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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 $H_n((X \cup CA)/X) = H_n(A)$.

I have seen this discussion in one of the post in mathstack but having no clue how to use this in the problem. The long exact sequence I can calculate but why $H_n(X \cup CA, CA) \approx H_n(X,A)$. And where suspension is used and how to split the long exact sequence in the direct sum?

Ri-Li
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3 Answers3

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Hatcher suggests to use the fact that $(X\cup CA)/X\simeq SA$: in order to do that you can consider what you obtained in the point (a) of the exercise.

From (a) you know that $A$ is contractible in $X$ iff $X$ is a retract of $X\cup CA$. Since $X$ is a retract of $X\cup CA$ you have that the following sequence splits: $$0\to \tilde H_n(X)\to \tilde H_n(X\cup CA)\to \tilde H_n(X\cup CA,X)\to 0,$$ hence $$\tilde H_n(X\cup CA)\approx \tilde H_n(X)\oplus \tilde H_n(X\cup CA,X).\label{a}\tag{1}$$ Now, $$\tilde H_n(X\cup CA,X)\approx \tilde H_n((X\cup CA)/X)\approx \tilde H_n(SA)\approx \tilde H_{n-1}(A).$$ In order to obtain the desired result you just need to recognize $\tilde H_n(X\cup CA)$ as $\tilde H_n(X,A)\approx H_n(X,A)$ and ($\ref{a}$) becomes $$H_n(X,A)\approx \tilde H_n(X)\oplus \tilde H_{n-1}(A).$$

Dario
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Being contractible in $X$ means that the inclusion $i: A \hookrightarrow X$ is nullhomotopic.

Homology being a homotopy invariant functor gives that the long exact sequence of that pair splits up into short exact sequences:

$$ 0\to H_iX \to H_i(X,A) \to H_{i-1}A \to 0 $$

Now we only need to show that this sequence splits. A splitting might be constructed by using that an element $\alpha$ in $C_{i-1}A$ is nullhomologous in $X$. Hence there is an element in $C_iX$ which gives $\alpha$ as boundary. This will define an element in $H_i(X,A)$ which maps to $\alpha$.

(it will also define an element in $H_iX$ but this will be zero, but we would expect that by exactness)

Daniel Valenzuela
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I don't know what notion of equivalence you're using when you write $SA \sim A$, but consider the case where $A = S^1$. Then $SA = S^2$, and this is not homotopy equivalent to $S^1$.

Perhaps this is where your argument goes off the rails.

The intuitive argument is this: an element of $H_n(X, A)$ is a chain in $X$ whose boundary maps to $A$. Since $A$ is contractile in $X$, we might as well regard this as a chain in $(X, pt)$, i.e., in the reduced homology of $X$.

But if $A$ is interesting, then it's possible that two maps into $A$ are fundamentally different...but these are maps of $n-1$-chains into $A$, and that gives you the $H_{n-1}(A)$ factor.

John Hughes
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  • Yeah that was wrong but the main problem in hatcher is right so how to derive it? – Ri-Li Mar 09 '15 at 19:11
  • And the argument given there was from another post regarding this problem. Here it was written but I can't understand it. – Ri-Li Mar 09 '15 at 19:14
  • My own inclination would be old school (as my intuitive answer suggests): write out everything at the level of cycles and boundaries and chain maps. Alternative, you could write out the long exact sequence of the pair and perhaps show that the contractibility of $A$ in $X$ implies a splitting of the sequence. It surely shows that $\tilde{H}_i(A) \to \tilde{H}_i(X)$ is the zero-map, so you get a short exact sequence from the long one at each $n$. – John Hughes Mar 09 '15 at 19:25