maps on quotients. I'm trying to define a map over a Klein bottle $\mathbb{K}^2$ but I'm not totally sure on how to do it the right way. My approach is to define over a fundamental domain (a square) and try extending it to the quotient.
The thing is anywhere I look to get further insight about how to do it, all I can find is graph theory for maps on non-orientable surfaces.
So my questions are the following:

*

*Can anyone recommend good literature about a geometrical approach to endomorphisms on quotient surfaces?

*In any case, what should I pay attention to in order to obtain a well defined endomorphism?

In order to provide a clearer picture of what I'm trying to do, consider a fundamental a fundamental domain of the klein bottle (let's call it a ''Mobius cylinder'' for a moment to get some perspective), so we have a middle points segment parallel to the sides with identical orientations, which makes it $S^1$ once I pass to the quotient. Let's say I want to define a map over all of $\mathbb{K}^2$ that restricted to that circle is $z^2$, I don't think I will be having any trouble with that passing to the quotient, but again, since it is my first time trying maps on non orientable surfaces I don't want to make any mistakes and hasn't been easy to find any good sources. So, thanks in advance for any help you can provide.
 A: Whenever we have a quotient object $X/\!\sim$, we will have the following universal property:

$f : X/\!\sim \to Y$ is a map if and only if $f \circ p : X \to Y$ is a map

here $p : X \to X/\!\sim$ is the projection map.
In the setting of topological spaces, this means that maps $f : X/\!\sim \to Y$ correspond exactly to maps $\tilde{f} : X \to Y$ with the bonus property that if
$x_1 \sim x_2$ then $f(x_1) = f(x_2)$.

Let's now apply this to your example. We want to understand all maps from $K$ to $K$, where the Klein Bottle $K$ is defined as
$$ [0,1] \times [0,1] \bigg/ \big \{(a,0)=(a,1), (0,b)=(1,1-b) \big \} $$
By the above universal property, these correspond exactly to maps $f : [0,1] \times [0,1] \to K$ so that $f(a,0) = f(a,1)$ and $f(0,b) = f(1,1-b)$.
This technique is extremely flexible, and gives you access to every possible map. I don't know of any references that explicitly discuss it, but this is probably because I haven't read many books on topology. That said, this example is used in many good category theory references, as it was one of the motivating examples for the more general notion of quotient.

Edit: I previously used $\theta \mapsto \theta^2$ as the example map, but $z^2$ in the complex plane really corresponds to $\theta \mapsto 2\theta$ in this setting. I've updated my answer to use this map instead.

To give a concrete example of this technique in action, say we want to give a map $K \to K$ that restricts to $2\theta$ on the circle that you've outlined. In this case, as you have noticed, it is easier to view $K$ as a quotient of a cylinder, so let's write
$$K = S^1 \times [0,1] \bigg / (\theta,0) = (-\theta,1)$$
Then a map $f : K \to K$ so that $f([(\theta, \frac{1}{2})]) = [(2\theta, \frac{1}{2})]$ corresponds to a map $\tilde{f} : S^1 \times [0,1] \to K$ with $\tilde{f}(\theta, \frac{1}{2}) = [(2\theta, \frac{1}{2})]$ with the bonus property that $\tilde{f}(\theta,0) = \tilde{f}(-\theta,1)$. Here we have written $[(\theta,x)]$ for the class of $(\theta,x)$ after quotienting.
But how do we map into $K$? Well if we map into $S^1 \times [0,1]$ then we can compose with the projection map $p : S^1 \times [0,1] \to K$!
So, to build our desired map $K \to K$, we follow this procedure:


*

*Build a map $\tilde{f} : S^1 \times [0,1] \to S^1 \times [0,1]$ so that

*

*$\tilde{f}(\theta, \frac{1}{2}) = (2\theta, \frac{1}{2})$

*$\tilde{f}(\theta, 0) = \tilde{f}(-\theta, 1)$



*Say the magic words "universal property of quotients!", which guarantees that this $\tilde{f}$ induces a function $f : K \to S^1 \times [0,1]$.


*Finally, look at the composition $p \circ f : K \to K$, and realize it does what we want.

But this is comparatively easy! If we can find a way to do $2\theta$ at every level, in a way that reverses orientation along the way, then we'll win. We can do this by linearly interpolating between the coefficients $1$ and $-1$ in a way that depends on $x$:
$$
\tilde{f}(\theta,x) = 
\begin{cases}
((4x-1)2\theta,x) & 0 \leq x \leq \frac{1}{2}\\
(2\theta,x)       & \frac{1}{2} \leq x \leq 1
\end{cases}
$$
To be extremely explicit, the universal property says that
$$f([(\theta,x)]) = \tilde{f}(\theta,x)$$
is well defined.
Then composing with $p$ gives
$$(p \circ f)([(\theta,x)]) = p(\tilde{f}(\theta,x)) = [\tilde{f}(\theta,x)]$$
which is the desired function from $K \to K$.

I hope this helps ^_^
