Given two functions $f$ and $g$, which are bijective $\mathbb{R} \rightarrow \mathbb{R}$, can $h(x) = f(x)g(x)$ also be bijective on $\mathbb{R} \rightarrow \mathbb{R}$?
I can prove no such $h$ exists if $f$ and $g$ are continuous: $f(x)$ and $g(x)$ must be $0$ for the same value of $x$ (else there would be two zeroes for $h(x)$).
Then for all $y \leq x$, $f(y)$ is either always positive, or always negative, and similarly for $g$. The functions must take the oppositve sign for all $y \geq x$. Then either $f$ and $g$ always have the same sign (so $h$ is always positive) or $f$ and $g$ always have oppposite signs (so $h$ is always negative).
However, this kind of argument doesn't get me anywhere once I let $f$ and $g$ be non-continous. This feels like it should be a common question, but I couldn't find an answer written anywhere.
