Prove a function $f: \mathbb{R}^{2} \to \mathbb{R}^{2}$ is surjective I'm given the map $f: (x,y) \mapsto (x+3,4-y): \mathbb{R}^{2} \to \mathbb{R}^{2}$; how do I prove this function is onto (surjective)?
So far I said that let $x=z$ and $y=k$,
therefore $f(x,y)=(z+3,4-k)$, that means $f(x,y)$ is onto (surjective).
I'm not sure if this is the way to prove a function is onto. Or does this mean that the function is not onto?? Thanks.

Let $(z_{1},z_{2}) \in \mathbb{R}^{2}$; then $z_{1} = x+3$ and $z_{2} = 4-y$ for some $(x,y) \in \mathbb{R}^{2}$ iff $x = z_{1}-3$ and $y = 4-z_{2}$; this shows that for every point $(z_{1},z_{2})$ of $\mathbb{R}^{2}$ there is some unique $(x,y) \in \mathbb{R}^{2}$ such that $(z_{1},z_{2}) = f(x,y)$, so $f$ is in fact bijective, and hence surjective.
