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I've recently started learning algebraic topology using Hatcher's book and had a little problem with one of the proofs. In the section on relative cohomology(pg 200), Hatcher discusses a duality relationship between the connecting homomorphisms in the long exact sequence of cohomology and homology groups. The sequence is

$ \cdots \rightarrow H^{n}(X,A:G) \xrightarrow{j^{*}} H^{n}(X;G) \xrightarrow{i^{*}} H^{n}(A;G) \xrightarrow{\delta} H^{n+1}(X,A;G) \rightarrow \cdots $ where all the maps are the obvious ones, $\delta$ is the connecting homomorphism.

Given the group $H^{n}(C;G)$, there is the natural map $h:H^{n}(C;G) \to Hom(H_{n}(C),G)$. Looking at the long exact sequence for relative homology where $A \subset X$, we get the connecting homomorphism $\partial:H_{n+1}(X,A) \to H_{n}(A) $. He shows that the connecting maps are dual in the sense that the following diagram commutes. Diagram

Essentially, we have to show that $h\delta=\partial^{*}h$. For some $\alpha \in H^{n}(A;G)$ represented by a cocycle $\psi \in C^{n}(A;G)$, $\psi$ can be extended to $\bar{\psi}$ by assigning the value 0 to singular simplices not contained in $A$, then composing with $\partial:C_{n+1}(X) \to C_{n}(X) $ gives the cochain $\bar{\psi}\partial \in C^{n+1}(X;G)$, now Hatcher says that this actually lies in $C^{n+1}(X,A;G)$. I do not understand why this is true. I feel like I'm missing something quite simple. Any help is appreciated.

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The cochain $\psi\in C^n(A)$ is a cocycle, so it vanishes on boumdaries of chains from A. That's why when you compose it with the boundary map, it becomes a relative cocycle (= one which vanishes on boundaries from $A$).

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  • $\begingroup$ Wow, that was quite stupid of me to miss that. Thank you for the response. $\endgroup$ Jan 10, 2020 at 12:09

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