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.

  • $\begingroup$ I see, it looks like I have to go back to some fundamentals. thank you. But assuming $[f] \in \pi_1(A,a)$, what does $j([f])$ look like? $\endgroup$
    – J. Kyei
    Jun 29 '18 at 16:42
  • $\begingroup$ Good question. Here, $f$ is a loop in the subspace $A$, so $j([f])$ "looks like" the same loop, just viewed inside the larger parent $X$. $\endgroup$
    – Randall
    Jun 29 '18 at 16:44
  • $\begingroup$ It is clear now, I only need to convince myself that if $g \circ f$ is identity, then $f$ must be injective and $g$ surjective. thank you $\endgroup$
    – J. Kyei
    Jun 29 '18 at 16:54

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 $r_* \circ i_*=id_*$.

  • $\begingroup$ @Andress Thank you, I think I was missing the part that $id_A$ is necessarily bijective. I followed the 'functor' link, it's a nice material. $\endgroup$
    – J. Kyei
    Jun 29 '18 at 16:51
  • $\begingroup$ @J.Kyei no problem. I hope that the issue is clarified. You can use this to prove the "no retract" from $D^2$ onto its boundary, and consequently Brouwers fixed point theorem if you like. Munkres has a weird treatment of the latter, see Hatcher for a standard proof. $\endgroup$ Jun 29 '18 at 16:52
  • $\begingroup$ That is actually the proof I was working on, to show that there is no retract of the disk to the circle, but part of the reasoning is based on this lemma and I wasn't convince enough. it is now clear. $\endgroup$
    – J. Kyei
    Jun 29 '18 at 16:56
  • $\begingroup$ well hopefully it is clear now: how can $r_*:0 \to \mathbb Z$ possibly be surjective! Glad everything is good. $\endgroup$ Jun 29 '18 at 16:57
  • $\begingroup$ Yes, Thank you :) $\endgroup$
    – J. Kyei
    Jun 29 '18 at 16:59

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