Show that $g$ is continuous.

Let $f:\Bbb R\to \Bbb R$ be a continuous function which is strictly increasing, let $g:\Bbb R\to \Bbb R$ be a function such that $f\circ g$ is continuous. Show that $g$ is continuous.

My try:

Let $a\in \Bbb R$. To show continuity of $g$ at $a$, let $g$ be not continuous at $a$. Then for each $n\in \Bbb N$ we have some $(x_n)_n$ such that $|x_n-a|<\dfrac{1}{n}$ but $$\left|g(x_n)-g(a)\right|\ge \dfrac{1}{n}$$

• Since $f\circ g$ is continuous then for every sequence $x_n\to a$ we have $(f\circ g)(x_n)\to (f\circ g)(a)$.
• Also, since $f$ is continuous then for every sequence $x_n\to a$ we have $f(x_n)\to f (a)$.

But I am failing to understand how to jot these equations to get the required proof.

• For $f$ not strictly increasing this does not hold, so I assume that you mean strictly increasing. Under these assumptions $f$ is invertible, so you may write $$g(x)=f^{-1}(f\circ g)(x)$$ It remains to show that $f^{-1}$ is continuous, but I am not sure how to do it. For example if $f(x)=x$ then $f^{-1}(x)=\frac1x$ with $f^{-1}$ not defined for $x=0$ but continuous everywhere else. – Jimmy R. Oct 17 '16 at 13:35
Assume g is not continuous for some $x\in\mathbb{R}$ now let $x_n \rightarrow x$.
Now what happens to $f(g(x_n))$ ? keep in mind that $f$ is increasing and that $max_{k\geq n} \{|g(x_k)-g(x)|\} > \delta >0$. Look for a contradicition