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Let $M,N$ be smooth manifolds (in the sense that all of their charts have partial derivatives of all orders). Let $F$ be a map from $M$ to $N$. Suppose that for any smooth $f: N\to \mathbb{R}$, $f\circ F$ is a smooth function. Then I would like to show, without using the inverse function theorem, why $F$ must be a smooth map between manifolds.

Any suggestions would be appreciated.

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It seems to come down to playing with charts $Phi$ for some p in M and $\Psi$ for F(p) in M, to show that $\Phi^{-1}oFo\Psi$ is a smooth map between R^N and R^M (the respective dimensions of N,M) – AQP Mar 4 '12 at 21:47
Yes, I see a way to do it like that using the inverse function theorem, but is it needed? – user21725 Mar 4 '12 at 21:48
What is the def. of smoothness of a map you're using? The one I know is that the composition of charts like in the formula is smooth. – AQP Mar 4 '12 at 21:50
An algebraic geometer would say that this is the definition of smooth. (Ignore this, it is irrelevant. Just thought it is interesting how differing branches of math take different points of view). – Matt Mar 4 '12 at 21:52
You'll probably need some partition of unity argument to show that the coordinate functions for a chart on $N$ can be extended in some suitable sense to all of $N$. – Dylan Moreland Mar 4 '12 at 21:55
up vote 1 down vote accepted

suggestion (requires: composition of smooth maps is smooth):

Suppose that for all $f:N\mathbb \rightarrow R$, $f\circ F$ is smooth.

Let $y:V\rightarrow \mathbb R^n$ be a chart for $N$. Then $y^i\circ F$ is smooth for all $i$ $\Rightarrow$ $y\circ F$ is smooth

Let $x:U\rightarrow \mathbb R^m$ be a chart for $M$, $U\subset F^{-1}(V)$. Then the map $(y\circ F) \circ x^{-1}$ is smooth.

EDIT: there is a glitch (see comment above) - $y^i$ is not defined on $N$, so the argument does not work! I think there is no way to avoid bump functions.

First one has to show: For all $V\subset N$ open and smooth $f:V \rightarrow \mathbb R$ the map $ f \circ (F|_{F^{-1}(V)}) $ is smooth. If $p\in V$ is a point, choose a smooth bump function $u\,$ s.t. $\text{supp}(u) \subset V$ and $u=1$ in an open neighbourhood of $p$, then $uf$ extends to a smooth function on $N$ and so on.

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This is a nice, simple argument. Thanks. – user21725 Mar 5 '12 at 2:10

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