about bijective function and continous functions . let $f,g:\mathbb{R} \to \mathbb{R} $ are continous functions and $f(g(x))$ is bijective .

prove $f$ and $g$ are also bijective .

i think because  $f(g(x))$ is bijective then $g$ is one to one and $f$ is surjective .
If $x \ne y$ but $g(x) = g(y)$ then $f(g(x)) = f(g(y))$; so $f \circ g$ injective $=> g$ injective.
If there is no $y$ such that $f(y) = z$ then there is no $g(x)$ such that $f(g(x)) = z$; so if $f \circ g$ is surjective, $f$ must be surjective.
so we must show $f$ is one to one and $g$ is surjective and use continuity of $f$ and $g$ .
 A: We know that $g$ is a continuous injective function. Therefore, $g$ is monotonic. Without loss of generality, take $g$ to be strictly increasing. Also, without loss of generality, take $f \circ g$ to be strictly increasing (since $f \circ g$ is also a continuous injective function).
We wish to show that the range of $g$ is $\mathbb{R}$.
To do this, we show that $\lim\limits_{x \to \pm \infty} g(x) = \pm \infty$ and apply the intermediate value theorem. I will just show that $\lim\limits_{x \to \infty} g(x) = \infty$.
In particular, note that $\lim\limits_{x \to \infty} g(x) = \sup\limits_{x \in \mathbb{R}} g(x)$, since $g$ is increasing. Thus, the limit is either finite or infinite.
Suppose the limit is finite. Then we have $\lim\limits_{x \to \infty} f(g(x)) = f(\lim\limits_{x \to \infty} g(x))$ by the continuity of $f$. But because $f \circ g$ is an increasing bijection, we must have $\lim\limits_{x \to \infty} f(g(x)) = \infty$. Contradiction. Therefore, the limit must be infinite.
Using the intermediate value theorem, we can therefore establish that the range of $g$ is $\mathbb{R}$ and therefore $g$ is a surjection, hence a bijection. Therefore, $f = (f \circ g) \circ g^{-1})$ is the composition of 2 bijections, hence a bijection.
