# Duality between homology and cohomology of complexes and the Kroenecker morphism

I am trying to understand an inetresting insight my professor gave in my Algebraic Topology course.

So, fix a commutative ring $$A$$, and the category of $$A$$-modules.
Now we can form two categories $$Ch(A)$$ of chain complexes and $$CoCh(A)$$ of cochain complexes of $$A$$-modules.

Then we can see these are dual notions, in the sense that we have a functor $$Hom(\cdot,A):Ch\rightarrow CoCh$$ which works 'pointwise' in the usual way: to any module it assigns its dual an to each morphism its transposed, so that from a chain complex $$\{C_q,\partial_q\}$$ we obatin a cochain complex $$\{C^q=Hom(C_q,A),\delta^q=\partial_q^t\}$$.

Q1 Does this functor give us an isomorphisms or an equivalence of categories?

We can now consider the two covariant homology $$H_q: Ch\rightarrow Ab$$ and cohomology $$H^q: CoCh\rightarrow Ab$$ functors. The canonical bilinear pairing between a moduled and its dual yields the following Kroenecker morphism, between the $$q$$ comohomology group of a cochain complex and the $$q$$ homology group of its dual chain complex. $$[f]\in H^q(C^\cdot)\mapsto K_q([f])\in Hom(H_q(C_\cdot), A)\quad st\quad K_q([f])([z])=f(z)$$

Now, in general this fails to be an isomorphism and from this my professor concludes that the $$Hom$$ duality somehow does not extend to the level of homology and cohomology groups.

Q2 Can we elaborate a bit more on this? Can't we find another isomorphism ? What is the consequence of this fact about the duality of $$H_q$$ and $$H^q$$? Which is the precise sense of the duality of homology and cohomology?

An equivalence of categories has to be covariant but $$\text{Hom}(-,A)$$ is contravariant, I'll just assume that you mean that the resulting functor $$Ch^{op} \rightarrow CoCh$$ is an equivalence.
In that case the functor does not give an equivalence of categories in general, take $$A = \mathbb Z$$ and let $$C_*$$ be any chain complex so that $$C_n$$ is nontrivial but only has torsion for all $$n$$. Then the only maps $$C_n \rightarrow \mathbb Z$$ will be $$0$$. And if $$F : CoCh \rightarrow Ch^{op}$$ was the inverse to $$\text{Hom}(-, A)$$ we have that equivalences preserve zero objects so $$F(0) = 0$$ Thus $$F\text{Hom}(C_*,\mathbb Z) = 0$$ but $$C_* \neq 0$$ so $$F$$ can't exist and $$\text{Hom}(-,\mathbb Z)$$ is not an equivalence.
If $$A = k$$ was a field however $$\text{Hom}(M,k)$$ where $$M$$ is some $$k-$$modules is the dual of $$M$$ and we know for a fact that the double dual of a vector space is naturally isomorphic to the original space. So in that case $$\text{Hom}(-,k)$$ would be an equivalence of categories.