Regarding bijectivity of $f(x)=\frac{x}{1-x^2}$ Consider the map $f:(-1,1)\to \mathbb R,$ $$f(x)=\frac{x}{1-x^2}$$
Munkres claims that it is an order preserving bijection.
To see it's order preserving, assume $x< y$. Then $x^2<y^2, 1-x^2 >1-y^2$, so $\frac{1}{1-x^2}<\frac{1}{1-y^2}$.
To show it is a bijection, I guess it's easiest to construct the inverse. 
We have $y=\frac{x}{1-x^2}\iff yx^2+x-y=0$. But this equation has two solutions for $x$; how to see which one we need?
 A: Suppose that $x \neq y$ but $f(x)=f(y)$ for some $x$ and $y$.
It follows that $\frac{x}{1-x^2}=\frac{y}{1-y^2}$, that is, $y(1+xy)=x(1+xy)$ and since $1+xy\neq 0$ it follows that $x=y$, but we assumed $x \neq y$ so when $x\neq y$ we have $f(x)\neq f(y)$
A: Here's a trick proof. 
Recall that
$$\tan 2t=\frac{2\tan t}{1-\tan^2 t}.$$
Therefore
$$\frac{x}{1-x^2}=\frac12\tan(2\arctan x).$$
This is valid for $x\in(-1,1)$. On this interval
$\arctan$ is increasing, and on the interval
$(-\pi/2,\pi/2)$, so is $\tan$.
The inverse function is then
$$y\mapsto\tan\left(\frac{\arctan(2y)}2\right).$$
A: $$f(x)=\frac x{1-x^2}=\frac{-x}{(x-1)(x+1)}=\frac x2\left(\frac 1{x+1}-\frac 1{x-1}\right)=-\frac 12\left(\frac 1{x+1}+\frac 1{x-1}\right)$$
Can you see now why $f$ is an order preserving bijection?


*

*$x\lt y\iff\dfrac 1{x\pm 1}\gt\dfrac 1{y\pm 1};$ so their sums preserve $\gt$; finally multiplying by $-1/2$ gives $f(x)\lt f(y)$

*Let $f(x)=f(y)$ and WLOG assume $x\leq y$. We need to show $x=y$ for bijectivity; so suppose not, ie $x\lt y$ and use the above argument to infer that $f(x)\lt f(y)$, a contradiction!
