How to show $f(x_1,x_2)=\begin{pmatrix} e^{x_1}+e^{x_2} \\ e^{x_1}-e^{x_2} \end{pmatrix}$ is surjective?

Question

Let $f: R^2 \to R^2$ be

$f(x_1,x_2)=\begin{pmatrix} e^{x_1}+e^{x_2} \\ e^{x_1}-e^{x_2} \end{pmatrix}$

I am trying to show it is invertible. So far I have shown it is injective, but I still need to show it is surjective. My understanding is that $f: R^2 \to A$ is surjective if the range of $f$ is the whole of $A$, therefore I would need to redefine the codomain to get global invertibility. Is this correct? How would you show surjectivity?

Answer 1

We have that

$$\begin{pmatrix} e^{x_1}+e^{x_2} \\ e^{x_1}-e^{x_2} \end{pmatrix}=\begin{pmatrix} a \\ b \end{pmatrix} \implies 2e^{x_1}=a+b \quad \land \quad 2e^{x_2}=a-b$$

and now, what about if $a+b\le0 \lor a-b\le 0$?

Then we need $a+b>0$ and $a-b >0$.

Answer 2

The function is certainly not surjective. The image contains only points whose first coordinate is positive.