Doubt in the proof of Lemma to prove No Retraction Theorem

Question

I have been reading Munkres, Topology. In Section 55, he goes on to prove No Retraction Theorem. To prove it, he uses the following lemma:

$$\text{If A is a retract of X, then the homomorphism of } \text{ fundamental groups induced by inclusion j: A }\to \text{ X is injective.}$$

After that he proves it as shown below:

$$\text {If r}: \text{ X }\to \text{ A is a retraction, then the composite map } r\circ \text{ j equals the identity map of A.} \text{ It follows that }r_* \circ j_* \text{ is the identity map of } \pi_1(A, a), \text{ so that } j_∗ \text{ must be injective.}$$

Some definitions : $\pi_1(A,a) $ is the first homotopy group.

$r_*$ is homomorphism map from one space to another group defined as: $$ \text{If h }: \text{X } \to \text{ Y}$$ is homomorphism map then: $$ h_*: \pi_1(X, x_0) \to \pi_1(Y,y_0) \text{ defined as }$$ $$ h_*[f]=[h\circ f]$$

My doubt in the proof is the last statement. How does he conclude that since $j_*$ is injective if $r_* \circ j_*$ equals identity? Please help me. It has been bugging me since last few hours.

Answer 1

Suppose $f\circ g={\rm Id}$ and assume that $g(x)=g(y)$. Then $$x={\rm Id}(x)=f\circ g(x)=f\circ g(y)={\rm Id}(y)=y.$$

Answer 2

In general, if $f\circ g$ is injective, then $g$ must be injective. Similarly, if $f\circ g$ is surjective, then $f$ is surjective.