Let $a<b, (a,b \in \mathbb{R})$
I want to show that there is an homeomorphism $f$ between $(a,b)$ and $\mathbb{R}$ such that $x < y \iff f(x) < f(y)$, $(x,y \in (a,b))$, but I don't want to use the $\text{tan}(x)$ function.
The reason for this is because I want to extend $f$ to an homeomorphism between $[a,b]$ and $[-\infty, \infty]$ that preserves the order in the most elementary way possible.
I saw on page 342 of Bourbaki's General Topology Part 1 that the following function appears to have the properties I'm looking for:
$f:(a,b) \rightarrow \mathbb{R}: x \mapsto f(x) = -\left (\frac{1}{x-a}+\frac{1}{x-b}\right)$
The problem is that Boubaki doesn't show that $f$ is strictly increasing, nor even that $f^{-1}$ is, and I'm having trouble to prove it by myself.
If you already know that f is strictly increasing, then by calculating its inverse you can easily conclude that $f$ is bijective, and $f$ is therefore an homeomorphism, as it is any monotonic (extended-real valued) bijective function with domain and codomain being intervals.
I would be very grateful, if anyone can help me show that for all $x,y \in (a,b) $: $$ x < y \iff - \left(\frac{1}{x-a}+\frac{1}{x-b}\right)<-\left(\frac{1}{y-a}+\frac{1}{y-b}\right)$$