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.

I don't understand the idea behind what you're doing. 
For surjective, you need to pick an arbitrary $(z, w) \in \Bbb R^2$ (the range $\Bbb R^2$) and show that there's some $(x, y) \in \Bbb R^2$ (the domain $\Bbb R^2$) with $f(x, y)$ equal to your chosen pair $(z, w)$.
If this can happen, it means $$\underbrace{f(x, y) = (x + 3, 4 - y)}_{\text{definition of $f$}} \underbrace{= (z , w)}_{\text{what we want}}.$$
Can we find $x$ and $y$ so that $x + 3 = z$ and $4 - y = w$?
