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Why is the homology of the pair (A,A) zero? $H_n(A,A)=0, n\geq0$

To me it looks like the homology of a point so at least for $n=0$ it should not be zero.

How do we see this?

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    A is bo$t$h a neighborhood and closed in A.2012-10-30

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Let us consider the identity map $i : A \to A$. This a homeomorphism and so induces an isomorphism on homology. Now consider the long exact sequence of the pair $(A,A)$: We get

$\ldots \longrightarrow H_n(A)\stackrel{\cong}{\longrightarrow} H_n(A) \stackrel{f}{\longrightarrow} H_n(A,A) \stackrel{g}{\longrightarrow} H_{n-1}(A) \stackrel{\cong}{\longrightarrow} H_{n-1}(A) \longrightarrow \ldots $

Now this tells you that $f$ must be the zero map because its kernel is the whole of the homology group. So the image of $f$ is zero. However this would mean that the kernel of $g$ is zero. But then because of the isomorphism on the right we have that the image of $g$ is zero. The only way for the kernel of $g$ to be zero at the same time as the image being zero is if

$H_n(A,A) = 0.$

Another way is to probably go straight from the definition: By definition the relative singular chain groups are $C_n(A)/C_n(A) = 0$. Now when you take the homology of of this chain complex it is obvious that you get zero because you are taking the homology of the chain complex

$\rightarrow C_n(A)/C_n(A) \rightarrow C_{n-1}(A)/C_{n-1}(A) \rightarrow \ldots$

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    Nice explanation, thanks!2012-10-30