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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.

Please give some hints.

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  • $\begingroup$ You seem to begin a contradiction proof. But where is the contradiction ? $\endgroup$ – Peter Oct 17 '16 at 13:24
  • $\begingroup$ 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. $\endgroup$ – Jimmy R. Oct 17 '16 at 13:35
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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

Let me know if this doesn't help you either

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  • $\begingroup$ No it does not help me at all $\endgroup$ – Learnmore Oct 17 '16 at 14:45

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