Proving that hyperbolic sinh is bijective I have to prove that sinh is bijective. So first i try to profe that it is 1-1:
$f(a)= f(b) => a=b$
I will use proof by contradiction: let f(a)= f(b) and $a$ doesn't equal to $b$.
After i multiply both sides by 2 i get:
$(e^a) - (1/e^a) = (e^b) - (1/e^b)$
$e^a - e^b = 1/e^2 - 1/e^b$
$e^a - e^b = (e^b - e^a)/(e^a*e^b)$
$e^a - e^b = -(e^a - e^b)/(e^a*e^b)$
$1=-1/(e^a*e^b)$
$e^a*e^b=-1$
Now this is a contradiction since exponential function doesn't have negative values.
There fore $a=b$ Is this valid proof?
And how do i prove that it is onto?
 A: Let $f(x) = \sinh(x)$, the hyperbolic sine function. Then $f'(x) = \cosh(x)$ which is strictly positive everywhere which implies that $f(x)$ is strictly increasing everywhere which implies it is one to one. The limits of $f(x)$ as $x$ goes to $+ \infty$ and $- \infty$ are $+ \infty$ and $- \infty$ respectively which implies surjective since the function is continuous (because it is differentiable everywhere).
A: You have made some mistakes in your algebra, what you should get is this:
If $f(a) = f(b)$, then 
$e^a-e^{-a} = e^b-e^{-b}$, and then
$e^a - e^b = 1/e^a - 1/e^b = (e^b - e^a)/e^ae^b$ so that
$e^ae^b(e^a - e^b) = (e^b - e^a)$
Then you can deduce that either $e^a = e^b$ (and we are done) or $e^ae^b = -1$, which is impossible.
To prove that it is onto, you can either derive a formula for the inverse function which works for any value in the codomain (it involves solving a quadratic in $e^x$), or you can argue that the function is continuous (and hence the image of $\mathbb R$ will be connected), and $\sinh x \to \infty$ as $x\to\infty$ and $\sinh x \to -\infty$ as $x\to-\infty$ so that the image must be the whole of $\mathbb R$.
Edit:
For the process of finding an explicit formula for the inverse function, you can to do this:
Let $x = \sinh^{-1} y$ so that $y = \sinh x$, and then
$y = \frac{e^x - e^{-x}}{2}$, giving $e^y - 2y - e^{-x} = 0$ or $(e^x)^2 - 2ye^x - 1=0$, and then use the quadratic equation formula (and since the discriminant $b^2-4ac$ is always positive, there is an answer for any value of $x$).
