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Do there exist two discontinuous surjective functions $f,g: \mathbb{R} \to \mathbb{R}$ such that the only one composition of them $f(g(x))$ or $g(f(x))$ is continuous?

I tried to use $x$ for rational and $-x$ for irrational numbers, and something similar but this is failed.

Matt24519
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1 Answers1

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Let

$f(x)=\begin{cases} {1\over x}\ \text{ if }x\neq0\\0 \text{ if }x=0\end{cases}$

$g(x)=\begin{cases}{1\over x-1}\ \text{ if }x\neq1\\0\ \text{ if }x=1\end{cases}$

$g\circ f(x)=\begin{cases}0\ \text{ if }x=1\\-1\ \text{ if }x=0\\ {x\over 1-x}\ \text{ if }0\neq x\neq1\end{cases}$

$f\circ g(x)=x-1$