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In the Lecture Notes in Algebraic Topology by Davis & Kirk, we consider the Kronecker pairing which is defined by

$$H^n(C;R) \times H_n(C;R) \to R \\ ([\varphi],[\alpha]) \mapsto \varphi(\alpha)$$

where $H^n(C;R)$ is the cohomology and $H_n(C;R)$ is the homology with coefficients over a ring $R$.

We know this defines a map $$H^n(C;R) \to \operatorname{Hom}(H_n(C;R),R) \\ [\varphi] \mapsto ([\alpha] \mapsto \varphi(\alpha)). $$

Davis & Kirk want to prove that cohomology is not the dual of homology in general and mentioned that The map $H^n(C;R) \to \operatorname{Hom}(H_n(C;R),R)$ does not need to be injective nor surjective.

My question: Why is it enough to consider this specific map?

If cohomology is the dual of homology, then $H^n(C;R) \simeq \operatorname{Hom}(H_n(C;R),R)$. So why can't there be another bijective map $H^n(C;R) \to \operatorname{Hom}(H_n(C;R),R)$ then?

Thank you in advance!

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    $\begingroup$ That the map under consider is an isomorphism is the definition of duality. Not simply that the two modules are isomorphic. $\endgroup$
    – Tyrone
    Mar 20, 2018 at 13:16
  • $\begingroup$ This map is always surjective. See hatcher's universal coefficient theorem for a discussion (he calls this map "h"). Failure of injectivity is exactly what Ext measures. $\endgroup$
    – Mr. Brown
    Aug 12, 2019 at 23:08
  • $\begingroup$ @Paul T don't we need to add that the complex is free to have surjectivity? $\endgroup$
    – Francesco Bilotta
    Feb 12, 2020 at 13:35

2 Answers 2

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My question: Why is it enough to consider this specific map?

There doesn't seem to be an immediate reason, although it could be true.

What I think the authors had in mind is that from an intuitive point of view, this map is the only natural map to look at. In a sense, if that map fails to be an isomorphism, then the existence or non-existence of another one that would be an isomorphism is not so interesting, because it would probably lack "good" properties.

However, their example (of the real projective plane) goes further and actually shows that homology and cohomology don't have to be dual.

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The idea for this map comes from the universal coefficient theorem for cohomology. That map is surjective and its kernel is $Ext^1(H_{n-1}(C),R)$.

https://en.wikipedia.org/wiki/Universal_coefficient_theorem

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