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I confused at one step of getting exact sequence of reduced homology groups for a good pair (ref. Algebraic topology by Hatcher).

Theorem 2.13 (p.114). Let $X$ be a space and $A$ a closed subspace which is a deformation retract of an open neighborhood in $X$. Then there is an exact sequence of reduced homology groups: $$\cdots \tilde{H}_n(A)\rightarrow \tilde{H}_n(X)\rightarrow \tilde{H}_n(X/A)\rightarrow\cdots\rightarrow \tilde{H}_0(X/A)\rightarrow 0.$$

Proof.

1. For $n\geq 1$, $\tilde{H}_n(Y)\cong H_n(Y)$ and for $Y\neq \phi$, and $n=0$, $H_0(Y)\cong \tilde{H}_0(Y)\oplus \mathbb{Z}$. (see para after defining reduced homology groups on p.110

2. For any pair $(X,A)$ the relative homology groups fit in exact sequence (end of p.115) $$\cdots H_n(A)\rightarrow H_n(X)\rightarrow H_n(X,A)\rightarrow\cdots\rightarrow H_0(X,A)\rightarrow 0.$$

3. As a consequence of excision theorem (prop.2.22, p.124), $$H_n(X,A)\cong \tilde{H}_n(X/A) \mbox{ for all } n.$$

4. Replacing relative homology groups by reduced homology of quotients in (2) we get exact sequence $$ \cdots \tilde{H}_n(A)\rightarrow \tilde{H}_n(X)\rightarrow \tilde{H}_n(X/A)\rightarrow\cdots\rightarrow H_0(A)\rightarrow H_0(X)\rightarrow \tilde{H}_0(X/A)\rightarrow 0.$$ Question. We have not replaced $H_0(A)$ and $H_0(X)$ by reduced homology groups, and we can not replace them directly from (1); how to proceed for this? Am I missing anything here?

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1 Answer 1

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It is possible to proceed as you have done, manually replacing $H_0(A)$ and $H_0(X)$ with their reduced versions and then explicitly working out that you can still get a long exact sequence. However, there's a much better way to do this: just work with the reduced chain complexes to begin with. In other words, there is a short exact sequence of chain complexes $$0\to \tilde{C}_*(A)\to \tilde{C}_*(X)\to C_*(X,A)\to 0$$ which in nonnegative degrees is just the usual short exact sequence $0\to C_n(A)\to C_n(X)\to C_n(X,A)\to 0$ but in degree $-1$ is $0\to\mathbb{Z}\to\mathbb{Z}\to 0\to 0$ where the map $\mathbb{Z}\to\mathbb{Z}$ is the identity. This then gives a long exact sequence on homology $$\dots\to\tilde{H}_n(A)\to\tilde{H}_n(X)\to H_n(X,A)\to\tilde{H}_{n-1}(A)\to\cdots$$ and then proceed with steps 3 and 4 of your argument as before.

Hatcher discusses this on page 118, though he seems to never explicitly mention that this is used in the proof of Theorem 2.13.

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    $\begingroup$ So is $H_1(X) \rightarrow H_1(X,A)$ Always a surjection if A is path connected? Because if it is, $\tilde{H}_0(A)$ will be zero, but $H_1(X,A)=\tilde{H}_1(X,A)$. So if we have a L.E.S of a homology pair, we can always change it to the reduced homology and still have a L.E.S.? $\endgroup$
    – user637978
    Apr 17, 2019 at 17:37
  • $\begingroup$ Yes, that's correct. $\endgroup$ Apr 17, 2019 at 18:07

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