A function mapping problem

Question

If $f(x,y)$ is a continuous on the $\mathbb{R}^2$, please show that $\exists g$ where $g$ is an injective function, and

$$g:\mathbb{R} \to\mathbb{R}^2 \quad s.t. \quad f\circ g= \text{ constant}.$$

This is a test of my entrance exams. I haven't thought about it for a long time. I hope you can give me some advice.

Thank you.

Answer 1

Saying $f$ is "a continuous on the $\Bbb R^2$" is unclear. Assuming you meant $f:\Bbb R^2\to\Bbb R$ is continuous:

If you don't require $g$ to be continuous you just need to show that there exists $c$ so that $S=f^{-1}(c)$ has the same cardinality as $\Bbb R$ (because then there is an injective $g:\Bbb R\to S$.)

So if $f$ is constant on some vertical line you're done. If not: Assume wlog for convenience that $f(0,1)>0$, $f(0,0)<0$. Choose $\delta>0$ so that $f(x,1)>0$ and $f(x,0)<0$ for every $x$ with $|x|<\delta$. Now for every $x\in(-\delta,\delta)$ there exists $y\in(0,1)$ so that $f(x,y)=0$.

Answer 2

This is wrong. If you take $f = Id_{\mathbb{R^2}}$, it is easy to see that there is no such $g$.