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

The boundary map $\partial : H_n (X, A)\rightarrow H_{n−1} (A)$ 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 $H_{n−1} (A)$ . 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_{n−1} $, Right hand side contains $\partial : C_n(A) \rightarrow C_{n-1}(A)$).

Franklin Pezzuti Dyer
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val11
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    Have you tried following Hatcher's suggestion? Take a careful look at the definition of the boundary homomorphism associated to a short exact sequence of chain complexes. Its definition, at the chain level, is pretty simple (then some work is required in order to see that it gives a well defined homomorphism between homology groups). The chain level definition is easy to interpret in the way you want. Giving different names to the different boundary maps involved will make your life simpler... – Dan Ramras Jan 14 '11 at 19:25
  • 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.

     – val11 Jan 14 '11 at 19:43

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