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If $M$ is a smooth manifold and $f$ is a smooth function on it, is it necessarily that there is an embedding $F$ of $M$ to $\mathbb R^k$ such that the first component of $F$ is $f$ ?

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Why do people vote such minimalistic and underspecified questions up? –  t.b. Nov 19 '11 at 10:34

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If you just want to look for an embedding $F$ with no restriction on $k$, then the answer is yes. Actually we can set $k=2m+1$ where $m$ is the dimension of the manifold $M$. By Whitney embedding theorem (see Whitney embedding theorem, there exists an embedding $G$ from $M$ to $\mathbb{R}^{2m}$. Now take $F$ to be $F=(f,G)$.

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In order to make $F$ an embedding, we should choose $G$ to be proper. –  Hezudao Nov 19 '11 at 11:23

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