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Let $(V, \partial)$ be a chain complex of $F$-vector spaces. I'm having trouble proving $\forall n \in \mathbb{Z}$: $$H^n(V^*, \partial^*) \cong H_n(V, \partial)^*$$ where the * denotes the dual vector space (i.e. $V^* = Hom(V, F)$).

Essentially this question is very similar to this 5 year old SE post. I understand the argument using the universal coefficient theorem (although one answer points out that this only works in the finitely generated case). Another answer suggests however, that the UCT is not needed here since the $Hom(\cdot, F)$ functor is exact in the case when $F$ is a field. This is what I want to understand. How does the argument with the exactness of the $Hom$-functor work? And under what circumstances do I still have an isomorphism in the non-finitely generated case?

I also found this question rather insightful, however I'm not sure how to translate it to the case of a contravariant functor (like $Hom(\cdot, F)$).

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Tom Bombadil
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    See this: http://math.stackexchange.com/a/1957785/10014 – Najib Idrissi Oct 20 '16 at 15:22
  • @NajibIdrissi Thank you for the reference! This answer seems to be independent of finite or non-finite generation. So, I'm a bit confused if this is then indeed a problem as pointed out in your answer here: http://math.stackexchange.com/a/1687448/124687. Would you be able to elaborate on this? – Tom Bombadil Oct 20 '16 at 15:33
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    So what's true is that $H^n(X;F) \cong \operatorname{Hom}_F(H_n(X;F), F)$. You need the finiteness when you want to prove that $\operatorname{Hom}(H_n(X;F), F) \cong H_n(X;F)$, because a vector space is isomorphic to its dual when it's finite-dimensional. – Najib Idrissi Oct 20 '16 at 15:36

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Thanks to @Najib Idrissi for linking to my answer! I hope it might be useful. It also explains why contravariance does not matter, but it does not matter for purely formal reasons: if you have a contravariant functor $F$, it is actually covariant functor $\mathbf{A}^\circ \to \mathbf{B}$ or $\mathbf{A} \to \mathbf{B}^\circ$, so it still preserves kernels, cokernels, and images, with the difference that kernels in the opposite category are cokernels, and vice versa. Note that the opposite of an abelian category is still abelian, but with all things dualized.

Let me just add a couple of points that you are asking about.

  • The contravariant functor $\operatorname{Hom}_k (-,k)\colon k\text{-Vect}^\circ \to k\text{-Vect}$ is exact by some linear algebra, and you can check this yourself. Alternatively ("Mathematics Made Difficult" :-), it is the same as saying that $k$ is an injective object in the category of $k$-modules. But over a field, every module is injective (hint: every short exact sequence of vector spaces splits), i.e. any functor $\operatorname{Hom}_k (-, W)$ is exact for any $k$-vector space $W$.

  • The assumption that the complex consists of finite dimensional vector spaces is not needed at any point to show that $$H^n (V_\bullet^\vee) \cong H_n (V_\bullet)^\vee.$$ It is needed to further say that $$H_n (V_\bullet)^\vee \cong H_n (V_\bullet),$$ since a finite dimensional vector space is isomorphic to its dual. However, this isomorphism is not canonical, so in my opinion it is morally wrong to use it (unless you are doing calculations and what you are interested in is that $\dim_k H^n (V_\bullet^\vee) = \dim_k H_n (V_\bullet)$).

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  • "this isomorphism is not canonical, so in my opinion it is morally wrong to use it" (???) – Pedro Oct 20 '16 at 17:26
  • @Pedro Tamaroff from a philosophical / categorical point of view. An isomorphism between two things is not very meaningful, unless it is natural. –  Oct 20 '16 at 17:33
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    Essentially the same, but perhaps more clear way to see exactness of $\operatorname{Hom}$ is to argue that all short exact sequences of vector spaces are split and all additive functors preserve biproducts. That is, all additive functors are exact in category of vector spaces. – Ennar Oct 20 '16 at 17:39
  • @Ennar Well, that's what I had in mind when I said that every $k$-vector space is injective over $k$. –  Oct 20 '16 at 17:43
  • Yes, of course, this is just rephrasing of your argument. – Ennar Oct 20 '16 at 17:45
  • @Alejo You'll find yourself quite unable to do anything if your disregard isomorphisms that are not canonical. – Pedro Oct 20 '16 at 17:56
  • @Pedro Tamaroff Let's agree to disagree. Everything depends on what you are doing, and it is always good to point out when some (iso)morphism is not canonical, especially at this elementary level. –  Oct 20 '16 at 19:02