# 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.

• Do you know any functions $f(x)$ that are continuous bijections on some open interval $(0,b)$ which has a pole at $b$? If so you can you change it to $f(bx)$ and it will work on the interval $(0,1)$. You probably know of one such function from trigonometry. Can you find it? Commented Mar 15, 2021 at 17:22
• In general it is a lot harder to prove that two spaces are homeomorphic than proving that two spaces are not homeomorphic. Commented Mar 15, 2021 at 18:39

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?

• Would it just be adjusting tan further, to $f(x)=tan (x* (pi / 2))$ or is this incorrect because there is a residual negative piece to the function when less than 0?
– User
Commented Mar 15, 2021 at 18:51
• Yes, $f(x) = \tan(x \pi/2)$ is a very nice simple $(0,1) \to (0, \infty)$. Commented Mar 15, 2021 at 23:19