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Show that if $A$ is contractible in $X$ then $H_n(X,A) \approx \tilde H_n(X) \oplus \tilde H_{n-1}(A)$

I know that $\tilde H_n(X \cup CA) \approx H_n(X \cup CA, CA) \approx H_n(X,A)$.

And $(X \cup CA)/X = SA$, where $SA$ is the suspension of $A$. So $H_n((X \cup CA)/X) = H_n(SA)$, where $SA$ is the suspension of $A$. But $SA \simeq A$, and homology is homotopic invariant, we have $H_n((X \cup CA)/X) = H_n(A)$.

The direct sum points me to Mayer-Vietoris sequence, and I guess I shall write as the direct sum of $X$ and $A$, or the homotopy equivalence respectively. But I am not sure how to meet the homology $H_{n-1}$ on the right hand side of the question?

1LiterTears
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1 Answers1

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Using the long exact sequence of a pair for $n \ge 2$, $$ H_n(A) \overset{0}{\to} H_n(X) \to H_n(X, A) \to H_{n-1}(A) \overset{0}{\to} H_{n-1}(X), $$ but one has to work a bit harder for the splitting.

Sammy Black
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  • In fact, the only place that splitting needs to be shown is the case $n=1$ because $H_{n-1}(A)=0$ for all $n\geq 2$ by contractibility of $A$ - hence $\tilde{H}{n-1}(A)=0$ for all $n\geq 2$. When $n=1$, $H{n-1}(A)$ is non-trivial. – Dan Rust Nov 18 '13 at 23:25
  • I'm not sure why you think the induced map $H_(A)\to H_(X)$ is always zero though. – Dan Rust Nov 18 '13 at 23:26
  • Because $A$ is contractable, so $H_*(A)$ is always zero map, I guess @DanielRust. But I still can't reconcile the question...? – 1LiterTears Nov 19 '13 at 01:03
  • When for $n=1$, we have $H_0(A)\to H_0(X)$ is a non-zero injection. – Dan Rust Nov 19 '13 at 01:09
  • You're right, @DanielRust. The last map is nonzero when $n = 1$. – Sammy Black Nov 19 '13 at 04:20
  • @DanielRust Is it also true that $H_*(A)=0$ when $A$ is contractable in $X$? This seems like a weaker condition than saying it is contractable. – Kaladin Apr 06 '15 at 15:41
  • @Kaladin Looking back at my comments, I have no idea why I assumed $A$ was contractible. Regardless, there was a duplicate question that received a good answer here. – Dan Rust Apr 06 '15 at 16:39
  • Oke, thank you very much. – Kaladin Apr 06 '15 at 18:53