Find a map $g: \mathbb{R}^2 \rightarrow \mathbb{R}$ to prove surjectivity for a given $f:\mathbb{R} \rightarrow \mathbb{R}^2 $

Question

When the following is given:

Let $f:\mathbb{R} \rightarrow \mathbb{R}^2 $ be given by $f(x)=(4x, -x)$ for all $x \in \mathbb{R}$

1. How to find a map $g: \mathbb{R}^2 \rightarrow \mathbb{R}$ such that $g\circ f$ is surjective and give proof that this holds.

2. Find a function $g: \mathbb{R}^2 \rightarrow \mathbb{R}$ such that $f\circ g$ is injective and give proof that g does satisfy the given condition.

What I did so far:

$f:\mathbb{R} \rightarrow \mathbb{R}^2 $ $f(x)=(4x, -x)$

$g: \mathbb{R}^2 \rightarrow \mathbb{R}$ =?

$g\circ f: \mathbb{R} \rightarrow \mathbb{R}$ so $g\circ f(x) = g(4x, -x)$ which must be surjective

If I use $g(x)= x-y =g(f(x))$ than $g\circ f(x)=4x+x=5x$ which is surjective

Can someone please tell me if this is right and help me to show how this holds for the second part? Thanks!

Answer 1

This is all well and good. There are a couple of possibilities for $g$ which might be even simpler to work with, though. For instance $g(x,y) =x/4$ or $g(x,y)=-y$, but it matters little.

As for the surjectivity, you just have to point out that for any $c\in\Bbb R$ there is an $x$ such that $$ c=g\circ f(x) =5x $$ and that $x$ is, for your $g$, given by $c/5$.