Retract and homology I have this problem in Hatcher's book :

Show that if $A$ is a retract of $X$ then the map $H_{n}(A)\rightarrow H_{n}(X)$ induced by the inclusion $A\subset X$ is injective.

I think I have problems with homology sequences, because (for me) this map is always injective.
Thank you.
 A: You don't need the homology exact sequence, you merely need functoriality of homology: $H(r\circ j)=H(r)\circ H(j)$ and $H(id)=id_H$. The map $H(A)\rightarrow H(X)$ is not always injective, but the map on the chain level $S(A)\rightarrow S(X)$ is always injective (and is often viewed as an inclusion). Let me guess why you think that the map $H(A)\rightarrow H(X)$ is always injective. Perhaps you think that since the inclusion $A\rightarrow X$ is always injective, it always has a left inverse $X\rightarrow A$, and so the argument with the functoriality always applies. Wait, this is WRONG: the whole point is that in the case of a retract you have a continuous left inverse (and only continuous maps induce maps in homology, as the functor $H$ is defined on $\mathbf{Top}$).
A: Suppose $\ker i_*\neq 0$. let $a\in\ker (i_*)\setminus \{0\}$. We have $0=\ker\mbox{Id}_{H_n(A)}=\ker(r_*\circ i_*)$. But note that $$(r_*\circ i_*)(a)=r_*(0)=0$$ so $a\in\ker\mbox{Id}_{H_n(A)}$. This is a contradiction by the non-triviality of $a$ so $\ker i_*=0$ hence $i_*$ is an injection. You can similarly show that $r_*$ is a surjection and you should try to show this yourself.
