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projective resolution

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Projective and injective resolutions

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 𝒜\mathcal{A} be an abelian category.

Definition

For X𝒜X \in \mathcal{A} an object, an injective resolution of XX is a cochain complex J Ch (𝒜)J^\bullet \in Ch^\bullet(\mathcal{A}) (in non-negative degree) equipped with a quasi-isomorphism

i:XJ i : X \stackrel{\sim}{\to} J^\bullet

such that J n𝒜J^n \in \mathcal{A} is an injective object for all nn \in \mathbb{N}.

Remark

In components the quasi-isomorphism of def. 1 is a chain map of the form

X 0 0 i 0 J 0 d 0 J 1 d 1 J n d n . \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

0Xi 0J 0d 0J 1d 1J 2d 2 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 X𝒜X \in \mathcal{A} an object, a projective resolution of XX is a chain complex J Ch (𝒜)J_\bullet \in Ch_\bullet(\mathcal{A}) (in non-negative degree) equipped with a quasi-isomorphism

p:J X p : J_\bullet \stackrel{\sim}{\to} X

such that J n𝒜J_n \in \mathcal{A} is a projective object for all nn \in \mathbb{N}.

Remark

In components the quasi-isomorphism of def. 2 is a chain map of the form

n J n n1 J 1 0 J 0 p 0 0 0 X. \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

J 2 1J 1 0J 0p 0X0 \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.

FF-Resolutions of an object

Projective and injective resolutions are typically used for computing the derived functor of some additive functor F:𝒜F \colon \mathcal{A} \to \mathcal{B}; see at derived functor in homological algebra. 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 here. 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 C𝒫𝒜C\in \mathcal{P} \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}.

For instance (Schapira, def. 4.6.5).

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 cochain 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. 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.

The first condition of def. 3 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. 3 use that there is the long exact sequence of derived functors prop. 7

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 by acyclic objects as in example 1 are called FF-acyclic resolutions.

Resolution of a chain complex

The above definition 2 of a projective resolution of an object has an immediate generalization to resolutions of chain complexes.

Definition

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

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

such that for each nn \in \mathbb{N} the component Q n,C nQ_{n,\bullet} \to C_n is a projective resolution of the object C nC_n, according to def. 2.

Remark

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

In other words, a projective resolution of a chain complex in an abelian category 𝒜\mathcal{A} is a projective resolution of an object in a category of chain complexes Ch (𝒜)Ch_\bullet(\mathcal{A}).

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

For any chain complex C C_\bullet, write Z Z_\bullet, B B_\bullet, and H 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 ,C Q_{\bullet,\bullet} \to C_\bullet of a chain complex C C_\bullet, def. 6, is called fully projective (or proper) if furthermore for all nn \in \mathbb{N} the induced sequence of (horizontal) cycles

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

and (horizontal) boundaries

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

and (horizontal) homology groups

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

are each projective resolutions, def. 2, 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 𝒜\mathcal{A} be an abelian category with enough injectives (such as RRMod for some ring RR).

Then every object X𝒜X \in \mathcal{A} has an injective resolution, def. 1.

Proof

Let X𝒜X \in \mathcal{A} be the given object. By remark 1 we need to construct an exact sequence of the form

0XJ 0d 0J 1d 1J 2d 2J n 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 J^\cdot are injective objects.

This we now construct by induction on the degree nn \in \mathbb{N}.

In the first step, by the assumption of enough enjectives we find an injective object J 0J^0 and a monomorphism

XJ 0 X \hookrightarrow J^0

hence an exact sequence

0XJ 0. 0 \to X \to J^0 \,.

Assume then by induction hypothesis that for nn \in \mathbb{N} an exact sequence

XJ 0d 0J n1d n1J n 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 J^\cdot are injective objects. Forming the cokernel of d n1d^{n-1} yields the short exact sequence

0J n1d n1J npJ n/J n10. 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 𝒜\mathcal{A} we may now again find a monomorphism J n/J n1iJ n+1J^n/J^{n-1} \stackrel{i}{\hookrightarrow} J^{n+1} into an injective object J n+1J^{n+1}. This being a monomorphism means that

J n1d n1J nd nipJ n+1 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

0XJ 0J n1d n1J nd nJ n+1 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. 2.

Proposition

Let 𝒜\mathcal{A} be an abelian category with enough projectives (such as RRMod for some ring RR).

Then every object X𝒜X \in \mathcal{A} has a projective resolution, def. 2.

Proof

Let X𝒜X \in \mathcal{A} be the given object. By remark 2 we need to construct an exact sequence of the form

2J 2 1J 1 0J 0X0 \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 J_\cdot are projective objects.

This we we now construct by induction on the degree nn \in \mathbb{N}.

In the first step, by the assumption of enough projectives we find a projective object J 0J_0 and an epimorphism

J 0X J_0 \to X

hence an exact sequence

J 0X0. J_0 \to X \to 0 \,.

Assume then by induction hypothesis that for nn \in \mathbb{N} an exact sequence

J n n1J n1 0J 0X0 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 J_\cdot are projective objects. Froming the kernel of n1\partial_{n-1} yields the short exact sequence

0ker( n1)iJ n n1J n10. 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 𝒜\mathcal{A} we may now again find an epimorphism p:J n+1ker( n1) p : J_{n+1} \to ker(\partial_{n-1}) out of a projective object J n+1J_{n+1}. This being an epimorphism means that

J n+1 nipJ n n1 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 nJ n n1 0J 0X0, 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 :X J f^\bullet : X^\bullet \to J^\bullet be a chain map of cochain complexes in non-negative degree, out of an exact complex 0 qiX 0 \simeq_{qi} X^\bullet to a degreewise injective complex J J^\bullet. Then there is a null homotopy

η:0f \eta : 0 \Rightarrow f^\bullet
Proof

By definition of chain homotopy we need to construct a sequence of morphisms (η n+1:X n+1J n) n(\eta^{n+1} : X^{n+1} \to J^{n})_{n \in \mathbb{N}} such that

f n=η n+1d X n+d J n1η n. f^n = \eta^{n+1} \circ d^n_X + d^{n-1}_J \circ \eta^n \,.

for all nn. We now construct this by induction over nn, where we take η 00\eta^0 \coloneqq 0.

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

First define now

g nf nd J n1η n g^n \coloneqq f^n - d_J^{n-1} \circ \eta^n

and observe that

g nd X n1 =f nd X n1d J n1η nd X n1 =f nd X n1d J n1f n1+d J n1d J n2η n1 =0+0 0. \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 ng^n factors as

X nX n/im(d X n1)g nJ n, 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 X^\bullet the morphism X n/im(d X n1)d X nX n+1X^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

X n/im(d X n1) d X n X n+1 g n η n+1 J n, \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 η n+1\eta^{n+1} may be found due to the defining right lifting property of the injective object J nJ^n against the top monomorphism.

Observing that the commutativity of this diagram is the chain homotopy condition involving η n\eta^n and η n+1\eta^{n+1}, this completes the induction step.

Remark

Without the assumption above that J 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 2 is the following.

Lemma

Let f :P Y f_\bullet : P_\bullet \to Y_\bullet be a chain map of chain complexes in non-negative degree, into an exact complex 0 qiY 0 \simeq_{qi} Y_\bullet from a degreewise projective complex X X^\bullet. Then there is a null homotopy

η:0f \eta : 0 \Rightarrow f_\bullet
Proof

This is formally dual to the proof of prop. 2.

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:XYf : X \to Y be a morphism in 𝒜\mathcal{A}. Let

i Y:YY i_Y : Y \stackrel{\sim}{\to} Y^\bullet

be an injective resolution of YY and

i X:XX 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 :X Y f^\bullet : X^\bullet \to Y^\bullet giving a commuting diagram

X X f f Y Y . \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 nY n) n(f^n : X^n \to Y^n)_{n \in \mathbb{N}} such that for all nn \in \mathbb{N} the diagrams

X n d X n X n+1 f n f n+1 Y n d Y n Y n+1 \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

X i X X 0 f f 0 Y i Y Y 0 \array{ X &\stackrel{i_X}{\to}& X^0 \\ \downarrow^{f} && \downarrow^{\mathrlap{f^0}} \\ Y &\stackrel{i_Y}{\to}& Y^0 }

commutes in 𝒜\mathcal{A} (which makes the full diagram in Ch (𝒜)Ch^\bullet(\mathcal{A}) commute).

We construct these f =(f n) nf^\bullet = (f^n)_{n \in \mathbb{N}} by induction.

To start the induction, the morphism f 0f^0 in the first diagram above can be found by the defining right lifting property of the injective object Y 0Y^0 against the monomorphism i Xi_X.

Assume then that for some nn \in \mathbb{N} component maps f nf^{\bullet \leq n} have been obtained such that d Y kf k=f k+1d X kd^k_Y\circ f^k = f^{k+1}\circ d^k_X for all 0k<n0 \leq k \lt n . In order to construct f n+1f^{n+1} consider the following diagram, which we will describe/construct stepwise from left to right:

X n X n/im(d X n1) d X n X n+1 f n g n h n f n+1 Y n d Y n Y n+1. \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 nf^n on the left is given by induction assumption and we define the diagonal morphism to be the composite

g nd Y nf n. g^n \coloneqq d^n_Y \circ f^n \,.

Observe then that by the chain map property of the f nf^{\bullet \leq n} we have

d Y nf nd X n1=d Y nd Y n1f n1=0 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 ng^n factors through X n/im(d X n1)X^n/im(d^{n-1}_X) via some h nh^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 X^{\bullet} is exact in positive degrees (being quasi-isomorphic to a complex concentrated in degree 0) and so a lift f n+1f^{n+1} as shown on the far right of the diagram exists by the defining lifting property of the injective object Y n+1Y^{n+1}.

The total outer diagram now commutes, being built from commuting sub-diagrams, and this is the required chain map property of f n+1f^{\bullet \leq n+1} This completes the induction step.

Proposition

The morphism f f_\bullet in prop. 4 is the unique one up to chain homotopy making the given diagram commute.

Proof

Given two chain maps g 1 ,g 2g_1^\bullet, g^2_\bullet making the diagram commute, a chain homotopy g 1 g 2 g_1^\bullet \Rightarrow g_2^\bullet is equivalently a null homotopy 0g 2 g 1 0 \Rightarrow g_2^\bullet - g_1^\bullet of the difference, which sits in a square of the form

X h X 0 f g 2 g 1 Y Y \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 f^\bullet is null-homotopic.

This we may reduce to the statement of prop. 2 by considering instead of f f^\bullet the induced chain map of augmented complexes

0 X h 0 X 0 d X 0 X 1 f 2=0 f 1=0 f 0 f 1 0 Y Y 0 d J 0 Y 1 , \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 h^\bullet is a quasi-isomorphism, the top chain complex is exact, by remark 1. 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 2 implies the existence of a null homotopy

0 X X 0 d X 0 X 1 f 2=0 η 1=0 f 1=0 η 0=0 f 0 η 1 f 1 0 Y Y 0 d Y 0 Y 1 \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 η 1=0\eta^{-1} = 0 and η 0=0\eta^{0 } = 0 (notice that the proof prop. 2 was formulated exactly this way), which works because f 1=0f^{-1} = 0. The de-augmentation {f 0}\{f^{\bullet \geq 0}\} of this is the desired null homotopy of f 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 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

  • 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. 2 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 \\ \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 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

  • 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 nn \in \mathbb{N}

B nA nC 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=(j A 0 j B 0):BA 0C 0 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 B 0:BCC 0 j^0_B : B \to C \to C^0

and in the first component we take

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

to be given by a lift in

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

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

Let the differentials be given by (…).

This establishes j 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 j^\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 (…).

Existence and construction of resolutions of complexes

Definition

If 𝒜\mathcal{A} has enough projectives, then every chain complex C Ch (𝒜)C_\bullet \in Ch_\bullet(\mathcal{A}) has a fully projective (proper) resolution, def. 7.

Proof

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

0B n(C)Z n(C)H n(C)0 0 \to B_n(C) \to Z_n(C) \to H_n(C) \to 0
0Z n(C)C nB n1(C)0. 0 \to Z_n(C) \to C_n \to B_{n-1}(C) \to 0 \,.

Now by prop. 2 we find for each nn \in \mathbb{N} projective resolutions of the objects H n(C)H_n(C) and B n(C)B_n(C):

H n, qiH n(C) H_{n,\bullet} \stackrel{\simeq_{qi}}{\to} H_n(C)
B n, qiB n(C). B_{n,\bullet} \stackrel{\simeq_{qi}}{\to} B_n(C) \,.

Moreover, by prop. 1 we find for each nn \in \mathbb{N} a projective resolution Z p,(C) qiZ n(C)Z_{p,\bullet}(C) \stackrel{\simeq_{qi}}{\to} Z_n(C) of the object Z p(C)Z_p(C) such that its fits into a short exact sequence of chain complexes with the previous two chosen resolutions:

0B n,(C)Z n,(C)H n,0(C)0. 0 \to B_{n,\bullet}(C) \to Z_{n,\bullet}(C) \to H_{n,0}(C) \to 0 \,.

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

0Z n,C n,B n+1,0. 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,} n\{C_{n,\bullet}\}_{n \in \mathbb{N}} arrange into a double complex by taking the horizontal differential to be the composite

    C n,kB n+1,kZ n+1,kC n+1,k; C_{n,k} \to B_{n+1,k} \hookrightarrow Z_{n+1,k} \to C_{n+1,k} \,;
  2. this double complex C ,C_{\bullet,\bullet} is indeed a fully projective resolution of C 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

𝒦 +(𝒜)𝒦(𝒜) \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

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

for the full subcategory on the degreewise injective complexes, and

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

for the full subcategory on the degreewise projective objects.

Theorem

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

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

together with a natural isomorphisms

H 0()Pid 𝒜 H^0(-) \circ P \simeq id_{\mathcal{A}}

and

H n1()P0. H^{n \geq 1}(-) \circ P \simeq 0 \,.
Proof

By prop. 2 every object X Ch (𝒜)X^\bullet \in Ch^\bullet(\mathcal{A}) has an injective resolution. Proposition 4 says that for XX X \to X^\bullet and XX˜ X \to \tilde X^\bullet two resolutions the there is a morphism X X˜ X^\bullet \to \tilde X^\bullet in 𝒦 +()\mathcal{K}^+() and prop. 5 says that this morphism is unique in 𝒦 +(𝒜)\mathcal{K}^+(\mathcal{A}). In particular it is therefore an isomorphism in 𝒦 +(𝒜)\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) P(X)^\bullet for each X X^\bullet.

Then for f:XYf : X \to Y any morphism in 𝒜\mathcal{A}, proposition 2 again says that it can be lifted to a morphism between P(X) P(X)^\bullet and P(Y) P(Y)^\bullet and proposition 4 says that there is a unique such image in 𝒦 +(𝒜)\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 𝒜\mathcal{A} has enough projectives then there exists a functor

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

together with a natural isomorphisms

H 0()Pid 𝒜 H_0(-) \circ P \simeq id_{\mathcal{A}}

and

H n1()P0. 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 10. 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: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 achive 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}(\mathcal{P}_{\mathcal{A}}) \hookrightarrow \mathcal{K}(\mathcal{A}) \stackrel{\mathcal{K}(F)}{\to} \mathcal{K}(\mathcal{A}') \,.

But the resulting composite lands in 𝒦(𝒜)\mathcal{K}(\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}(\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. 10 and def. 11 below do.

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 : \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

  • PP is the injective resolution functor of theorem 1;

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

  • H n()H^n(-) is the nn-chain homology functor. Hence

(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 : \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 : \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

  • QQ is the projective resolution functor of theorem 2;

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

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

(L nF)(X )H n(F(Q(X) )). (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:𝒜F \colon \mathcal{A} \to \mathcal{B} a left exact functor 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, of 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 1 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) \,.
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. 10

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 a functor FF 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

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 \,.

But by construction of the resolutions and by def. 10 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. 6.

Remark

Prop. 7 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)0 \to F(A) \to F(B) \to F(C) \to is a short exact sequence itself precisely if R 1F(A)0R^1 F(A) \simeq 0.

In fact we even have the following.

Let FF be an additive functor which is an exact functor. Then

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.

We now discuss how the derived functor of an additive functor FF may also be computed not necessarily with genuine injective/projective resolutions, but with (just) FF-injective/FF-projective resolutions, such as FF-acyclic resolutions, as defined above.

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. 10. Finally let 𝒜\mathcal{I} \subset \mathcal{A} be a subcategory of FF-injective objects, def. 3.

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. 3 we have that all the im(d n)im(d^n) are in \mathcal{I}. Then the third condition in def. 3 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. 5, 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. 10 of FF evaluated on AA and the cochain cohomology of FF applied to the FF-injective resolution I F I_F^\bullet.

Proof

By prop. 2 we can also find an injective resolution A qiI A \stackrel{\simeq_{qi}}{\to} I^\bullet. By prop. 4 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 Hom-functor/ExtExt-functor and extensions

Consider the derived functor of the hom functor.

Definition

For A𝒜A \in \mathcal{A}, write

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

for the right derived functor, def. 10.

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

Definition

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

0AG^G0. 0 \to A \to \hat G \to G \to 0 \,.

Two extensions G^ 1\hat G_1 and G^ 2\hat G_2 are called equivalent if there is a morphism f:G^ 1G^ 2f : \hat G_1 \to \hat G_2 in 𝒜\mathcal{A} such that we have a commuting diagram

G^ 1 A f G G^ 2. \array{ && \hat G_1 \\ & \nearrow && \searrow \\ A &&\downarrow^{\mathrlap{f}}&& G \\ & \searrow && \nearrow \\ && \hat G_2 } \,.

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

Remark

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

Definition

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

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

from the group of extensions, def. 13, to the first Ext functor group as follows. Choose any projective resolution Y qiGY_\bullet \stackrel{\simeq_{qi}}{\to} G, which exists by prop. 2. Regard then AG^G0A \to \hat G \to G\to 0 as a resolution

0 0 A G^ 0 0 0 G \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 GG, by remark 2. By prop. 4 there exists then a commuting diagram of the form

Y 2 0 1 Y Y 1 c A 0 Y Y 0 G^ G id G \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 GG two a chain map between the two resolutions.

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

Proposition

The construction of def. 14 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 c˜\tilde c of the identity. By prop. 5 any other choice c˜\tilde c fitting into a commuting diagram as above is related by a chain homotopy to cc.

Y 2 0 1 Y η 1=0 Y 1 cc˜ A 0 Y η 0 Y 0 G^ G G. \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

cc˜=η 0 0 Y c - \tilde c = \eta_0 \circ \partial^{Y}_0

and hence that in Hom(Y ,N)Hom(Y_\bullet,N) we have that dη 0=cc˜d \eta_0 = c - \tilde c is a coboundary. Therefore for the given choice of resolution Y Y_\bullet we have obtained a well-defined map

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

If moreover Y qiGY'_\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 GG to a chain map between these resolutions in both directions, by prop. 4, establishes an isomorphism between the resulting maps, and hence the construction is independent also of the choice of resolution.

Definition

Define a function

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

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

Y 2 0 Y 1 c A Y 0 G \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

Y 1/Y 2 c A Y 0 G, \array{ Y_1/Y_2 &\stackrel{c}{\to}& A \\ \downarrow \\ Y_0 \\ \downarrow \\ G } \,,

where now the left column is itself an extension of GG by the cokernel Y 1/Y 2Y_1/Y_2 (because by exactness the kernel of Y 1Y 0Y_1 \to Y_0 is the image of Y 2Y_2 so that the kernel of Y 1/Y 2Y 0Y_1/Y_2 \to Y_0 is zero). Form then the pushout of the horizontal map along the two vertical maps. This yields

Y 1/Y 2 c A Y 0 Y 0 Y 1/Y 2A G id G. \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 GG, 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 AY 0 Y 1/Y 2Y 0GA \to Y_0 \coprod_{Y_1/Y_2} Y_0 \to G is an element in Ext(G,A)Ext(G,A) which we assign to cc.

Proposition

The construction of def. 10 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 ExtExt-element.

Proof

The coproduct Y 0 Y 1/Y 2AY_0 \coprod_{Y_1/Y_2} A is equivalently

coker(Y 1/Y 2(incl,c)Y 0A). coker(Y_1/Y_2 \stackrel{(incl,-c)}{\to} Y_0 \oplus A) \,.

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

Y 1 c˜c A 0 λ Y 0. \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

Y 1/Y 2 id Y 1/Y 2 (id,c) (id,c˜) Y 0A (id 0 λ id) Y 0A. \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(0,id)Y 0AY 0 Y 1/Y 2AA \stackrel{(0,id)}{\hookrightarrow} Y_0 \oplus A \to Y_0 \coprod_{Y_1/Y_2} A. It also manifestly respects the projection to GG. Therefore this defines a morphism and hence by remark 7 even an isomorphism of extensions.

Proposition

The functions

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

from def. 14 to def. 10 are inverses of each other and hence exhibit a bijection between extensions of GG by AA and Ext 1(G,A)Ext^1(G,A).

Proof

By straightforward unwinding of the definitions.

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

Y 1 c A Y 0 Y 0 Y 1/Y 2 cA G id G \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 cc as the cocycle extracted from the extension AY 0 Y 1/Y 2 cAGA \to Y_0 \coprod^c_{Y_1/Y_2} A \to G.

Conversely, when starting with an extension AG^GA \to \hat G \to G then extracting a cc 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 7 is an isomorphism of extensions, hence an equality in Ext(G,A)Ext(G,A).

Relation to syzygies

(…) syzygy (…)

Examples

Length-1 resolutions

Proposition

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

0F 1F 0A0 0 \to F_1 \to F_0 \to A \to 0

with F 1F_1 and F 0F_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:F 0Ap \colon F_0 \to A the surjection out of the free group F 0F(U(A))F_0 \coloneqq F(U(A)) on the underlying set of AA, setting F 1ker(p)F_1 \coloneqq ker(p) yields the desired short exact sequence.

The same argument holds true for RR any principal ideal domain.

Projective resolutions adapted to abelian group cocycles

(…)

Projective resolutions adapted to general group cohomology

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

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

Definition

Write

ϵ:[G] \epsilon \colon \mathbb{Z}[G] \to \mathbb{Z}

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

ϵ:r gGr g. \epsilon \colon r \mapsto \sum_{g \in G} r_g \,.

This is called the augmentation map.

Definition

For nn \in \mathbb{N} let

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

be the free module over the group ring [G]\mathbb{Z}[G] on nn-tuples of elements of GG (hence Q 0 u[G]Q^u_0 \simeq \mathbb{Z}[G] is the free module on a single generator).

For n1n \geq 1 let n1:Q n uQ n1 u\partial_{n-1} \colon Q^u_n \to Q^u_{n-1} be given on basis elements by

n1(g 1,,g n)g 1[g 2,,g n]+ i=1 n1(1) i[g 1,,g ig i+1,g i+2,,g n]+(1) n[g 1,,g n1], \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 1G[G]g_1 \in G \hookrightarrow \mathbb{Z}[G] times the basis element [g 2,,g n][g_2, \cdots, g_n] in F(U(G) n1)F(U(G)^{n-1}).

In particular

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

Write furthermore Q nQ_n for the quotient module Q n uQ nQ^u_n \to Q^n which is the cokernel of the inclusion of those elements for which one of the g ig_i is the unit element.

Proposition

The construction in def. 16 defines chain complexes Q uQ^u_\bullet and Q Q_\bullet of [G]\mathbb{Z}[G]-modules. Moreover, with the augmentation map of def. 15 these are projective resolutions

ϵ:Q u qi \epsilon \colon Q^u_\bullet \stackrel{\simeq_{qi}}{\to} \mathbb{Z}
ϵ:Q qi \epsilon \colon Q_\bullet \stackrel{\simeq_{qi}}{\to} \mathbb{Z}

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

n 0[g 1,,g n]g 1[g 2,,g n] \partial^0_n [g_1, \cdots, g_n] \coloneqq g_1 [g_2, \cdots, g_n]
n i[g 1,,g n][g 1,,g ig i+1,g i+2,,g n]for1in1 \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
n[g 1,,g n][g 1,,g n1] \partial_n [g_1, \cdots, g_n] \coloneqq [g_1, \cdots, g_{n-1}]

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

That the augmentation map is a quasi-isomorphism is equivalent, by remark 2, to the augmentation

2[G] 2 1[G]ϵ0 \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 s_\bullet. To that end, let

s 1:Q 0 u s_{-1} \colon \mathbb{Z} \to Q^u_0

be given by sending 11 \in \mathbb{Z} to the single basis element in Q 0 u[G][*][G]Q^u_0 \coloneqq \mathbb{Z}[G][*] \simeq \mathbb{Z}[G], and let for nn \in \mathbb{N}

s n:Q n uQ n+1 u s_n \colon Q^u_n \to Q^u_{n+1}

be given on basis elements by

s n(g[g 1,,g n])[g,g 1,,g n]. s_n(g[g_1, \cdots, g_n]) \coloneqq [g, g_1, \cdots, g_n] \,.

In the lowest degrees we have

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

because

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

and

0s 0+s 1ϵ=id Q 0 u \partial_0 \circ s_0 + s_{-1}\circ \epsilon = id_{Q^u_0}

because for all gGg \in G we have

0(s 0(g[*]))+s 1(ϵ(g[*])) = 0([g])+s 1(1) =g[*][*]+[*] =g[*]. \begin{aligned} \partial_0 (s_0(g[*])) + s_{-1}(\epsilon(g[*])) & = \partial_0( [g] ) + s_{-1}(1) \\ & = g[*] - [*] + [*] \\ & = g[*] \end{aligned} \,.

For all remaining n1n \geq 1 we find

ns n+s n1 n1=id Q n u \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 uQ^u_\bullet consists by construction degreewise of free modules hence in particular of a projective module, it is a projective resolution.

Propoition

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

H Grp n(G,A) Ext [G] n(,A) H n(Hom [G](Q n u,A)) H n(Hom [G](Q n,A))., \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[G]A \in \mathbb{Z}[G]Mod.

Proof

By the free functor adjunction we have that

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

is the set of functions from nn-tuples of elements of GG to elements of AA. It is immediate to check that these are in the kernel of Hom [G]( n,A)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 [G]( n1,A)Hom_{\mathbb{Z}[G]}(\partial_{n-1}, A). This establishes the first equivalences.

Similarly one finds that H n(Hom(F n,A)))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 kG = C_k be a cyclic group of finite order kk, with generator gg. We discuss the group cohomology of GG, as discussed at group cohomology - In terms of homological algebra.

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

N1+g+g 2++g k1N \coloneqq 1 + g + g^2 + \ldots + g^{k-1}
\,
Dg1,D \coloneqq g - 1,

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

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

NGDGNGDG0\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 G\mathbb{Z}G \to \mathbb{Z} is induced from the trivial group homomorphism G1G \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)H^n(C_k, A) for a C kC_k-module AA are periodic of order 2:

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

for n1n \geq 1. More precisely,

Proposition

For G=C kG = C_k, we have

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

  • H 2j+1(G,A)=ker(N)/im(D)H^{2 j + 1}(G, A) = \ker(N)/im(D) for j0j \geq 0,

  • H 2j(G,A)=ker(D)/im(N)H^{2 j}(G, A) = \ker(D)/im(N) for j1j \geq 1.

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

Theorem

Suppose KK be a finite Galois extension of a field kk, with a cyclic Galois group G=gG = \langle g \rangle of order nn. Regard the multiplicative group K *K^\ast as a GG-module. Then H 1(G,K *)=0H^1(G, K^\ast) = 0.

Proof

Let σG\sigma \in \mathbb{Z}G, and denote the action of σ\sigma on an element βK\beta \in K by exponential notation β σ\beta^\sigma. The action of the element NGN \in \mathbb{Z}G is

β N=β 1+g++g n1=ββ gβ g n1\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(β)N(\beta). We are to show that if N(β)=1N(\beta) = 1, then there exists αK\alpha \in K such that β=α/g(α)\beta = \alpha/g(\alpha). By the lemma that follows, the homomorphisms 1,g,,g n1:K *K *1, g, \ldots, g^{n-1}: K^\ast \to K^\ast are, when considered as elements in a vector space of KK-valued functions, KK-linearly independent. It follows in particular that

1+βg+β 1+gg 2++β 1+g++g n2g n11 + \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 θK *\theta \in K^\ast such that the element

α=θ+βθ g+β 1+gθ g 2++β 1+g++g n2θ g n1\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(β)=1N(\beta) = 1, one may easily calculate that βα g=α\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 KK be a field, let GG be a monoid, and let χ 1,,χ n:GK *\chi_1, \ldots, \chi_n \colon G \to K^\ast be distinct monoid homomorphisms. Then the functions χ i\chi_i, considered as functions valued in KK, are KK-linearly independent.

Proof

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

(1)a 1χ 1++a nχ n=0a_1 \chi_1 + \ldots + a_n \chi_n = 0

where a iKa_i \in K, and assume nn is as small as possible. In particular, no a ia_i is equal to 00, and n2n \geq 2. Choose gGg \in G such that χ 1(g)χ 2(g)\chi_1(g) \neq \chi_2(g). Then for all hGh \in G we have

a 1χ 1(gh)++a nχ n(gh)=0a_1 \chi_1(g h) + \ldots + a_n \chi_n(g h) = 0

so that

(2)a 1χ 1(g)χ 1++a nχ n(g)χ n=0.a_1 \chi_1(g) \chi_1 + \ldots + a_n \chi_n(g)\chi_n = 0.

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

(a 2χ 2(g)χ 1(g)a 2)χ 2+=0(a_2\frac{\chi_2(g)}{\chi_1(g)} - a_2)\chi_2 + \ldots = 0

which is a contradiction.

Related concepts

References

For instance section 4.5 of

or sections 3.1 and 4.2 in

or section 4 of

Revised on June 23, 2014 23:05:01 by andrew hubery? (78.94.114.199)