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$M$ is a smooth manifold and $A$ is a subset of $M$. If a function $f$ on $M$ is $C^\infty$ on $A$(that is, for every point $x\in A$, there is a open set $V_x$ and a $C^\infty$ function $f_x$ such that $f_x=f$ on $A \cap {V_x}$), then is there a $C^\infty$ function $f_1$ on $M$ such that $f_1=f$ on $A$?

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    Think about $f(x) = 1/x$ on $M = (-1, 1) \backslash \{0\}$.2012-05-01
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    If $A$ is closed, this is [Whitney's extension theorem](http://en.wikipedia.org/wiki/Whitney_extension_theorem), [original paper](http://dx.doi.org/10.2307%2F1989708)2012-05-01

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