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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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For a good pair $(X,A),$ i.e. if $A$ nonempty closed subspace which is a deformation retract of some neighborhood in $X,$ then $H_n(X,A) \cong \tilde{H}_n(X/A).$ In particular, $H_n(A,A)$ is equal to the reduced homology of a point and is therefore zero. – jspecter Oct 30 '12 at 12:56
@jspecter Why is $(A,A)$ a good pair? – user38268 Oct 30 '12 at 13:11
A is both a neighborhood and closed in A. – jspecter Oct 30 '12 at 13:30
up vote 2 down vote accepted

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! – Lola Oct 30 '12 at 14:48

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