Let $f,g : \mathbb R \to\mathbb R$. Suppose that $f\circ g$ is a monotone growing function and $f$ is bijective.
Can I conclude from those details that $g$ is a monotone function? I've managed to show that:
$x_1 < x_2 \implies g(x_1) \neq g(x_2) $, but I don't how if I can continue from there.
No. Consider the case $f(x)=g(x)=\begin{cases}x+1&\text{if }x\in\Bbb Z\wedge 2\nmid x\\x-1&\text{if }x\in\Bbb Z\wedge 2\mid x\\ x&\text{if }x\notin\Bbb Z\end{cases}$
It holds $f(g(x))=g(g(x))=x$ (hence, $g$ is bijective), but $g$ is not monotone, because $g(2)<g(3/2)< g(1)<g(5/2)$.