If $f: M \longrightarrow N$ is a differentiable function between manifolds and $A$ is a submanifold of $M$, can I conclude that $f_{|_A}$ is a differentiable function? It seems that the answer should be "Yes." (we are restricting a differentiable function to the good sets for that property). The statement is clearly true in the case $A$ is open (the charts of $A$ are the charts of $M$ restricted to A), but how about in the general case?
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Submanifold in what sense? An embedded submanifold, or an immersed one? – Nov 25 '14 at 00:12
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$A$ is a manifold with the induced topology of $M$. – ted Nov 25 '14 at 00:14
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1Let $i:A\hookrightarrow M$ denote the inclusion. Then the restriction of $f$ to $A$ is just $f\circ i$. But by definition of either an immersed or embedded submanifold, $i$ is smooth. As $f:M\to N$ was smooth, the composition is then smooth. – Matt Nov 25 '14 at 00:29
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@MikeMiller I guess this depends on your definition of immersed submanifold? My default definition is to say that $A$ is an immersed submanifold of $M$ if $A$ is a topological manifold (of possibly a different topology than the subspace topology) and is endowed with a smooth structure so that $i:A\hookrightarrow M$ is a smooth immersion. – Matt Nov 25 '14 at 00:42
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@MattRobinson I had confused the issue (thinking of restricting the codomain rather than the domain). I deleted my previous comment when I realized what I'd done :) – Nov 25 '14 at 00:44
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Maybe I'm confused, but if the definition of submanifold is just a subset which is a manifold with the induced topology, then is not so trivial that the inclusion map between the submanifold and the manifold is differentiable. – ted Nov 26 '14 at 17:51