(I am working out of Hatcher. This is theorem 2.13.)

I am brewing up some confusion about the long exact sequence of the homology groups of $A \subset X$ and $X / A$. (For (X,A) a good pair, which is to say, that A is a closed subspace of X contained in a open set V that deformation retracts to A.)

I understand how a short exact sequence of chain complexes produces (naturally) a long exact sequence in homology, and also how one can use that and excision to show that $\tilde H_n(X/A) \cong H_n(X/A,A/A) \cong H_n(X,A)$ ($\tilde{}$ denotes the reduced homology.)

What I don't understand is how the following long exact sequence is derived from these facts:

$\ldots \to \tilde H_n(A) \to \tilde H_n(X) \to \tilde H_n(X/A) \to \tilde H_{n-1}(A) \ldots \to \tilde H_0(X/A) \to 0$

Of course, I see how this result follows above the zeroth homology, just by substituting the previously described identity into the long exact sequence of the pair. At the zeroth homology I am confused. Is it somehow the case that $\tilde H_n(X/A) \cong \tilde H_n(X,A)$ - that would fix this, but it seems likely to be false... (since reducing homology removes one copy of $Z$ from the zeroth homology).

Can someone help me?

  • $\begingroup$ Reduced and unreduced homology are equal in the relative case for $A$ nonempty, i.e. $\tilde{H}_0(X,A)=H_0(X,A)$. If $X$ is connected, this is zero. $\endgroup$ – Peter Franek May 17 '14 at 7:08
  • $\begingroup$ @user1119629 The difference cancels in the quotient induced by the long exact sequence of the pair? - $H_0(X,A) \cong H_0(X) / i_*(H_0(A)) \cong \tilde H_0(X) / i_*(\tilde H_0(A)) \cong \tilde H_0(X,A)$? $\endgroup$ – Lorenzo May 17 '14 at 17:27
  • $\begingroup$ The identity is ok. Both $H_0(X,A)$ and $\tilde{H_0}(X,A)$ are generated by components of $X$ that don't intersect $A$ (if it is nonempty). Formally, the $-1$-chain group $C_{-1}(X,A)\simeq \mathbb{Z}/\mathbb{Z}\simeq 0$ for $A$ nonempty. $\endgroup$ – Peter Franek May 17 '14 at 17:57

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