If $f\circ g$ is smooth and $f$ is smooth, does it follow that $g$ is smooth? Note that I cannot simply take the inverse of $f$. Do I have to use implicit function theorem?
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2If $f$ is constant, $g$ can be as wild as it wishes. If $f$ is a local diffeomorphism (or immersion), then $g$ must be smooth too. – Daniel Fischer Oct 03 '13 at 02:09
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Duh, thank you. Got too involved into messy details and lost track of simple facts... – D. Huang Oct 03 '13 at 02:11
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@DanielFischer That looks like an answer to me ... :) – Neal Oct 03 '13 at 02:12
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A composition of functions can smooth out irregular behaviour of either partner. If $f$ is constant, you cannot say anything about $g$. If $f$ is locally invertible, however (e.g. a local diffeomorphism), and $g$ is continuous, then the smoothness of the composition implies the smoothness of $g$.
Daniel Fischer
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