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The question is clear (every composition etc. is defined). I personally think this is not true. If this were true then the proof of $f$ is continuous $\Rightarrow$ $f^{-1}$ is continuous would be much simpler by noting that $f^{-1}(f(x))=x$. If it is not true then provide a counterexample. If it is true then provide a hint to the $\epsilon-\delta$ proof.

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    $X$ is arbitrary and $f,g$ can possibly have discontinuities2012-11-13

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Take $X = (-\infty, 0] \cup \mathbb N \subset \mathbb R$ and $a = 0$. Now $g(x) = \begin{cases} x &\text{ if } x\leq 0 \\ x^{-1} &\text{otherwise}\end{cases}$ and $f(x) = \begin{cases} x &\text{ if } x \leq 0 \\ 1 &\text{otherwise}\end{cases}$ give a counterexample.

Note however if we assume that there is some $\epsilon > 0$ such that $A = (a- \epsilon, a+\epsilon) \subset X$ and $g\vert_A$ is continuous and injective, then the statement becomes true.

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    Well at least $g$ restricted to some neighborhood of $0$ does not notice :D2012-11-13