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.

This is a general fact about set-maps, even: if $g \circ f$ is the identity map, then $f$ must be injective (and $g$ surjective).  The proof is simple:  if $f(a_1) = f(a_2)$ then hit this with $g$ on the left to get $g(f(a_1)) = g(f(a_2)$.  But $g \circ f$ is the identity so $a_1=a_2$.  Done.
Since you have $r \circ j = 1_A$ by definition of retract, apply $\pi_1$ to get
$(r \circ j)_* = 1_*$, which is $r_* \circ j_* = 1$ by functoriality business.  Now apply the previous fact.
