How does one just find and come up with homeomorphisms? For example how would I find a homeomorphism from $(0, 1)$ to the reals and one from $(0, 1)$ to $(0, ∞)$?
I know the definition of a homeomorphism is if a function is both one-to-one, onto, continuous, and has a continuous inverse. However, I do not know how to craft my own to fit these standards from the maps above.
 A: There are some basic real homeomorphisms which we can use as building blocks to construct more with various desired properties. The most useful are:

*

*Linear functions $f(x) = ax+b$ map interval endpoints to different interval endpoints.

*$\exp : \mathbb{R} \to \mathbb{R}^{+}$ and its inverse $\ln$ go between the full set and one-ended open intervals.

*$\tan : (-\frac{\pi}{2}, \frac{\pi}{2}) \to \mathbb{R}$ and its inverse $\arctan$ go between bounded and unbounded open intervals. (For a one-ended open interval, note $\tan : (0, \frac{\pi}{2}) \to \mathbb{R}^{+}$.)

Composing any two homeomorphisms gives another homeomorphism. (If you haven't seen the proof, it's straightforward from definitions.)
For the first example, to map from $(0,1)$ to the reals, $\tan$ has similar properties, but the wrong interval endpoints. So let's compose it with a linear map $L(x) = ax+b$, and see if we can get $f = \tan \mathrel{\circ} L$ with the desired domain and range.
$$ \lim_{x \to 0^{+}} f(x) = \lim_{x \to 0^{+}} \tan L(x) = \lim_{x \to 0^{+}} \tan(ax+b) $$
so we would like $a(0) + b = -\frac{\pi}{2}$. Similarly, for the limit as $x \to 1^{-}$, we would like $a(1) + b = \frac{\pi}{2}$. This gives $b= -\frac{\pi}{2}$, $a = \pi$, $L(x) = \pi(x - \frac{1}{2})$, and finally
$$ f(x) = \tan \left[ \pi \left(x - \frac{1}{2}\right) \right] $$
(Optionally, then notice $f(x) = -\cot(\pi x)$. Possibly drop the negative sign from this if you don't mind it decreasing rather than increasing. These usually won't matter much for further use.)
Can you write a $(0,1) \to (0,\infty)$ homeomorphism now?
