Why is $H_i(E_+^n,S^{n-1})\cong H_{i-1}(S^{n-1})?$

Question

Let $E_n^+\subset S^n$ be the upper hemisphere of $S^n$. Why does the fact that $E_+^n$ is contractible imply that in homology $H_i(E_+^n,S^{n-1})\cong H_{i-1}(S^{n-1})?$ I know that if a space is contractible then its homology is $0$ except in $i=0$ but that doesn't really help, no?

Answer 1

Ted Shifrin's suggestion of looking at the LES of a pair allows you to deduce the result by contracitiblity, because writing the sequence down yields

$$H_{i}(E_+) \to H_{i}(E_+,S^{n-1}) \to H_{i-1}(S^{n-1}) \to H_{i-1}(E_+)$$ which will give you the result since $\tilde{H}_i(E_+)$ is trivial for all $i \geq 0$.

A different geometric reason to see this is that since if $(X,A)$ are a good pair and in particular, if $X$ is a $CW$ complex and $A$ is a subcomplex, then $H_i(X,A) \cong \tilde{H}_i(X/A)$. Here, you can give $S^n$ a $CW$ structure inductively. Then $H_i(E_+,S^{n-1}) \cong H_i(E_+/S^{n-1}) = \tilde{H}_i(S^n)=\tilde{H}_{i-1}(S^{n-1})$.

The last equality above may be deduced by observation if you know the the homology of spheres, but more importantly, using professor Shifrin's argument together with the fact that $H_i(E_+,S^{n-1}) \cong H_i(E_+/S^{n-1}) = \tilde{H}_i(S^n)$, you can deduce the homology of a sphere.