Bijective continuous function between $(0 ,\infty)$ and $\mathbb{R}$. I've been working on some problems in topology, and it is odd to start
working again on sets, so I've needed some help. I want to build a
bijective continuous function $f$ such that its inverse is continuous
from $(0,\infty)$ and $(-\infty,\infty).$ I have a feeling that this
problem is similar to the Hilbert's Hotel problem, but I am having
issues figuring out the trick to make this surjective. I understand that
I need to negate the negatives, but how to differentiate the mapping of
the negatives and the positives to $\mathbb{R}^+$ seems peculiar to me.
 A: Let $f:(0,\infty)\to(-\infty,\infty)$ be defined by:
$$f(x)=x-\frac1x$$
It's inverse $f^{-1}:(-\infty,\infty)\to(0,\infty)$ is defined by:
$$f^{-1}(x)=\frac x2+\sqrt{\frac{x^2}4+1}$$
The function $x\mapsto\ln x$ is simpler, in my opinion.
A: Another option (which has germs of generality) is the following.
Consider the function $f: \mathbb{R} \rightarrow (-1,1)$ given by
$$x \mapsto \frac{x}{1+|x|}.$$
The inverse is the map $g(x)=\frac{x}{1-|x|}$, as easily verified. Both are obviously continuous. A homeomorphism between $(-1,1)$ and $(0,1)$ is simple (translation+dilation). A homeomorphism between $(0,1)$ and $(1,\infty)$ is also simple (given by $\frac{1}{x}$). Translation now gives a homeomorphism between $(1,\infty)$ and $(0,\infty)$. We then have the chain
$$\mathbb{R} \leftrightarrow (-1,1)\leftrightarrow (0,1) \leftrightarrow(1,\infty) \leftrightarrow (0,\infty).$$
The "germs of generality" are imbued in the function $f$, which is easily generalizable for a homeomorphism between the open unit ball of a Banach space and the Banach space itself (in particular, the open ball of $\mathbb{R}^n$ and $\mathbb{R}^n$).
A: How about $f(x)=e^x$? It's inverse is $f^{-1}(x)=\ln(x)$ which has the desired properties.
