# Connection between Kronecker pairing and duality of cohomology/homology

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?

• That the map under consider is an isomorphism is the definition of duality. Not simply that the two modules are isomorphic. Mar 20, 2018 at 13:16
• 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. Aug 12, 2019 at 23:08
• @Paul T don't we need to add that the complex is free to have surjectivity? Feb 12, 2020 at 13:35

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