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 $f$ ?
The embedding of manifold
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differential-geometry
differential-topology
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2Why do people vote such minimalistic and underspecified questions up? – 2011-11-19
1 Answers
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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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0In order to make $F$ an embedding, we should choose $G$ to be proper. – 2011-11-19