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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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    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 $N$.2012-03-04

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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.2012-03-05