Derived Functors

October 27, 2009, 8:43 am
Filed under: Category Theory, Homological Algebra | Tags: Abelian Categories, complexes, Derived Functors, Dummit & Foote, Functors, Weibel

While studying commutative algebra and module theory, you’ll eventually be introduced to projective and injective modules. Then, suddenly, people will start throwing around words like derived functors, chain complexes, Ext and Tor. It can all be pretty daunting, especially if homological algebra and category theory is only taught as a ‘side-dish’ on a need to know basis, supplementary to whatever material is typically meant to be covered.

On the other hand, you could also be studying Algebraic Topology, where chain complexes first came about. Either way, we’ll be approaching the topic in a more or less abstract manner (hopefully) appropriate for both situations.

So, where do you go when many homological algebra books don’t talk about derived functors until the second half of the book? I’d recommend either An Introduction to Homological Algebra by C.A. Weibel or, my personal preference, Section 17.1 of Abstract Algebra by Dummit & Foote, which most students already own. Both start out from (co)chain complexes and go straight to derived functors.

The only downside to using Dummit & Foote is that they do proofs using specific functors ( $\textnormal{Hom}(\cdot,\cdot), \cdot \otimes \cdot$ ) as opposed to arbitrary left/right exact functors as Weibel does it.

Here I’ll give a basic introduction on how to get from an understanding of commutative algebra with a little category theory (including: functors, objects, (mono/epi)morphisms, exact sequences) to understanding what a derived functor is and how it is derived from left/right exact functor. Giving proof outlines and/or citations as needed.

Projective and Injective Objects

Likely, if you’ve been introduced to Projective and Injective modules, you were shown that one way to characterize them is in terms of maps into or out of the module. These same characterizations are used to define a Projective or Injective objects in terms of morphisms without the need for notions of direct or internal summands.

I find it convenient and memorable to define both injective and projective objects in a using the following single diagram.

diagram1

An object $P$ in a category $\mathfrak{C}$ is defined to be Projective if for every epimorphism $\psi$ and $f\in \textnormal{Hom}(P,N)$ , there exists an $f' \in \textnormal{Hom}(P,M)$ such that this diagram commutes (i.e. $f = \psi f'$ ).

An object $P$ in a category $\mathfrak{C}$ is defined to be Injective if for every monomorphism $\phi$ and $g\in \textnormal{Hom}(L,Q)$ , there exists an $g' \in \textnormal{Hom}(M,Q)$ such that this diagram commutes (i.e. $g = g'\phi$ ).

Note: Only three of the objects and morphisms from the diagram are needed to describe a projective (resp. injective) object. Combining the definitions into a single diagram gives the added effect of showing the duality of these objects as well as incorporate the familiar exact sequence.

This diagram is valid in any category, but the exact sequence only has significance in an abelian category (one with kernels and cokernals).

Chain and Cochain Complexes

In this section we’ll introduce chain and cochain complexes explicitly using modules. Typically, (co)chain complexes are made up of either modules or abelian groups( $\mathbb{Z}$ -modules), but techincally, they can be constructed in any abelian category (correct me if I’m wrong on this).

A chain complex $\mathcal{C}$ is a family $\{C_n\}_{n\in \mathbb{Z}}$ of modules, together with morphisms $d_n: C_n \to C_{n-1}$ such that $d_n \circ d_{n-1} = 0$ . These morphisms $d_n$ are called differentials of $\mathcal{C}$ .

The kernel of $d_n$ is the module of $n$ -cycles, denoted $Z_n = Z_n(\mathcal{C})$ . The image of $d_n$ is the module of $n$ -boundaries, denoted $B_n = B_n(\mathcal{C})$ .

A chain complex’s $n\textsuperscript{th}$ homology module is the quotient $H_n(\mathcal{C}) = \ker d_{n}/\textnormal{image }d_{n-1}$ .

Note: Abelian groups are just $\mathbb{Z}-modules$ , so the word module can be replaced with group at every point as a special case.

diagram2

The dual notion is that of a cochain complex, in which the morphisms are between modules in increasing order. In this case, we use superscript instead of subscript notation.. i.e. talking about cocycles and coboundaries when refering to $Z^n = Z^n(\mathcal{C})$ and $B^n = B^n(\mathcal{C})$ , respectively. The cohomology modules are defined as $H^n(\mathcal{C}) = \ker d_{n}/\textnormal{image }d_{n-1}$ . diagram3

Note: When describing properties of chain or cochain complexes in generality, we will use chain complexes in the diagrams and details, however the notion is easily translated to cochain complexes.

morphisms

As with all algebraic objects, there are morphisms between them, (co)chain complexes are no exception.

A morphism of (co)chain complexes $\alpha: \mathcal{A}\to \mathcal{B}$ is a family of group or module homomorphisms $\alpha_n: A_n \to B_n$ (resp. with superscripts for cochain complexes) such that the following diagram commutes.

diagram4

Note: It should be noted that the differentials $d_n$ on the top and bottom row are not identical, they refer to the differentials in their corresponding chain complexes.

Exercise: to show that such a morphism induces a module homomorphism from $H_n(\mathcal{A})$ to $H_n(\mathcal{B})$ .

hint: show that (co)boundaries go to (co)boundaries and (co)cycles go to (co)cycles

A short exact sequence of (co)chain complexes

diagram5

is a sequence of module homomorphisms such that

diagram6

is a short exact sequence for each $n$ .

Harder Exercise: Prove that a short exact sequence of (co)chain complexes induces a long exact sequence of (co)homology groups.

diagram7

Hint: The difficult part of the exercise is the $\delta_n$ map. For this, apply the snake lemma in conjunction with a short proof that cycles are mapped to cycles with the snake lemma’s connecting map.

If unfamiliar with the snake lemma, see exercise 2 of section 17.1 in Dummit & Foote for a guide on constructing $\delta_n$ or exercise 3 of the same section for a guided proof to the snake lemma.

Alternatively, Addendum 1.3.3 provides a guide in Weibel’s book.

Resolutions

Though it is convenient to think of modules or abelian groups in some cases (i.e. snake lemma), for the remainder of this post, we will speak in terms of categories, objects and morphisms. The previous material has categorical analogs as well, but it is more easily understood through module theory.

When talking about categories, they will always be abelian, since kernels and cokernels are an integral part of this theory.

Now, onto resolutions, where we see the return of projective and injective objects.

For any object $A$ , A projective resolution of $A$ is an exact sequence

diagram8

such that each $P_i$ is a projective object.

Note: A category is said to have enough projectives, if every object has a projective resolution. It is sufficient for every object to have an epimorphism from a projective object map onto it. Whenever we reference a projective resolution, it’s category is assumed to have enough projectives.

We can always construct a projective resolution for any object $A$ given the condition above.

proof: Let $P_0$ be a projective object with $\epsilon$ to be the morphism mapping $P_0$ onto $A$ . Next, let $P_1$ be the projective object mapping onto $\ker \epsilon$ in $P_0$ via the map we will define as $d_1$ . Proceeding inductively by finding projective objects that map onto each $\ker d_{n-1}$ , we get the objects $P_n$ and maps $d_n$ .

For any object $A$ , A injective resolution of $A$ is an exact sequence

diagram9

such that each $I_i$ is a projective object.

An impressive and useful fact is that every module is contained in an injective module(more on this in a later post), this allows us to construct an injective resolution for any module $A$ . Otherwise, we will say a category has enough injectives if every object has an injective resolution.

Note: Whenever we reference a injective resolution, it’s category is assumed to have enough injectives.

It doesn’t take much to notice that projective and injective resolutions are examples of chain and cochain complexes (resp.), however since they are exact, they have trivial (co)homologies.

Derived Functor

Let $A$ be an object of an abelian category $\mathfrak{A}$ , with projective resolution $\mathcal{P}$

where $P_n$ are projective objects of $\mathfrak{A}$ .

If $F: \mathfrak{A} \to \mathfrak{B}$ is a right exact functor between abelian categories. Then applying $F$ to the projective resolution and removing the leading term gives us

diagram10

a (not necessarily exact) chain complex we will call $F(\mathcal{P})$ . To simplify notation, we denoted the induced maps from the projective resolution $\mathcal{P}$ again by $d_n$ and $\epsilon$ .

If $\mathfrak{A}$ has enough projectives, we can construct the left derived functors $L_iF$ of $F$ as follows. If $A$ is an object of $\mathfrak{A}$ , choose any projective resolution of $A$ and define

$\displaystyle L_i F(A) = H_i(F(\mathcal{P}))$

Note: The map out of $F(P_0)$ in the chain complex has kernel $F(P_0)$ . And since $F(P_1) \xrightarrow{\;\; d_1 \;\;} F(P_0) \xrightarrow{\;\; \epsilon \;\;}F(A) \xrightarrow{\;\; \;\;} 0$ is exact, we have $\textnormal{image }d_1 = \ker \epsilon$ . By the first isomorphism theorem, the quotient $F(P_0)/\ker \epsilon$ is isomorphic to $F(A)$ .

You may have noticed that we didn’t make any specific choice of projective resolution (there may be many), as it turns out, the resulting homologies do not depend on the choice of resolution. We will spend the rest of this post proving just this fact.

Lemma 1

If $f \in \textnormal{Hom}(A,A')$ and $A,A'$ have projective resolutions, then $f$ induces a morphism of chain complexes from the projective resolution of $A$ to the projective resolution of $B$ . That is, $f$ induces a family of morphisms $\{f_n\}$ such that this diagram commutes.

diagram11

Proof: Since $P_0$ is a projective object, we may lift the map $f\epsilon: P_0 \to A'$ to a map $f_0: P_0 \to P'_0$ such that $\epsilon'f_0 = f\epsilon$ .

Proceeding inductively, the commutativity of the diagram shows us that $\textnormal{image }f_nd_{n+1} \subseteq \ker d'_n$ , hence $P'_{n+1}$ maps surjectively onto the image of $P_{n+1}$ in $P'_n$ . Hence, we may lift map $f_nd_{n+1}: P_{n+1} \to P'_n$ to a map $f_{n+1}: P_{n+1} \to P'_{n+1}$ .

Note: the lifts $\{f_n\}$ may not be unique for a given $f$ .

Lemma 2

Let $f \in \textnormal{Hom}(A,A')$ , $A,A'$ have projective resolutions, and $F$ be a right-exact functor, then $f$ induces a family of morphisms $\phi_n : L_nF(A) \to L_nF(A')$ on the homology groups obtained from the resolutions. The maps $\phi_n$ do not depend on the choice of lifts $f_n$ .

[17.1.5 in Dummit & Foote or 2.2.7 in Weibel, this proof is a mix of notation and methods of both.]

Proof: The existence of such maps $\phi_n$ is a direct result of our first exercise which showed a morphism of (co)chain complexes induces morphisms between respective (co)homologies.

The only difficult part is showing that the maps $\phi_n$ does not depend on the choice of lifts. This is equivalent to showing if $f$ is the zero map, then the induced maps $\phi_n$ are also zero.

Since each $P_n$ is projective, we inductively define maps $s_n: P_n \to P'_{n+1}$ as follows. For $n \leq 0$ set $s_n = 0$ . When $n = 0$ , because $f$ is the zero map, we have $\epsilon'f_0 = f \epsilon = 0$ , so $f_0(P_0) \subset \ker \epsilon' = d'_1(P'_1)$ . Hence, we may lift $f_0$ to a map $s_0$ , such that $f_0 = d'_1s_0 = d'_1s_0 + s_{-1}d_{0}$ .

diagram12 Proceeding inductively, suppose we are given maps $s_i (i < n)$ so that $d'_{i}s_{i-1} = f_{i-1} - s_{i-2}d_{i-1}$ . Now consider the map $f_n - s_{n-1}d_{n}$ from $P_n$ to $P'_n$ . Applying $d'_n$ we get the following

diagram13 Which implies that $f_n - s_{n-1}d_n$ maps into $\ker d'_n = \textnormal{image }d'_{n+1}$ . Hence we may lift map $f_n - s_{n-1}d_n$ to a map $s_n : P_n \to P'_{n+1}$ .

By construction, this collection of maps $\{s_n\}$ satisfies the following property

$\displaystyle f_n = d'_{n+1}s_n + s_{n-1}d_n$

for all $n$ .

Note: The collection of maps $\{s_n\}$ is called a chain homotopy between the chain morphism $\{f_n\}$ and the zero chain morphism.

From here, it is straightforward to check that the $f_n$ maps $\ker d_n$ to $\textnormal{image } d'_{n+1}$ . Hence, the induced map $\phi_n: L_nF(A) \to L_nF(A')$ is the zero map.

Theorem

The homologies $L_nF(A)$ depend only on $F$ and $A$ , i.e. they are independent of the choice of projective resolution of $A$ .

Proof: In the same notation as the previous lemma, let $A' = A$ and let $f: A \to A'$ be the identity morphism, and similarly let $g: A' \to A$ be the identity morphism ‘going the other way’ with $A$ and $A'$ having differing projective resolutions and lifts.

diagram14 As per our previous lemma’s, lifts $f_n$ induce morphisms $\phi_n: L_n F(A) \to L_nF(A')$ and similarly lifts $g_n$ induce morphisms $\psi_n: L_nF(A') \to L_nF(A)$ . It is sufficient for us to show that $\phi_n\psi_n = i_{H_n(\mathcal{A'})}$ and $\psi_n \phi_n = i_{H_n(\mathcal{A})}$ .

As it stands, the maps $g_nf_n$ are now also lifts of the identity map on $A$ , who’s induced maps are $\psi_n \phi_n$ . However, the identity maps on $P_0$ are also choices of lifts from the first row in the diagram to itself inducing the identity map on the homologies. By Lemma 2, the choice of lifts from the first row to itself is independent of the induced maps on the homologies, hence $\psi_n \phi_n = i_{H_n(\mathcal{A})}$ .

A similar argument shows that $\phi_n\psi_n = i_{H_n(\mathcal{A'})}$ . Hence $L_nF(A) \cong L_nF(A')$ , independent of the projective resolutions.

Corollary 1

Let $0 \to L \to M \to N \to 0$ be a short exact sequence in an abelian category. Then there is a long exact sequence

Corollary 2 [Weibel 2.5.1]

If $F: \mathfrak{A} \to \mathfrak{B}$ is a left exact functor between two abelian categories. If $\mathfrak{A}$ has enough injectives, then we can construct right derived functors $R^iF(i\geq 0)$ as follows. Choose an injective resolution $\mathcal{A}$ and define

$\displaystyle R^i F(A) = H^i(F(\mathcal{A})).$

Since $0 \to F(A) \to F(I^0) \to F(I^1)$ is exact, we always have $R^0F(A)\cong(A)$ . And we may proceed in construction as we did in the projective case, however there is also a nicer way.

The left exact functor $F$ can be seen as a right exact functor $F^{op}: \mathfrak{A}^{op} \to \mathfrak{B}^{op}$ , where $\mathfrak{A}^{op}$ has enough projectives. We can construct left derived functors $L_iF^{op}$ , and since the injective resolution in $\mathfrak{A}$ becomes a projective resolution in $\mathfrak{A}^{op}$ , we see that

$\displaystyle R^iF(A) = (L_iF^{op})^{op}(A).$

So, similarly, the right derived functors are independent of the choice of injective resolution.

Conclusion

We gave a fairly complete construction of derived functors, in a later post I will likely give examples and applications of specific algebraically important derived functors such as $Ext^n_R(\cdot, \cdot)$ and $Tor_n^R(\cdot, \cdot)$ .

There are a handful of additional results just on abstract derived functors involving direct limits, products, coproducts, and left/right adjoints, Weibel’s book has these (starting around page 55 and on). I’d like to write some more on those, but this is already long enough to bore anybody to death.

My only gripe with this kind of treatment of derived functors is that I end up using kernels and images instead of the inherent, yet identical, notion of (co)cycles and (co)boundaries which has a higher meaning from the topological perspective.

Credits:

Diagrams created using latex diagrams.sty package, which can be found at http://www.paultaylor.eu/diagrams/. Theorem structure modeled against Dummit & Foote’s treatment, while applying the notation and abstract treatment of Weibel.

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