## Fundamental Group Induced Homomorphisms.

### April 3, 2011

"Why can’t you have a retract of the disk onto its boundary circle?"

This is the question we will attempt to answer in this post.  Along the way, we will climb treacherous mountains and dive deep into dark waters — most frightening, perhaps: we will find things that we didn’t even know we were looking for.

(Note.  I will be going through, albeit at a much slower and detailed pace, Hatcher’s pages 34 – 35.  Much of the material here is not difficult and is beautifully spread out in the usual Hatcher style.  I write this up for three reasons: one, to compound my learning of the subject; two, to emphasize that this topic actually has some neat applications; three, for the sake of clarifying things that I felt were not so trivial.  As always, I highly recommend reading through Hatcher’s book.)

For those of you who are not familiar with the definition of a retract, you’re in luck.

Definition.  Let $X$ be a topological space and $A$ is a subspace.  Then a continuous map $r:X\rightarrow A$ is a retraction if $r|_{A}$ the restriction of $r$ to $A$ is the identity.  If such a retraction exists, $A$ is called the retract.

Some examples are: a big disk retracts onto a smaller disk; any disk retracts onto a point; an annulus retracts onto a circle.  Now, let’s bring up the question again.  Why can’t we make a disk retract onto its boundary circle?  Straight from the definition, it is not obvious — what turns out to be tricky is the "continuous" part.  Intuitively, we’d need to "rip a hole" in the disk, which gets us in all kinds of continuity trouble.

At this point, it might occur to you that there is a fundamental difference between the disk and the circle: the fundamental group!  The former is trivial, the latter is ${\mathbb Z}$.  It we had some kind of continuous retraction, we’d assume that we’d preserve some of the structure of the space; maybe the fundamental groups would be the same?  Well, maybe not — but how much different could they be?

### They’re Not That Different.

Let’s work in general here, because a lot of what we’ll say applies more generally.  Suppose we have some continuous map $f:X\rightarrow Y$ where $X,Y$ are spaces.  Since we’re working carefully, we also might want to mention that these are based spaces; that is, if we want to work with the fundamental group, our spaces might not be nice and path-connected, so we better pick a basepoint.  We will write

$f:(X,x_{0})\rightarrow (Y,y_{0})$

to mean, "$f$ is a continuous map from $X$ to $Y$ such that $f(x_{0}) = y_{0}$."

Now suppose we have some loop $\gamma:I\rightarrow X$ based at $x_{0}$, then what does our $f$ do?  Well, we really want to have that $[\gamma]$ in the fundamental group go to $[f(\gamma)]$; that is, a loop in $X$ should map via $f$ to a loop in $Y$, and it should be the case that if we take $f$ of any loop in the equivalence class $[\gamma]$ (remember, these are all loops homotopic to $\gamma$ by definition; this is an element of the fundamental group!) to some loop in the equivalence class $[f(\gamma)]$; again, in other, other words, if $\gamma$ is homotopy equivalent to some other loop $\delta$ in $X$ then $f(\gamma)$ and $f(\delta)$ should be homotopy equivalence in $Y$.  This picture attempts to demonstrate this (base points in red, and the blue and green loops are homotopic in $X$, and we’d really like them to be homotopic in $Y$):

In fact, this definition of the induced map works and it is the canonical map to define in some sense (for those of you who are familiar with functors, since $\pi_{1}$ is a functor, this is exactly the induced map) — but, the fact that this induced map works (in the sense of this preserving homotopic paths that we talked about in the lat paragraph) is not entirely obvious.  Let’s be formal for a moment.

Definition.  Let $X,Y$ be based spaces with bases $x_{0}, y_{0}$ respectively and $f:(X,x_{0})\rightarrow (Y,y_{0})$ be a continuous map.  The induced map on fundamental groups denoted $f_{\ast}:\pi_{1}(X,x_{0})\rightarrow \pi_{1}(Y,y_{0})$ is defined to be $f_{\ast}([\gamma]) = [f(\gamma)]$ where $\gamma: I\rightarrow X$ is an arbitrary loop based at $x_{0}$ in $X$

This is a nice definition, but what have we said?  We’ve simply defined a map — but, we don’t know anything about it!  We don’t know if this is even well-defined!  Let’s just sketch a few facts about this map quickly.

The induced map is well-defined.  If we have some homotopy $g_{t}$ of loops based at $x_{0}$ will give us a composed homotopy $f(g_{t})$ as $f:X\rightarrow Y$ was assumed to be continuous to begin with.  Recall that $g_{0}$ is our "starting loop" and $g_{1}$ is the "ending loop" in the homtopy: this composition gives us that $f_{\ast}[g_{0}] = [f(g_{0})] = [f(g_{1})] = f_{\ast}[g_{1}]$ which gives us that $f$ is well-defined.

The induced map is a homomorphism.  Informally speaking, given two loops $\gamma_{0}, \gamma_{1}$, recall that composition of loops given by $\gamma_{0}\cdot\gamma_{1}$.  We have that $f(\gamma_{0}\cdot \gamma_{1}) = f(\gamma_{0})\cdot f(\gamma_{1})$ since $f$ is continuous, and this gives us that

$f_{\ast}([\gamma_{0}]\cdot [\gamma_{1}]) = f_{\ast}([\gamma_{0}\cdot\gamma_{1}]) = [f(\gamma_{0}\cdot\gamma_{1})]$

$= [f(\gamma_{0})\cdot f(\gamma_{1})] = [f(\gamma_{0})]\cdot [f(\gamma_{1})] = f_{\ast}([\gamma_{0}]) \cdot f_{\ast}([\gamma_{1}]).$

We have that $(f\circ g)_{\ast} = f_{\ast}\circ g_{\ast}$.  This is not hard to see; write this one out for yourself!  You only need associativity here: that $(f\circ g)(\gamma) = f(g(\gamma))$

The induced map of the identity is the identity.  If $id:X\rightarrow X$ is the identity, then $id_{\ast}([\gamma]) = [id(\gamma)] = [\gamma]$.  Cute.

Now, what if two spaces were topologically "the same"?  That is, what if two spaces $X$ and $Y$ were homeomorphic?  We’d expect the homotopy classes of loops to be the same as well since we’re just kind of "squishing out spaces around" under the homeomorphism.  Let’s state this formally.

Theorem.  If $f:X\rightarrow Y$ is a homeomorphism, then the induced map $f_{\ast}$ is an isomorphism on fundamental groups.

Proof.  This is not hard to show, and it’s actually kind of fun.  Let $f$ be the homeomorphism and let the inverse be $g$.  Then note

$f_{\ast}\circ g_{\ast} = (f\circ g)_{\ast} = id_{\ast} = id$

and similarly

$g_{\ast}\circ f_{\ast} = (g\circ f)_{\ast} = id_{\ast} = id$

which gives us that $f_{\ast}$ is both injective and surjective (as well as a homomorphism) which implies that $f_{\ast}$ is a isomorphism.  $\Box$

Theorem.  We have that $\pi_{1}(S^{n}) = 0$ if $n\geq 2$

Proof.  Take any based loop $\gamma$ in $S^{n}$.  The image of the loop on $S^{n}$ is disjoint from some point $x\in S^{n}$ if $n\geq 2$ (see the note at the end of the proof).  This should intuitively make sense to you: if you wrap yarn around a balloon so that it covers up everything, we should be able to "blow up the balloon" more so that a little bit of balloon shines through.  Either way, pick this point is misses and remove it: now our $S^{n} - \{x\}$ is homeomorphic to ${\mathbb R}^{n}$, but this is nullhomotopic and so we may contract our loop to a point (here is where we use the previous theorem!).  One-point-compactifying ${\mathbb R}^{n}$ again, we will have that our loop is still trivial.  Therefore, we get the isomorphism $\pi_{1}(S^{n}) = \pi_{1}({\mathbb R}^{n}) = 0$$\Box$

Note that there is some difficulty here in considering loops: we may be able to take a loop that seems like a "space filling curve" so that it is impossible to find a point that the loop does not hit.  In fact, this does not happen — but it is not quite trivial.  The argument is not difficult, but I feel it would deter us from our induced map discussion and so I will write about it another day.

### Back To Retracts.

But I started this post talking about retracts, so why don’t we get back to them?  As we noted, retracts are a "special kind" of continuous maps.  Before we state the theorem, let’s note the following property of retracts:

Let $A\subseteq X$ be a subspace and $r:X\rightarrow A$ a retract.  If $i:A\rightarrow X$ is the inclusion mapping, then we have $r\circ i = id$.  In other words, if we have the diagram

$A\stackrel{i}{\hookrightarrow} X \stackrel{r}{\rightarrow} A$

then this composition is the identity.  Using this fact we can easily prove the following theorem about retracts which turns out to be pretty useful.

Theorem.  If $X$ retracts onto $A$ via some retract $r:X\rightarrow A$, then we have that $i_{\ast}:\pi_{1}(A, x_{0})\rightarrow \pi_{1}(X,x_{0})$ is injective where $i_{\ast}$ is the map induced from the inclusion map $i:A\rightarrow X$

Proof.  If $r:X\rightarrow A$ is a retraction, we have $r\circ i = 1$ by above and so $r_{\ast}\circ i_{\ast} = 1$.  From this, let’s show that $i_{\ast}$ is injective.  If $i_{\ast}(a) = i_{\ast}(b)$, then $a = r_{\ast}(i_{\ast}(a)) = r_{\ast}(i_{\ast}(b)) = b$ and so we have that $i_{\ast}$ is injective.  $\Box$

Note that in order for $i_{\ast}$ to be surjective as well, we need that this retraction be what’s called a deform retraction, which we will discuss in some other post.  For now, this theorem gives us enough to play with a little bit.  In particular, we can now prove that a number of things do not retract to a number of other things.  I’ll give three similar examples, but I urge the reader to look at some of the Hatcher exercises which will give some other applications of this theorem.

### Examples!

The disk does not retract to its boundary circle.  Recall, the disk is denoted $D^{2}$ and the circle is denoted $S^{1}$.  If such a retract existed, then the inclusion would induce an injective map

$i_{\ast}:\pi_{1}(S^{1}) \rightarrow \pi_{1}(D^{2})$

but we know these fundamental groups!  The disk is contractible, and the circle is just ${\mathbb Z}$.  So we have that there would be an injective map

$i_{\ast}:{\mathbb Z}\rightarrow 0$

but this is clearly impossible.  If you don’t see this immediately, think of what you’d have to send the element 0 to.  Now what do you have to send 1 to?

Two disks connected at one point (in a filled-in figure 8) do not retract to the wedge of two circles (the figure 8) which forms its boundary.  If so, we’d have an injective map

$i_{\ast}:\pi_{1}(S^{1}\vee S^{1}) \rightarrow \pi_{1}(D^{2}\vee D^{2})$

but by the property of the wedge product in the fundamental group, we have that this would be an injective map

$i_{\ast}:{\mathbb Z}\ast {\mathbb Z} \rightarrow 0\ast 0 = 0$

where the left hand side is the free product.  We run into the same problem we had last time.  Sad.

The solid torus does not retract to the torus "shell" which forms its boundary.  Another way to say this: a donut (which is all "filled in") does not retract to its skin.  If so, then we’d have an injective map

$i_{\ast}:\pi_{1}(S^{1}\times S^{1}) \rightarrow \pi_{1}(D^{2}\times D^{2})$

but by the property of cartesian products in the fundamental group, we have that this would be an injective map

$i_{\ast}:{\mathbb Z}\times {\mathbb Z} \rightarrow 0\times 0 = 0$

which is nearly the same exact problem as we had the last two times.

Are you seeing the pattern here?  Many problems can be solved this way.  Unfortunately, higher dimensional spheres all have trivial fundamental group (which is kind of upsetting) and so we have to consider other invariants that this type of mapping might play nicely with.  For example, using the theorem above can you prove that $D^{3}$ does not retract to its boundary $S^{2}$

This idea of wanting "higher" invariants to differentiate higher-dimensional disks from their sphere shells motivates the idea of wanting "higher dimensional" homotopy groups and homology groups.  But this is certainly a topic for another day.

Let me leave you with a question.  Can we retract $S^{1}\vee S^{1}$ onto $S^{1}$?   If so, can you think of such a retract?  If so, can you think of another?  If so, can you think of (countably) infinitely more?  Hm.

### 4 Responses to “Fundamental Group Induced Homomorphisms.”

1. LINDA said

i love it when you talk dirty.

2. A guy studying topology now said

Yo thanks for this, it was useful.

3. Anonymous said

You explained retracts so much more clearly than my professor. Thanks so much!

4. MPitts said

Great stuff. I’m working thru Hatcher on my own, so this really helps.