The hyperbolic sine function is bijective on the set $\mathbb{R}$.

To demonstrate bijectivity, we'll show that for any $x \in \R$ it admits a unique solution $y$ in its arrival set, in this case $\R$.

We'll now show that the function $f(x)=y$ has a single solution with $y \in \R$.

Let's solve the equation $f(x)=y$, with unknown $x \in \R$. To do this, let set $X=e^x$.
As a reminder, $e^x>0$, so $X>0$. This point is important for what follows.

$$f(x)= \frac{e^x-e^{-x}}{2} \Longleftrightarrow X-\frac{1}{X}=2y$$
$$X^2 - 2Xy -1 =0$$

Using the relation $(X-y)^2 = X^2 + y^2 -2Xy$, we can write :

$$(X-y)^2 -y^2 - 1 = 0$$
$$(X-y)^2 = y^2+1$$

$$X = y \pm \sqrt{y^2 + 1}$$

There are therefore two solutions for $X$ :

$$\begin{align*} X_1 &=& y + \sqrt{y^2 + 1} \\ X_2 &=& y - \sqrt{y^2 + 1} \end{align*}$$

Analysis of $X_2$

For all $y \in \R$, $X_2 = y - \sqrt{y^2 + 1} <0$, since $y \leq \sqrt{y^2} < \sqrt{y^2+1}$ ).
$X_2$ can't be a solution because $X=e^x >0$, or $X_2<0$.

Analysis of $X_1$

Similarly, $y+\sqrt{y^2+ 1} > 0$, so $X = y + \sqrt{y^2 + 1}$.
So $X_1$ is a solution of $f(x)=y$, because $X_1>0$ .

Conclusion

So, for all $y \in \R$, the equation has a unique solution (defined on $\R$ ):

$$x = ln(y + \sqrt{y^2+1})$$

In conclusion, $f$ is bijective and its reciprocal function is $f^{-1} : \begin{cases} \R \rightarrow \R \\ x \rightarrow ln(y + \sqrt{y^2+1}) \end{cases}$

To sum up, the function $sinh$ is bijective on the set $\mathbb{R}$.