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In in the category of finite-dimensional vector spaces, say $\textbf{FinVect}_\mathbb{F}$, there exists and isomorphism between homology and cohomology: Let $(C_\bullet,\partial)$ be a chain complex and $(C^\bullet,\delta) = \text{Hom}(C_\bullet,\mathbb{F})$ the cochaincomplex such that $\delta_p = \partial_p^*$ is the vector space dual of the boundary map $\partial_p$. Using the two isomorphic definitions of homology we have

$$ H_p(C_\bullet) = \text{ker}(\text{coker}(\partial_{p+1}) \to \text{im}(\partial_p)) ,\qquad H_p(C^\bullet) = \text{coker}(\text{im}(\delta_{p}) \to \text{ker}(\delta_{p+1})) $$

Since $\text{Hom}(-,\mathbb{F})$ is an exact contravariant functor it turns kernels into cokernels and vice versa and the isomorphism between homology and cohomology $H_p(C_\bullet) \cong H_p(C^\bullet)$ becomes apparent.

We may also express homology as $H_p(C_\bullet) \cong \text{coker}(\text{im}(\partial_{p+1}) \to \text{ker}(\partial_p))$. So there must be an isomorphism $\text{ker}(\partial_p) \cong \text{ker}(\delta_{p+1})$ as well as $\text{im}(\partial_{p+1}) \cong \text{im}(\delta_p)$.

My question is by which map these two isomorphisms are given. If the map is not canonical, I still would like to see how one defines it by means of a basis.
I suspect it is given by the dual $\text{Hom}(-,\mathbb{F}) = (-)^*$, i.e. an element $z \in \text{ker}(\partial_p)$ maps to $z^* \in \text{ker}(\delta_{p+1})$. However I can't quite come up with why $z^*$ should lie in the kernel of $\delta_{p+1}$. How could one see this if it is indeed true?

elfeck
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  • Does this answer your question, or are you looking for something more computational? Nothing goes wrong, so $H^n \cong \text{Hom}(H_n, \mathbb{F})$. Of course, in the case of finite dimensional homology, these are (noncanoncially!) isomorphic. I suspect it is this noncanonicity that you're struggling with. – HallaSurvivor Nov 18 '21 at 17:58
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    I saw that one, but no that does not answer my question. I would like to know how the isomorphism $H_p(C_\bullet) \cong H_p(\text{Hom}(C_\bullet,\mathbb{F}))$ "acts" on cycles and boundaries. If need be (i.e. it is not canonical) I would still like to see what happens by the means of some basis. So I guess in that sense I do want something more computational. – elfeck Nov 18 '21 at 18:05
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  • If you want dual definitions in your first line, you need to replace $\mathrm{im}(\delta_p)$ with $\mathrm{coim}(\delta_p)$. 2. As HallaSurvivor says, you only get an isomorphism $H^p\cong H_p^{\ast}$ by your argument, not necessarily $H^p\cong H_p$. 3. Just because the cokernels are isomorphic does not allow you to conclude $\ker(\partial_p)\cong\ker(\delta_{p+1})$ (nor the dual claim). Indeed, consider the complex $0\rightarrow F\rightarrow F^2\rightarrow0$, where the non-trivial map is inclusion in the first factor. So, I'm not sure what of the question remains.
    • – Thorgott Nov 20 '21 at 03:04
  • Regarding 1 and 2, thanks for pointing it out. 3 answers my question I suppose, thanks for the counter-example. I follow it up with one more question that motivates my post: We have a split-short exact sequence $0 \to Z_p \to C_p \to B_p \to 0$ that is preserved by $\text{Hom}(-,\mathbb{F})$. Can we conclude anything about $Z^p$ and/or $B^p$ by this dualization? – elfeck Nov 20 '21 at 06:36