# Thread: Algebraic Topology - Retractions and Homomorpisms Induced by Inclusions

1. ## Algebraic Topology - Retractions and Homomorpisms Induced by Inclusions

I am reading Munkres book on Topology, Part II - Algegraic Topology Chapter 9 on the Fundamental Group.

On page 348 Munkres gives the following Lemma concerned with the homomorphism of fundamental groups induced by inclusions":

" Lemma 55.1. If A is a retract of X, then the homomorphism of fundamental groups induced by inclusion $\displaystyle j: A \rightarrow X$ is injective"

I am struggling with the proof - not so much intuitively - but in formulating a formal and explicit proof.

Because explaining my postion requires diagrams I have set out my problem in an attachment - see the attachment "Retractions and Induced Homomorphisms.

I have also provided an attachement of the relevant pages of Munkres book

I would like as much as anything a confirmation that my reasoning in the attachment "Problem ... ... " is correct. I would also be most interested to see how to formulate a formal and explicit proof of the Lemma

Peter

2. ## Re: Algebraic Topology - Retractions and Homomorpisms Induced by Inclusions

well, let's look at an actual retraction, so we can see what happens.

let's let X = {z in C: 1 ≤ |z| ≤ 2}, so we have an annulus in C. we'll give C the standard topology (the metric topology induced by |.|), and X the relative topology. and let's say A = S1, the unit circle.

now, to define the retraction, r, we'll set r(z) = z/|z|. if |z| = 1, then r(z) = z, so r is the identity map on the circle (we'll assume as given the somewhat messy details that |.|, complex multiplication, and inversion are all continuous maps).

now let's look at what r* does. by defintion, r*([f]) = [rof].

now, we need to choose base-points for our two spaces, so let's choose 1 = 1+i0 for both. so suppose we have a loop f in X that "goes outside of A", like a circle of radius 1/2, centered at 3/2. explicitly:

f(t) = (1/2)(cos(2πt + π) + i sin(2πt + π)) + 3/2

well, let's look at rof(t).

rof(0) = r(f(0)) = r((1/2)(-1 + i(0)) + 3/2) = r(-1/2 + 3/2 + i0) = r(1+i0) = (1+i0)/√(12+02) = 1/1 = 1

clearly, rof(1) = rof(0) = 1, as well. what happens is rof "arcs down the circle" as t goes from 0 to 1/4, comes back to 1 as t goes from 1/4 to 1/2, "arcs up the circle" as t goes from 1/2 to 3/4, and then returns to 1 as t goes from 3/4 to 1.

in this example, it's clear that [rof] = [e], the "constant loop" at 1 (rof isn't the constant loop, but it's homotopic to the constant loop, we can "shrink" it to the constant loop continuously, without ever leaving S1).

(note r isn't the ONLY possible retraction of X, we could pull elements of X to the circle along parallel curves, instead of straight lines, like the shutter of a camera lens).

in fact, we're going to have 2 kinds of loops in X: those that go "around the hole" (corresponding to their winding number in π1(S1), and those that don't: which will correspond to arcs in S1 we can shrink to the point 1 (that is, equivalent to loops in R, which are trivial). if a loop in X "leaves A", then r "pulls it back" when we form the corresponding loop in A (no matter what element x of X we choose, there's some element of A (namely r(x)) that corresponds to it. the continuity of r keeps our loop rof "connected").

so all r* does, is turn "loop homotopy equivalence classes in X" to "loop homotopy equivalence classes" in A. that is: suppose f~g, that is [f] = [g].

this means we have some continuous function F:[0,1]2→X with F(0,t) = f(t), F(1,t) = g(t).

now roF is continuous, being the composition of two continuous functions.

roF(0,t) = r(F(0,t)) = r(f(t)) = rof(t)
roF(1,t) = r(F(1,t)) = r(g(t)) = rog(t), which shows that [rof] = [rog] (since roF is a homotopy of rof with rog).

this shows that r* only depends on [f], and not on the particular representative f.

so now let's look at r*oj* two different ways:

first, let's apply j*, and then r* to an arbitrary loop class in A, [f].

that is: r*oj*([f]) = r*(j*([f]))

= r*([jof]) = [ro(jof)] = [(roj)of] = [f] (because f lies entirely within A).

now, let's use the fact that r*oj* = (roj)*

since roj = 1A, (roj)*[f] = [1Aof] = [f].

now, "ignore the stars" for a second. suppose you have two functions, h:X→Y and k:Y→Z. and suppose you know that koh is injective.

this means that if koh(x) = koh(x'), then x = x'.

that is, if k(h(x)) = k(h(x')), then x = x'.

so if h(x) = h(x'), then (since k is a FUNCTION), k(h(x)) = k(h(x')), whence x = x', so h is injective.

you can think of it this way: as long as we stay in koh(X), koh has a unique inverse. if h didn't have this same property,

say h(x) = h(x') for two different x,x', then koh would map that same pair to the same image in Z, and so koh wouldn't have a unique image:

we would have {x,x'} contained in (koh)-1(koh(x)).

why does this work? because k is injective on h(X). k doesn't have to be injective, in general, as long as h(X) is "small enough".

and that's why we get injectivity of j* here. r isn't injective, in fact r is usually very non-injective. but it's injective on A, which is all we care about.

(i chose the example of an annulus, so you could see where the name "retraction" comes from: we "retract the ring to a circle" by "squeezing" its thickness". other standard examples deal with "flattening" a cylinder to a circle, which is really the same thing, since an annulus and a cylinder are homeomorphic (say, by using sterographic projection)).