) for n1n \geq 1. Dually if NN is an injective object.

For proving the basic property of derived functors below in prop. 4 which continues these basis statements to higher degree, in a certain way, we need the following technical lemma.

Lemma

For 0AiBpC00 \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

0 A B C 0 f g h 0 A i B p C 0 \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

and in addition

Proof

By prop. \ref{ExistenceOfInjectiveResolutions} we can choose f f_\bullet and h h_\bullet. The task is now to construct the third resolution g 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 nn \in \mathbb{N}

B nA nC 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=((j 0) A (j 0) B):A 0C 0B 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 0C 0A 0f 0AiB (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 0C 0C 0ζB (j_0)_C : A_0 \oplus C_0 \to C_0 \stackrel{\zeta}{\to} B

to be given by a lift in

B ζ p C 0 h 0 C, \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 0C_0 (since C C_\bullet is a projective resolution) against the epimorphism p:BCp : B \to C of the short exact sequence.

In total this gives in degree 0

A 0 A 0C 0 C 0 f 0 ((g 0) A,(g 0) C) ζ h 0 A i B p C. \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 B_\bullet be given by

d k B =(d k A (1) ke k 0 d k C ):A k+1C k+1A kC k, 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}\{e_k\} are constructed by induction as follows. Let e 0e_0 be a lift in

A 0 e 0 f 0 ζd 0 C : C 1 A B \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 1C_1 is a projective object and A 0AA_0 \to A is an epimorphism by A A_\bullet being a projective resolution. Here we are using that by exactness the bottom morphism indeed factors through AA as indicated, because the definition of ζ\zeta and the chain complex property of C C_\bullet gives

pζd 0 C =h 0d 0 C =0h 1 =0. \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 n1e_{n-1} has been been found satisfying the chain complex property, let e ne_n be a lift in

A n e n d n1 A e n1d n C : C n+1 ker(d n1 A )=im(d n1 A )) A n1, \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+1C_{n+1} is projective. That the bottom morphism factors as indicated is the chain complex property of e n1e_{n-1} inside d n1 B d^{B_\bullet}_{n-1}.

To see that the d B d^{B_\bullet} defines this way indeed squares to 0 notice that

d n B d n+1 B =(0 (1) n(e nd n+1 C d n A e n+1) 0 0). 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 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 g_\bullet is indeed a quasi-isomorphism, consider the homology long exact sequence associated to the short exact sequence of cochain complexes 0A B C 00 \to A_\bullet \to B_\bullet \to C_\bullet \to 0. In positive degrees it implies that the chain homology of B B_\bullet indeed vanishes. In degree 0 it gives the short sequence 0AH 0(B )B00 \to A \to H_0(B_\bullet) \to B\to 0 sitting in a commuting diagram

0 A H 0(B ) C 0 = = 0 A B C 0, \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 1 is the following.

Lemma

For 0ABC00 \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

0 A B C 0 0 A B C 0 \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

and in addition

Long exact sequence of a derived functor

The central general fact about derived functors to be discussed here is now the following.

Proposition

Let 𝒜,\mathcal{A}, \mathcal{B} be abelian categories and assume that 𝒜\mathcal{A} has enough injectives.

Let F:𝒜F : \mathcal{A} \to \mathcal{B} be a left exact functor and let

0ABC0 0 \to A \to B \to C \to 0

be a short exact sequence in 𝒜\mathcal{A}.

Then there is a long exact sequence of images of these objects under the right derived functors R F()R^\bullet F(-) of def. 4

0 R 0F(A) R 0F(B) R 0F(C) δ 0 R 1F(A) R 1F(B) R 1F(C) δ 1 R 2F(A) 0 F(A) F(B) F(C) \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 2 we can find an injective resolution

0A B C 0 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 nA^n is an injective object for all nn, its component sequences 0A nB nC n00 \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 FF (and also since FF is additive it even preserves the direct sum structure that is chosen in the proof of lemma 1) and so it follows that

0F(A˜ )F(B˜ )F(C˜ )0 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. \ref{HomologyLongExactSequence}:

H n1(F(A ))H n1(F(B ))H n1(F(C ))δH n(F(A ))H n(F(B ))H n(F(C ))δH n+1(F(A ))H n+1(F(B ))H n+1(F(C )). \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. 4, this is equal to

R n1F(A)R n1F(B)R n1F(C)δR nF(A)R nF(B)R nF(C)δR n+1F(A)R n+1F(B)R n+1F(C). \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)F(B)F(C)F(A) \to F(B) \to F(C) is given by prop. 2.

Remark

Prop. 4 implies that one way to interpret R 1F(A)R^1 F(A) is as a “measure for how a left exact functor FF fails to be an exact functor”. For, with ABCA \to B \to C any short exact sequence, this proposition gives the exact sequence

0F(A)F(B)F(C)R 1F(A) 0 \to F(A) \to F(B) \to F(C) \to R^1 F(A)

and hence 0F(A)F(B)F(C)00 \to F(A) \to F(B) \to F(C) \to 0 is a short exact sequence itself precisely if R 1F(A)0R^1 F(A) \simeq 0.

Dually, if FF is right exact functor, then L 1F(C)L_1 F (C) “measures how FF fails to be exact” for then

L 1F(C)F(A)F(B)F(C)0 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)0L_1F(C) \simeq 0.

Notice that in fact we even have the following statement (following directly from the definition).

Proposition

Let FF be an additive functor which is an exact functor. Then its left and right derived functors vanish in positive degree:

R 1F=0 R^{\geq 1} F = 0

and

L 1F=0. L_{\geq 1} F = 0 \,.
Proof

Because an exact functor preserves all exact sequences. If Y AY_\bullet \to A is a projective resolution then also F(Y) F(Y)_\bullet is exact in all positive degrees, and hence L n1F(A))H n(F(Y))=0L_{n\geq 1} F(A) ) H_{n \geq}(F(Y)) = 0. Dually for R nFR^n F.

Conversely:

Definition

Let F:𝒜F \colon \mathcal{A} \to \mathcal{B} be a left or right exact additive functor. An object A𝒜A \in \mathcal{A} is called an FF-acyclic object if all positive-degree right/left derived functors of FF are zero on AA.

Via acyclic resolutions

We now discuss how the derived functor of an additive functor FF may also be computed not necessarily with genuine injective/projective resolutions as in def. 4, but with (just) “FF-injective”/“FF-projective resolutions”.

While projective resolutions in 𝒜\mathcal{A} are sufficient for computing every left derived functor on Ch (𝒜)Ch_\bullet(\mathcal{A}) and injective resolutions are sufficient for computing every right derived functor on Ch (𝒜)Ch^\bullet(\mathcal{A}), if one is interested just in a single functor FF 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 FF-projective resolutions and FF-injective resolutions, respectively, we discuss now. A special case of both are FF-acyclic resolutions.

Let 𝒜,\mathcal{A}, \mathcal{B} be abelian categories and let F:𝒜F \colon \mathcal{A} \to \mathcal{B} be an additive functor.

Definition

Assume that FF is left exact. An additive full subcategory 𝒜\mathcal{I} \subset \mathcal{A} is called FF-injective (or: consisting of FF-injective objects) if

  1. for every object A𝒜A \in \mathcal{A} there is a monomorphism AA˜A \to \tilde A into an object A˜𝒜\tilde A \in \mathcal{I} \subset \mathcal{A};

  2. for every short exact sequence 0ABC00 \to A \to B \to C \to 0 in 𝒜\mathcal{A} with A,B𝒜A, B \in \mathcal{I} \subset \mathcal{A} also C𝒜C \in \mathcal{I} \subset \mathcal{A};

  3. for every short exact sequence 0ABC00 \to A \to B \to C \to 0 in 𝒜\mathcal{A} with A𝒜A\in \mathcal{I} \subset \mathcal{A} also 0F(A)F(B)F(C)00 \to F(A) \to F(B) \to F(C) \to 0 is a short exact sequence in \mathcal{B}.

And dually:

Definition

Assume that FF is right exact. An additive full subcategory 𝒫𝒜\mathcal{P} \subset \mathcal{A} is called FF-projective (or: consisting of FF-projective objects) if

  1. for every object A𝒜A \in \mathcal{A} there is an epimorphism A˜A\tilde A \to A from an object A˜𝒫𝒜\tilde A \in \mathcal{P} \subset \mathcal{A};

  2. for every short exact sequence 0ABC00 \to A \to B \to C \to 0 in 𝒜\mathcal{A} with B,C𝒫𝒜B, C \in \mathcal{P} \subset \mathcal{A} also A𝒫𝒜A \in \mathcal{P} \subset \mathcal{A};

  3. for every short exact sequence 0ABC00 \to A \to B \to C \to 0 in 𝒜\mathcal{A} with C𝒜C\in \mathcal{I} \subset \mathcal{A} also 0F(A)F(B)F(C)00 \to F(A) \to F(B) \to F(C) \to 0 is a short exact sequence in \mathcal{B}.

With the ,𝒫𝒜\mathcal{I},\mathcal{P}\subset \mathcal{A} as above, we say:

Definition

For A𝒜A \in \mathcal{A},

  • an FF-injective resolution of AA is a cochain complex I Ch ()Ch (𝒜)I^\bullet \in Ch^\bullet(\mathcal{I}) \subset Ch^\bullet(\mathcal{A}) and a quasi-isomorphism

    A qiI A \stackrel{\simeq_{qi}}{\to} I^\bullet
  • an FF-projective resolution of AA is a chain complex Q Ch (𝒫)Ch (𝒜)Q_\bullet \in Ch_\bullet(\mathcal{P}) \subset Ch^\bullet(\mathcal{A}) and a quasi-isomorphism

    Q qiA. Q_\bullet \stackrel{\simeq_{qi}}{\to} A \,.

Let now 𝒜\mathcal{A} have enough projectives / enough injectives, respectively.

Example

For F:𝒜F \colon \mathcal{A} \to \mathcal{B} an additive functor, let Ac𝒜Ac \subset \mathcal{A} be the full subcategory on the FF-acyclic objects, def. 6. Then

  • if FF is left exact, then Ac\mathcal{I} \coloneqq Ac is a subcategory of FF-injective objects;

  • if FF is right exact, then 𝒫Ac\mathcal{P} \coloneqq Ac is a subcategory of FF-projective objects.

Proof

Consider the case that FF is right exact. The other case works dually. Then the first condition of def. 7 is satisfied because every injective object is an FF-acyclic object and by assumption there are enough of these.

For the second and third condition of def. 7 use that there is the long exact sequence of derived functors prop. 4

0ABCR 1F(A)R 1F(B)R 1F(C)R 2F(A)R 2F(B)R 2F(C). 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 AA and BB and definition of FF-acyclic object we have R nF(A)0R^n F(A) \simeq 0 and R nF(B)0R^n F(B) \simeq 0 for n1n \geq 1 and hence short exact sequences

00R nF(C)0 0 \to 0 \to R^n F(C) \to 0

which imply that R nF(C)0R^n F(C)\simeq 0 for all n1n \geq 1, hence that CC is acyclic.

Similarly, the third condition is equivalent to R 1F(A)0R^1 F(A) \simeq 0.

Example

The FF-projective/injective resolutions (def. 9) by acyclic objects as in example 1 are called FF-acyclic resolutions.

Let 𝒜\mathcal{A} be an abelian category with enough injectives. Let F:𝒜F \colon \mathcal{A} \to \mathcal{B} be an additive left exact functor with right derived functor R FR_\bullet F, def. 4. Finally let 𝒜\mathcal{I} \subset \mathcal{A} be a subcategory of FF-injective objects, def. 7.

Lemma

If a cochain complex A Ch ()Ch (𝒜)A^\bullet \in Ch^\bullet(\mathcal{I}) \subset Ch^\bullet(\mathcal{A}) is quasi-isomorphic to 0,

X qi0 X^\bullet \stackrel{\simeq_{qi}}{\to} 0

then also F(X )Ch ()F(X^\bullet) \in Ch^\bullet(\mathcal{B}) is quasi-isomorphic to 0

F(X ) qi0. F(X^\bullet) \stackrel{\simeq_{qi}}{\to} 0 \,.
Proof

Consider the following collection of short exact sequences obtained from the long exact sequence X X^\bullet:

0X 0d 0X 1d 1im(d 1)0 0 \to X^0 \stackrel{d^0}{\to} X^1 \stackrel{d^1}{\to} im(d^1) \to 0
0im(d 1)X 2d 2im(d 2)0 0 \to im(d^1) \to X^2 \stackrel{d^2}{\to} im(d^2) \to 0
0im(d 2)X 3d 3im(d 3)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. 7 we have that all the im(d n)im(d^n) are in \mathcal{I}. Then the third condition in def. 7 says that all the sequences

0F(im(d n))F(X n+1)F(im(d n+1))0 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

0F(X 0)F(X 1)F(X 2) 0 \to F(X^0)\to F(X^1) \to F(X^2) \to \cdots

is exact, hence that F(X )F(X^\bullet) is quasi-isomorphic to 0.

Theorem

For A𝒜A \in \mathcal{A} an object with FF-injective resolution A qiI F A \stackrel{\simeq_{qi}}{\to} I_F^\bullet, def. 9, we have for each nn \in \mathbb{N} an isomorphism

R nF(A)H n(F(I F )) R^n F(A) \simeq H^n(F(I_F^\bullet))

between the nnth right derived functor, def. 4 of FF evaluated on AA and the cochain cohomology of FF applied to the FF-injective resolution I F I_F^\bullet.

Proof

By this prop. we may also find an injective resolution A qiI A \stackrel{\simeq_{qi}}{\to} I^\bullet. By this prop there is a lift of the identity on AA to a chain map I F I I^\bullet_F \to I^\bullet such that the diagram

A qi I F id f A qi I \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 (𝒜)Ch^\bullet(\mathcal{A}). Therefore by the 2-out-of-3 property of quasi-isomorphisms it follows that ff is a quasi-isomorphism

Let Cone(f)Ch (𝒜)Cone(f) \in Ch^\bullet(\mathcal{A}) be the mapping cone of ff and let I Cone(f)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))Cone(F(f))F(Cone(f)) \simeq Cone(F(f));

  2. Cone(f)Ch ()Ch (𝒜)Cone(f) \in Ch^\bullet(\mathcal{I})\subset Ch^\bullet(\mathcal{A}) (because FF-injective objects are closed under direct sum).

The first implies that we have a homology exact sequence

H n(I )H n(I F )H n(Cone(f) )H n+1(I )H n+1(I F )H n+1(Cone(f) ). \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 f^\bullet a quasi-isomorphism Cone(f )Cone(f^\bullet) is quasi-isomorphic to 0. Therefore the second item above implies with lemma 3 that also F(Cone(f))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(R nF(A)H n(I )H n(I F )0. 0 \to (R^n F(A)\coloneqq H^n(I^\bullet) \stackrel{\simeq}{\to} H^n(I_F^\bullet) \to 0 \,.

Derived adjoint functors

Observation

If

(FG):𝒜FG (F \dashv G) : \mathcal{A} \stackrel{\overset{G}{\leftarrow}}{\underset{F}{\to}} \mathcal{B}

is a pair of additive adjoint functors, then

(…)

Preservation of further limits and colimits

Proposition

Let F:𝒜F \colon \mathcal{A} \to \mathcal{B} be an additive right exact functor with codomain an AB5-category.

Then the left derived functors L nFL_n F preserves filtered colimits precisely if filtered colimits of projective objects in 𝒜\mathcal{A} are FF-acyclic objects.

See here.

Examples

(…)

References

A standard textbook introduction is chapter 2 of

A systematic discussion from the point of view of homotopy theory and derived categories is in chapter 7 of

The above text is taken from

Revised on July 18, 2016 05:02:46 by Urs Schreiber (131.220.184.222)