I have some trouble understanding the map on homology which is induced by a quotient map. If $(X,A)$ is a good pair, and $q:X\to X/A$ is the quotient map, is the induced map $q_*$ then given by the map $H_n(X)\to H_n(X,A)$ in the long exact chain, composed with the isomorphism $H_n(X,A)\cong H_n(X/A)$? If yes, why exactly? If no, then what is the best way to compute the induced map on homology of such a quotient map?
Thank you!
You do not have an isomorphism $H_n(X,A) \approx H_n(X/A)$, but an isomorphism $H_n(X,A) \approx \tilde H_n(X/A)$. See relative homology is the reduced homology of the quotient. This shows that by excision $\tilde{H_n}(X ∪ CA) = H_n(X ∪ CA,CA) = H_n(X ∪ CA − \{p\},CA − \{p\}) = H_n(X,A)$. The answer invokes that the quotient map $X \cup CA \to X \cup CA / CA = X/A$ is a homotopy equivalence for good pairs.
Okay, for $n > 0$ we have $\tilde H_n(Y) = H_n(Y)$, but nevertheless it is conceptually essential that reduced homology gropus are involved. However, we can avoid the use of reduced homology via replacing $\tilde H_n(X/A)$ by the isomorphic group $H_n(X/A,*)$. The above considerations show that for good pairs $q : (X,A) \to (X/A,*)$ induces isomorphisms $q_* : H_n(X,A) \to H_n(X/A,*)$. But clearly the diagram $\require{AMScd}$ \begin{CD} H_n(A) @>{}>> H_n(X) @>{}>> H_n(X,A) \\ @V{q_*}VV @V{q_*}VV @V{q_*}VV \\ H_n(*) @>{}>> H_n(X/A) @>{}>> H_n(X/A,*) \end{CD} commutes. For $n > 0$ the right bottom horizontal arrow is an isomorphism. For $n = 0$ it is an epimorphism with kernel isomorphic to $H_0(*) = \mathbb Z$. This should answer your question.
