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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?

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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.

As for your second question, you should look up the universal coefficient theorem for cohomology, there you will find a partial answer for when the kronecker morphism is an isomorphism.

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