Proving that there are only 2 ways to glue pairs of sides of a fundamental polygon

Question

Given a fundamental polygon of your choice (I'll stick with a square), suppose you were to topologically identify a pair of sides. It's generally understood that there are two "distinct" ways to do this, which i'll call "vanilla" (cylinder-like) and "goofy" (mobius-like) respectively: see diagrams

So I believe that these are, up to some continuous "local deformations" the only ways of identifying sides of a fundamental polygon. I want to turn that subjective statement into a concrete one and prove it.

Here is my attempt:

Really identifying maps between sides can be boiled down to looking at maps from the unit line-segment to itself. We want to show that there are only 2 such distinct classes of ways to do this:

Consider the set of all invertible continuous maps $ \Phi =\lbrace \phi: [0, 1] \rightarrow [0,1] \rbrace $ I consider two elements $\phi_1, \phi_2$ homotopic if there exists a $\theta: [0, 1] \rightarrow \Phi$ such that $\theta(0) = \phi_1$ and $\theta(1) = \phi_2$ and that $\theta$ is continuous in a certain sense. I make this explicit by selecting the following definition of continuity:

$\forall x,y \in [0,1] \forall \epsilon > 0 \exists \delta \in [0,1] $ such that $| \theta(y)[x] - \theta(\delta)[x]| < \epsilon$ (I believe this is called "pointwise continuity")

If we call this relation $H$ (for homotopic) then i'm trying to show that $\Phi / H$ has two equivalence classes.

Now what to do: A skeleton could be:

1. Show that "vanilla" and "goofy" aren't homotopic

2. Show that every transformation is either homotopic to "vanilla" or homotopic to "goofy"

3. Show that homotopy is an equivalence relation (for good measure)

(2.) Is causing me the most stress, since I don't know any good way to quantify over all possible maps, and don't know any good properties to identify with all maps that differentiates between vanilla and goofy.

Answer 1

I suppose you want to say $\phi_1\sim\phi_2$ if there exists continuous $\theta\colon[0,1]\times[0,1]\to[0,1]$ such that $\theta(0,\cdot)=\phi_1$ and $\theta(1,\cdot)=\phi_2$ and $\theta(t,\cdot)\in\Phi$ for all $t$.

Let $\phi\in\Phi$. There exists unique $a,b\in[0,1]$ with $\phi(a)=0$ and $\phi(b)=1$. Suppose $0<a<1$. Then $\phi(0)>0$ and $\phi(1)>0$. By the IVT, there exist $x_1\in[0,a)$ and $x_2\in(a,1]$ woth $\phi(x_1)=\phi(x_2)=\min\{\phi(0),\phi(1)\}$, contradicting injectivity of $\phi$. Hence $a\in\{0,1\}$, likewise $b\in\{0,1\}$. In other words, $\phi|_{\{0,1\}}$ is a bijection $\{0,1\}\to\{0,1\}$.

If $\theta $ is a homotopy betawwn $\phi_1$ and $\phi_2$, then $t\mapsto \theta(t,0)$ is acontinuous map $[0,1]\to\{0,1\}$, hence constant. In partivular, $x\mapsto 1-x$ is not homotopic to the identity.

The main work perhaps is:

Claim. If $\phi|_{\{0,1\}}$ is the identity, then $\phi$ is homotopic to the identity.

Proof. For such $\phi$, define $\theta\colon[0,1]\times[0,1]\to[0,1]$ as $\theta(t,x)=t\phi(x)+(1-t)x$. Clearly, $\theta$ is continuous and $0\le \theta(t,x)\le 1$ for all $t,x\in[0,1]$. Also $\theta(0,x)=x$ and $\theta(1,x)=\phi(x)$. For good measure, we also note that $\theta(t,\cdot)\in\Phi$ for all $t\in[0,1]$. Indeed, for $0<t<1$, $\theta(t,\cdot)$ is continuous and onto (because $\theta(t,0)=0$ and $\theta(t,1)=1$). Assume $\theta(t,x)=\theta(t,x')$ with $x<x'$. Then $\phi(x)>\phi(x')$, and by another IVT-argument as above, we arrive at a contradiction with the injectivity of $\phi$. $\square$

Likewise, if $\phi|_{\{0,1\}}$ permutes the two points, $\phi$ is homotopic to $x\mapsto 1-x$ (just note that $x\mapsto \phi(1-x)$ is homotopic to the identity).

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Remark: If we drop the requirement that $\theta(t,\cdot)\in\Phi$ (or is at least injective) for all $t$, then all elements $\in\Phi$ may become homotopic because we can pass through a constant map (or a map that zig-zags a bit).