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Let $A$ be a finite dimensional k-algebra, where k is a fixed field. Suppose $$ X^{\bullet}:\cdots \rightarrow X_{-1} \stackrel{d_{-1}}{\longrightarrow} X_{0} \stackrel{d_0}{\longrightarrow} X_1 \stackrel{d_1}{\longrightarrow} \cdots $$ is a complex of A-modules. $H^i$ is the cohomology functor,i.e. $H^i(X^{\bullet})=Ker d_i/Im d_{i-1}$. $D=Hom_k(-,k)$. I want to know wether $H^i(D(X^{\bullet})) \cong DH^{-i}(X^{\bullet})$ for any integer i?

I have done the following: $$ D(X^{\bullet}): \cdots \rightarrow DX_{1} \stackrel{Dd_{0}}{\longrightarrow} DX_{0} \stackrel{Dd_{-1}}{\longrightarrow} DX_{-1} \stackrel{Dd_{-2}}{\longrightarrow} \cdots $$ Since $Dd_{-1}Dd_0=D(d_{0}d_{-1})=0$, $D(X^{\bullet})$ is still a complex. And I get $H^0(D(X^{\bullet}))=Ker D(d_{-1})/Im D(d_0)$ and $D(H^0(X^{\bullet}))=D(Ker d_0/Im d_{-1})$.

How to get $H^0(D(X^{\bullet}))=D(H^0(X^{\bullet}))$?

Xiaosong Peng
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2 Answers2

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Your question basically boils down to the following: what happens to the (co)homology of a (co)chain complex of $k$-vector spaces after applying the dualization functor $$(-)^\vee = \mathrm{Hom}_k (-, k).$$ The universal coefficient theorem in the other answer is a total overkill: the above functor is exact, and any exact functor commutes with taking (co)homology. That's it.

The whole point of the universal coefficient theorem is the study of the situation when you apply some inexact functor $\mathrm{Hom}_R (-, M)$.

Edit: since it seem to cause some confusion, let me sketch a proof of the fact that exact functors preserve (co)homology. So let $F$ be a contravariant functor (so that you don't think that contravariance causes any troubles; the only issue is the numeration of the complex with reversed arrows, but that is why you deal with the cohomology in degree $0$). Let $C^\bullet$ be a cochain complex $$\cdots \to C^{-1} \xrightarrow{d^{-1}} C^0 \xrightarrow{d^0} C^1 \to \cdots$$ After applying $F$, we obtain a chain complex $$\cdots \to F (C^1) \xrightarrow{F (d^0)} F (C^0) \xrightarrow{F (d^{-1})} F (C^{-1}) \to \cdots$$ The cohomology of $C^\bullet$ in degree $0$ is given by $$H^0 (C^\bullet) = \operatorname{coker} (\operatorname{im} d^{-1} \rightarrowtail \ker d^0) = \ker (\operatorname{coker} d^{-1} \twoheadrightarrow \operatorname{im} d^0).$$

A useful diagram

Now let us see what is $F (H^0)$. Since $F$ is exact contravariant, it converts kernels to cokernels and cokernels to kernels (and images and coimages are canonically isomorphic); you can check (by applying $F$ to the above diagram) that the canonical monomorphism $\operatorname{im} d^{-1} \rightarrowtail \ker d^0$ will corresond to the canonical epimorphism $\operatorname{coker} F (d^0) \twoheadrightarrow \operatorname{im} F (d^{-1})$ and the canonical epimorphism $\operatorname{coker} d^{-1} \twoheadrightarrow \operatorname{im} d^0$ will correspond to the canonical monomorphism $\operatorname{im} F (d^0) \rightarrowtail \operatorname{ker} F (d^{-1})$. Therefore, $$F (H^0 (C^\bullet)) = \ker (\operatorname{coker} F (d^0) \twoheadrightarrow \operatorname{im} F (d^{-1})) = \operatorname{coker} (\operatorname{im} F (d^0) \rightarrowtail \operatorname{ker} F (d^{-1})).$$ But this is precisely $H_0 (F (C^\bullet))$.

  • I know $Hom_k(-,k)$ is an exact functor, but how to use it to get that it commutes with chomology? – Xiaosong Peng Oct 07 '16 at 09:08
  • Actually $Hom_k(-,k)$ is a contravariant functor, so you can see $Im Hom_k(d_{i-1},k) \not \subseteq Ker Hom_k(d_i,k)$. – Xiaosong Peng Oct 07 '16 at 09:23
  • Yes, my formula was for the covariant case. I erased it since it might be confusing here. But the covariance does not matter really: a contravariant exact functor converts kernels to cokernels and cokernels to kernels (which is the same as preserving (co)kernels in the opposite category $A^\circ$, since $F$ is a covariant functor $A^\circ \to B$). Try to write down yourself the proof of the fact that an exact functor commutes with (co)homology, it is rather straightforward. You may treat the contravariant case separately and reverse all the arrows. –  Oct 07 '16 at 09:42
  • Ok, I will try it. Thank you for your help – Xiaosong Peng Oct 07 '16 at 09:47
  • The functor is exact when $k$ is a field. That assumption wasn't present in the version of the question I answered. It's not very unusual for a ring to be called $k$. (PS: Showing that $D$ is exact is equivalent to showing that $k$ is self-injective.) – Najib Idrissi Oct 07 '16 at 11:27
  • @Penson I added a proof of the fact that exact functors preserve (co)homology. It is a standard exercise, but in this case the reversal of arrows indeed might cause extra confusion. (But strictly speaking, there is no need to treat the contravariant case separately.) –  Oct 07 '16 at 11:30
  • @Najib Idrissi I posted my answer precisely because you wrote in the comments "So if $k$ is a field then what you want is indeed true. I don't know if it can be proven more easily than by applying the universal coefficient theorem though". The UCT is indeed unnecessary. –  Oct 07 '16 at 11:36
  • Ah, right. Okay, fair enough :) (though honestly the proof of the UCT isn't atrocious, so proving that $D$ is exact + proving that an exact functors commutes with taking cohomology isn't very much shorter than just the UCT, IMO). – Najib Idrissi Oct 07 '16 at 11:38
  • @Najib Idrissi Exactness of $\operatorname{Hom}_k (-,k)$ is a rather obvious/fundamental fact (and the fact that every module over a field is injective), it is basic linear algebra. Preservation of (co)homology by exact functors is also a fundamental fact that is used in many, many situations. Of course, the UCT is also something very important and it is not that complicated, but I still believe it is rather misused here :-| –  Oct 07 '16 at 11:53
  • @Najib Idrissi To me, this particular problem is a motivation for the UCT: "look, if we are over a field, we can dualize everything, and nothing interesting happens... so what if we work over a ring and apply $\operatorname{Hom}_R (-,R)$?" –  Oct 07 '16 at 11:56
  • @Najib Idrissi, Alejo Thank you for all of your help. I think all your methods are very good. If we can get the result we want, why do we have to mind using what way? – Xiaosong Peng Oct 07 '16 at 11:57
  • @Penson Well, it is nice to understand that your particular problem is just some linear algebra, and also see what really happens to your complex when you dualize it. Later you might come across a more general situation when you are over a ring, not a field, and then you will learn the UCT. Which is still not that complicated to invoke it as a blackbox without knowing where the Ext comes from. –  Oct 07 '16 at 13:49
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No, that's far from being true when $k$ is not a field (the assumption that $k$ is a field was added later, please see the last paragraph of my answer). Instead you have the Universal Coefficient Theorem. In the Wikipedia article this is written for topological spaces, but more generally this applies to chain complexes (the topological case being the general theorem applied to the singular chains of a space).

For example you can look at Theorem 3.6.5 in Weibel's book An introduction to homological algebra. It says that if $P$ is a chain complexes of projective $k$-modules such that $d(P_n)$ is projective for all $n$, then you have a short exact sequence for all $n$ and for all $k$-modules $M$: $$0 \to \operatorname{Ext}^1_k(H_{n-1}(P), M) \to H^n(\operatorname{Hom}_k(P,M)) \to \operatorname{Hom}_k(H_n(P), M) \to 0.$$

Moreover this exact sequence is split hence you have a (noncanonical) isomorphism: $$H^n(\operatorname{Hom}_k(P,M)) \cong \operatorname{Hom}_k(H_n(P), M) \oplus \operatorname{Ext}^1_k(H_{n-1}(P), M).$$

Since you're working with cochain complexes, you probably need to adapt the statement wrt. the indices a bit. If you don't have projective components for your chain complexes, then things go crazy and as far as I know the best you can hope for is a universal coefficient spectral sequence.

For a simple example, consider the chain complex such that $P_0 = P_1 = \mathbb{Z}$, $P_n = 0$ for $n \neq 0,1$, and $d : P_1 \to P_0$ is given by $d(x) = 2x$. Then $H_1(P) = 0$, but you have $$H^1(\operatorname{Hom}_\mathbb{Z}(P, \mathbb{Z})) = \mathbb{Z}/2\mathbb{Z} \neq \operatorname{Hom}_\mathbb{Z}(H_1(P), \mathbb{Z}) = 0.$$

Note that if $k$ is a field (or more generally a self-injective ring), however, then the Ext functor vanishes and so you have the relation you want. But it's a rather special case. For a direct proof without the UCT see Alejo's answer.

Najib Idrissi
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  • Sorry I haven't descirbe it clear. Here $D$ is the k-duality $Hom_k(-,k)$, where k is a fixed field. But I can't make sure $X^{\bullet}$ is a complex of projective modules. – Xiaosong Peng Oct 07 '16 at 08:30
  • @Penson Well if $k$ is a field then all $k$-modules are projective (and in fact free, as you probably know every vector space has a basis). So if $k$ is a field then what you want is indeed true. I don't know if it can be proven more easily than by applying the universal coefficient theorem though. – Najib Idrissi Oct 07 '16 at 08:37
  • Actually $X_i$ are $A$-modules, where A is a finite dimensional k-algebra. But I can see them as k-modules. To get my result, I also take $M$=k. But there is a place I don't know: why the Ext functor vanishes when k is a field? – Xiaosong Peng Oct 07 '16 at 08:49
  • @Penson Because every module over a field is projective, so in particular $H_{n-1}(P)$ is projective, and if the first argument of the Ext functor is projective then the Ext vanishes. The fact that the $X_i$ are $A$-modules does not really matter here, since you're taking the dual wrt. the $k$-module structure. – Najib Idrissi Oct 07 '16 at 08:53
  • Yes, get it. Thank you. – Xiaosong Peng Oct 07 '16 at 08:58