This an example from a book I am reading. The author claims that there are two ways to see that the bijection $f : (-1,1) \rightarrow \mathbb{R}$ defined by $f(x) = \frac{x}{1-x^2}$ is a homeomorphism. The simpler one is to note that both $f$ and its inverse are continuous functions, according to basis principles of calculus. The second way is to note that $f$ takes basis elements to basis elements and vice versa, where $(-1,1)$ and $\mathbb{R}$ are both endowed with the order topology.
How does one show that? What does he mean by "vice-versa," that the inverse function $f^{-1}$ takes a basis element to a basis element? I have tried many things, but I could not figure it out. E.g., I tried showing that $f((a,b)) = (f(a),f(b))$, but I had difficulty proving this.
EDIT
Okay I have a few related follow up questions. The proof seems to rely on the following claim:
Let $(X, \tau_X)$ and $(Y,\tau_Y)$ be topology spaces generated by the bases $\mathcal{B}$ and $\mathcal{C}$, respectively. If $f : X \rightarrow Y$ is a bijection such that $f(B) \in \mathcal{C}$ and $f^{-1}(C) \in \mathbb{B}$ for all $B \in \mathcal{B}$ and $C \in \mathcal{C}$, then $f$ is a homeomorphism.
Is this right?
My next question is,
If the function $g$ is an order preserving bijection between the ordered topogical spaces $X$ and $Y$, will $g$ always map open intervals to open intervals? Do we have some sort of generalized intermediate value theorem for maps between general ordered spaces?
You have the right idea: you want to show that both $f$ and $f^{-1}$ take basic open sets $(a,b)$ to basic open sets. One way to show this is to note that your function is monotone (necessarily, lest it not be invertible!). Because it is monotone increasing, it means it is 'order preserving' so $a<b$ if and only if $f(a) < f(b)$.
So you should show two things (this is the vice-versa): $f^{-1}\big((a,b)\big)$ is some interval of the form $(a', b')$, and $f\big((c',d')\big)$ is some interval $(c,d)$. Moishe's comment is a great way to do this.