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


*

*Show that "vanilla" and "goofy" aren't homotopic 

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

*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. 
 A: 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).

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). 
