Suppose one has a smooth manifold with boundary M and compact on top of it. Is it true that it can always be embedded in an upper half plane such that the boundary is embedded in the hiperplane $x_n=0$? Or are there obstructions to that? If yes, what are those obstructions? Thank you!
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3$\begingroup$ Uhhh, if you Whitney-embed M into a large enough Euclidean space $R^n$, then embed that into $R^n\times R$, doesnt this do it? assuming the hyperplane is part of the "upper half space". $\endgroup$– Chris GerigFeb 8, 2012 at 23:39
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1$\begingroup$ As Chris mentions, there's no obstruction. You can embed an $n$-dimensional manifold with boundary in a Euclidean half-space of dimension $2n$ "neatly" as you describe. And embeddings are generic if the half-space has dimension $k > 2n$. $\endgroup$– Ryan BudneyFeb 8, 2012 at 23:49
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5$\begingroup$ Daniel probably wants to know that the interior can be mapped in the (strictly) upper half space. This is a consequence of the collaring theorem: the boundary of M has a neighborhood homeo (or diffeo) to $\partial M\times [0,1)$. This can extended over $M$ and used as a last coordinate. $\endgroup$– PaulFeb 9, 2012 at 0:46
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1$\begingroup$ Thank you all for the comments. That is exactly what I wanted, Paul! Shall I understand that the proof of Whitney's theorem goes thru without any headaches, other than what you just said? $\endgroup$– DanielFeb 9, 2012 at 2:41
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1$\begingroup$ That's correct. It's the same proof, just with one (small) extra layer of complexity that doesn't get in the way of any essential steps. $\endgroup$– Ryan BudneyFeb 9, 2012 at 2:44
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See Theorem 1.4.3. of
M. Hirsch: Differential Topology, Springer Verlag, 1976
answered Feb 9, 2012 at 16:53