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Assume $C_*$ is a chain complex of free $R$-modules where $R$ is a ring.

Following [Davis, Kirk, Lecture Notes in Algebraic Topology, Exercise 11 - p. 14], we can define then a $\textit{Kronecker pairing}$ $\langle \hspace{0.25cm}, \hspace{0.22cm}\rangle_R$ in the following way: $$ \langle \hspace{0.25cm}, \hspace{0.22cm}\rangle_R:H^k(C_*;R)\times H_k(C_*;R)\longrightarrow R $$ $$ \langle[\varphi],[\Sigma]\rangle_R=\varphi(\Sigma), $$ where $k\in\{0,\ldots,\}$ and $H^k(C_*;R)$, $H_k(C_*;R)$ are $k$-th cohomology and comology of $C_*$ with coefficients in $R$ respectively and $\varphi\in C^k(C_*;R)$, $\Sigma\in C_k(C_*;R)$.

Let $k\geq 0$ and $\varphi\in C^k(C_*;\mathbb{Z})$, $\Sigma\in C_k(C_*;\mathbb{Z})$.

For $n\geq 2$ define $\varphi_n\in C^k(C_*;\mathbb{Z}_n)$ and $\Sigma_n\in C_k(C_*;\mathbb{Z}_n)$ as follows $$ \varphi_n(x)=\varphi(x)(\bmod n), $$ $$ \Sigma_n=\Sigma\otimes_{\mathbb{Z}}b, $$ where $b$ is the generator of $\mathbb{Z}_n$.

Is it true that $$ \langle[\varphi_n],[\Sigma_n]\rangle_{\mathbb{Z}_n}=\langle[\varphi],[\Sigma]\rangle_{\mathbb{Z}}(\bmod n)? $$

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Here is a counter example. (Or at least something to think about.) Let $K$ be the Klein bottle. Then $H_2(K, \mathbb{Z})=0$ and $H^{2}(K,\mathbb{Z})=\mathbb{Z}_2$. Thus the pairing here is zero.

However, $H_2(K, \mathbb{Z}_2)=\mathbb{Z}_2$ ( as the Klein bottle is orientable mod $2$) and $H^{2}(K,\mathbb{Z})=\mathbb{Z}_2$. And you can convince yourself that the pairing here is nontrivial.

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  • $\begingroup$ Yeah. It works. What about if we assume that $C_*$ is a chain complex of an orientable manifold? $\endgroup$ Nov 17, 2017 at 9:26

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