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Prove (or disprove):

Suppose that $f$ and $g$ are $1$-$1$ on $\mathbb{R}$. If $f$ and $g\circ f$ are continuous on $\mathbb{R}$, then $g$ is continuous on $\mathbb{R}$.

There are similiar questions floating around, but these all lack the "$1$-$1$" requirement; none of the counterexamples for those statements will work for this one.

This statement seems intuitively true (not that that means much), but I'm having a hard time with the proof.

Theorems from the preceding chapter, that may or may not be relevant:

Let $I$ be a nondegenerate interval and suppose that $f:I\to\mathbb{R}$ is $1$-$1$. If $f$ is continuous on $I$, then $J:=f(I)$ is an interval, $f$ is strictly monotone on $I$, and $f^{-1}$ is continuous and strictly monotone on $J$.

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Let $I$ be an open interval and $f:I\to\mathbb{R}$ be $1$-$1$ and continuous. If $b=f(a)$ for some $a\in I$ and if $f'(a)$ exists and is nonzero, then $f^{-1}$ is differentiable at $b$ and $(f^{-1})(b)=1/f'(a)$.

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    $\begingroup$ I changed the minus sign in $1-1$ to a hyphen: $1$-$1$. If it is felt that a minus sign should be used but without the spacing normally used in things like $3-5$, one can write $1{-}1$ instead of $1-1$, coded as 1{-}1. $\endgroup$ – Michael Hardy Jun 8 '13 at 22:23
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    $\begingroup$ Well, if we knew that $f^{-1}$ is continuous, then we are ready, as $g=g\circ f\circ f^{-1}$. $\endgroup$ – Berci Jun 8 '13 at 22:26
  • $\begingroup$ @Berci ... Note that this only gives continuity of $g$ on the image of $f$ ... $\endgroup$ – martini Jun 8 '13 at 22:45
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    $\begingroup$ Think of $f = \arctan$, $g$ being equal to the identity on $(-\infty, 10]$ and equal to the identity ${}+ 1$ on $(10, \infty)$. Then $g \circ f = \arctan$. $\endgroup$ – martini Jun 8 '13 at 22:59
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Hint: $(g \circ f) \circ f^{-1} = g$ on $f(\mathbb R)$ implies that $g$ is continuous on $f(\mathbb R)$. Think of a non onto $f$ and $g$ which is discontinuous somewhere outside of $f(\mathbb R)$.

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  • $\begingroup$ Note that $f(\Bbb R)=\Bbb R$ since $f$ is 1-1. Can $g$ be discontinuous somewhere outside of $f(\Bbb R)$? $\endgroup$ – Paul Jun 9 '13 at 1:24

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