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I was reading through Guillemin and Pollack and was having trouble verifying this for myself. Given $M \subset N$ and $N \subset P$, where $M$ is a submanifold of $N$, and $N$ a submanifold of $P$, how would one show that $M$ is also a submanifold of $P$?

mm-aops
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Sky123
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    This should follow from the more general principle that a composition of embeddings is also an embedding. – Cheerful Parsnip Sep 15 '14 at 21:42
  • Is there a simple way to do it using the definition of smooth submanifolds through coordinate charts directly – Sky123 Sep 15 '14 at 21:46
  • Ah, well, you didn't use the word smooth at first! I was speaking to the topological case. But yes you can do it. The point is that $M\subset N$ looks locally like $\mathbb R^m\subset\mathbb R^n$ and $N\subset P$ looks locally like $\mathbb R^n\subset \mathbb R^p$. Now use the induced inclusion $\mathbb R^m\subset\mathbb R^p$ to see that $M\subset P$ looks locally like $\mathbb R^m\subset\mathbb R^p$. – Cheerful Parsnip Sep 15 '14 at 22:09
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    How exactly would I do that using the direct definition that if $M$ is a submanifold of $N$, then $\forall x \in M, \exists U $ open in $N$ containing $x$ and $\exists$ an $h : U \to \mathbb{R}^{n}$ with $h(U \cap \mathbb{R}^{m}) = h(U) \cap \mathbb{R}^{m}$ where we have $dim(M) = m, dim(N) = n$. Sorry, but I'm just having trouble establishing this relation between the first and third objects in the inclusion. – Sky123 Sep 16 '14 at 04:18
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    Take your $h\colon U\to\mathbb R^n$ and compose it with the inclusion $\mathbb R^n\subset \mathbb R^p$ to get a map $\hat{h}\colon U\to \mathbb R^p$. – Cheerful Parsnip Sep 16 '14 at 20:47

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