1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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
0
$\begingroup$

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

$\endgroup$
5
  • $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.