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How do you prove the isomorphism $\tilde{H}_{n}(X/A,A/A)\cong\tilde{H}_{n}(X/A)$ for reduced homology groups? Is it just a matter of seeing that chains into the point $A/A$ end up being trivial in the latter group?

SihOASHoihd
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    You should tell us which definition of reduced homology groups (absolute and relative) is used here. Why do you consider $\tilde H_n (Z, {z}) \approx \tilde H_n (Z)$ only for $Z = X/A$? – Paul Frost Mar 26 '20 at 17:20

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It comes from the reduced long exact sequence of the pair $(X/A,A/A)$. In fact you have that $\widetilde{H_n}(A/A)\cong 0$ for all $n$.

Watanabe
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