Homology of pair (A,A) 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?

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