$\frac{x}{x^2-1}$ bijection from $(-1,1)$ to $\mathbb{R}$ I am trying to show that the function $f:(-1,1) \rightarrow \mathbb{R}$, $f(x) = \frac{x}{x^2-1}$ is bijective. I have the following two questions:
1) Proving it using calculus: First I show that $f$ is injective. Since $f'(x) = -\frac{x^2+1}{(x^2-1)^2} < 0$ for all $x \in (-1, 1)$, then $f$ is monotonic and hence injective. 
I am having trouble showing that it is surjective using the Intermediate Value Theorem (IVT). The IVT states that if $f: [a,b] \rightarrow \mathbb{R}$ is continuous and $L$ is a real number satisfying $f(a) < L < f(b)$ or $f(b) < L < f(a)$, then there exists a $c \in (a,b)$ where $f(c) = L$. Clearly the function at hand is continuous on $(-1, 1)$, but to use the aforementioned theorem, we need to have a closed interval $[a,b]$ (whereas $(-1, 1)$ is not a closed interval...). I "sort of" have an idea: Since $f$ is continuous on $(-1, 1)$, $f(-1)$ isn't defined but it is basically $+\infty$ and $f(1)$ is "like" $-\infty$, so if $L$ is a real number $-\infty< L < +\infty$, i.e., $L \in \mathbb{R}$, then there exists a $c \in (-1, 1)$ such that $f(c) = L$, hence $f$ is surjective. But I feel this is wrong because $f(-1)$ and $f(1)$ aren't defined and the IVT needs a closed interval.
2) How can I prove bijection without using calculus? E.g., to show injection, let $\frac{x_1}{x_1^2-1} = \frac{x_2}{x_2^2-1}$ and I want to reach the conclusion that $x_1 = x_2$, but I am stuck. Also, how can I prove surjection?
 A: You can show it is surjective by solving the quadratic equation.  Given $y \in \Bbb R$ we want to find $x$ such that $\frac x{x^2-1}=y$, or $yx^2-x-y=0, x=\frac 1{2y}(1\pm \sqrt{1+4y^2})$ and we want the signs of $x$ and $y$ to be opposite, so we take the minus sign.  The fact that you get a unique answer shows it is injective, so answers your second part as well.
A: $$=>x_1x_2^2-x_1=x_2x_1^2-x_2$$
$$=>x_1x_2(x_2-x_1)=-(x_2-x_1)$$
$$(x_1-x_2)(1+x_1x_2)=0$$
$$\text{So, either} \hspace{.2cm}x_1=x_2 \hspace{.2cm} or  \hspace{.2cm} x_1=-\frac{1}{x_2}.$$
The last equality is not possible due to the domain $(-1,1)$.
A: $f(x) = \frac{x}{x^2-1} = \frac{1}{2} (\frac{1}{1+x} - \frac{1}{1-x}).$
$-1 < x < y < 1 \implies \frac{1}{1+x} - \frac{1}{1+y} = \frac{y - x}{(1+x)(1+y)} > 0.$
$-1 < x < y < 1 \implies -\frac{1}{1-x} - (-\frac{1}{1-y}) = \frac{-(1-y)+(1-x)}{(1-x)(1-y)} = \frac{y-x}{(1-x)(1-y)} > 0.$  
$\therefore$
$-1 < x < y < 1 \implies \frac{x}{x^2-1} > \frac{y}{y^2-1}.$  
$\therefore f$ is injective.
Let $y \in \mathbb{R}$.  
$$\lim_{x \rightarrow -1+0} f(x) = +\infty.$$
$$\lim_{x \rightarrow 1-0} f(x) = -\infty.$$
There is $\delta_1 > 0$ such that $-1 < x < -1 + \delta_1 \implies f(x) < y$.  
There is $\delta_2 > 0$ such that $1 - \delta_2 < x < 1 \implies y < f(x)$.  
Let $-1 < a < -1+\delta_1$ and $1-\delta_2 < b < 1$.  
Then $f(a) < y < f(b)$.
$f$ is continuous on $[a, b]$.  
So by the intermediate-value theorem, there exists $c \in (a, b)$ such that $f(c) = y$.  
$\therefore f$ is surjective.
