<s>[INST] 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?
 [/INST] 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 $f(x)\neq f(y)$
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