If $A$ is retract of $X$, then the homomorphism of fundamental groups induced by inclusion $j:A \rightarrow X$ is injective
This Lemma in Munkres has about two lines of proof as below,
If $r:A \rightarrow X$ is a retraction , then the composite map $r \circ j$ equals the identity map of A. it follows that $r_* \circ j_*$ is the identity map of $\pi_1(A,a)$ so that $j_*$ must be injective.
I don't seem to get the argument well, I was hoping someone could break it down for me.
I know given the retraction $r:A \rightarrow X$, we can find and inclusion map $j:A \rightarrow X$ (Which will be the inverse of the retraction map ) such that for any point $a\in A$ $$(r \circ j)(a)=r(j(a))=a$$ My first question is, does this setup necessarily make the map $r$ surjective? and why?
The maps $r$ and $j$ (being continuous) induces the homomorphisms (functorials) $$r_*:\pi_1(X,a) \rightarrow \pi_1(A,a)$$ and $$j_*:\pi_1(A,a) \rightarrow \pi_1(X,a)$$ respectively.
Using the notion of loops, why is $r_* \circ j_*$ an identity?
and how does that make $j_*$ injective?
Any help will be appreciated. Thank you.