We have the linear map $f:\mathbb{R}^3\rightarrow \mathbb{R}^2, \ \begin{pmatrix}x_1 \\ x_2 \\ x_3\end{pmatrix}\mapsto \begin{pmatrix}x_3 \\ x_1\end{pmatrix}$.
I have shown that this map is not injective, since the kernel doesn't contain only the zero vector.
Now I want to show that this map is surjective.
For that we want that for each $z\in \mathbb{R}^2$ there is vector in $\mathbb{R}^3$ such that $f\begin{pmatrix}x_1 \\ x_2 \\ x_3\end{pmatrix}=z$.
So $z$ is in the form $\begin{pmatrix}x_3 \\ x_1\end{pmatrix}$. Therefore we already have the values $x_1$ and $x_3$ and we can arbitrarily choose $x_2$.
Is that corect toshow the surjectivity of $f$ ?