retraction induced homomorphism is surjective

Question

I am having a hard time proving this although it looks trivial...

Let $r:X\to A$ be a retraction between a topological space $X$ and $A\subset X$ such that $r(a_0)=a_0$ for $a_0\in A$ then the induced homomorphism $r_*:\pi_1(X,a_0)\to \pi_1(A,a_0)$ is surjective.

I tried to prove it as follows:

I showed that if $g$ is a loop in $A$ based at $a_0$ then it is a loop also in $X$ based at $a_0$

So, given $[g]\in \pi_1(A,a_0)$, let us take $[g]\in \pi_1(X,a_0)$ (which will stand for a different homotopy class)

So we get: $$r_*([g])=[r\circ g]=[id_A\circ g]=[g]$$ that's since $Im(g)\subset A$ and $r$ is a retraction.

I believe that something is off in that proof in the part with the homotopy classes, so help please

Answer 1

Consider $i$ canonical injection $i\colon A \to X$. The fact that $r$ is a retract means $$r \circ i = \mathrm{id}_A$$ From this we get $$r_* \circ i_* = \mathrm{id}_{\pi_1(A)}$$ and this implies $r_*$ surjective since it has a right inverse.

Answer 2

this follows by functoriality of $\pi_1$, i.e. the fact that $f_*g_*=(fg)_*$ and $(id_A)_* = id_{\pi_1(A)}$.

Just note that a retract $r$ is a map $X \to A$ sucht that $A \hookrightarrow X \to A$ is the identity. This means that the composition $\pi_1(A,a_0) \to \pi_1(X,a_0) \stackrel {r_*} \to \pi_1(A,a_0)$ induces actually the identity. This means that $r_*$ has a right inverse induced by the inclusion, which implies surjectivity of $r_*$.

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To comment on your solution: you shouldn't give $[g]$ two different meanings. Call $g$ a loop in $A$ and $[g]$ its equivalence class in $\pi_1(A,a_0)$ and $i_*([g])$ its equivalence class in $\pi_1(X,a_0)$.

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The following is a clean-up of your proof:

We want to show that an arbritary element in $x \in \pi_1(A,a_0)$ is in the image. We know that this element is represented by a loop $g$ based in $a_0$, ie. $[g]=x$. We want to consider the element $i_*([g]) \in \pi_1(X,a_0)$. The claim is, that this element hits $x$. Indeed, by definition of induced maps (which is well-defined, hence the equality signs): $r_*(i_*([g])) = r_*([i\circ g]) = [r \circ i \circ g]= [id_A \circ g] = [g] =x$.