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In Allen Hatcher's book on page 117 (bottom) he says the following:

The boundary map $\partial : H_n (X, A)\rightarrow H_{nāˆ’1} (A)$ has a very simple description: If a class $[\alpha] \in H_n(X,A)$ is represented by a relative cycle $\alpha$, then $\partial[\alpha]$ is the class of the cycle $\partial \alpha$ in $H_{nāˆ’1} (A)$ . This is immediate from the algebraic definition of the boundary homomorphism in the long exact sequence of homology groups associated to a short exact sequence of chain complexes.

I can see that $\partial \alpha \in C_{n-1}(A)$ and that it must be a cycle since $\partial\partial = 0$. But how is the relation $\partial[\alpha] = [\partial \alpha]$ established (Left hand side contains the $\partial : H_n (X, A)\rightarrow H_{nāˆ’1} $, Right hand side contains $\partial : C_n(A) \rightarrow C_{n-1}(A)$).

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    Can you verify or correct the following reasoning? I don't think it is very nice working with the "same" element in different groups, without giving it different names.. But: $\alpha \in C_n(X,A)$ can be considered as an element $\alpha' \in C_n(X)$, for which it holds that $j(\alpha')=\alpha$. Using the boundary map on chain level, we get $\partial \alpha'$, which is uniquely determined as an element of $C_{n-1}(A)$, since $i$ is an isomorphism by exactness. Since essentially $\alpha'=\alpha$, the desired result holds. – 2011-01-14

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