Let $A \subset X$; A retraction of $X$ onto $A$ is a continuous 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.

The first step is to show that $\pi_1:\mathrm{Top}_* \to \mathrm{Grp}$ is a functor which can be found in Munkres as well.
The next step is the definition of a retraction.  A retraction $r:X \to A$ is a map so that $r \circ i=id_A$. Since $id_A$ is bijective, $r$ must be surjective and likewise $i$ must be injective.
The axioms of a functor tell that $id_*:\pi_1(A) \to \pi_1(A)$ is necessarily identity and that $(r \circ i)_*=r_* \circ i_*$ from which we gather that $r_* \circ i_*=id_*$.
