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For a smooth manifold $M$, the locally Euclidean property goes:

Even though $M$ itself is not Euclidean space, We have that for each $p \in M$, there exists a positive integer $n_p$ such that there exists a neighborhood $U_p$ of $p$ in $M$ such that $U_p$ is diffeomorphic to an open subset $V$ of $\mathbb R^{n_p}$ (If $M$ has a dimension $n$, since in some textbooks, not all manifolds have uniform dimension, then $n=n_p = n_q$ for all $p, q \in M$).

If we view $U_p$ and $V$ as regular/embedded smooth submanifolds with codimension 0 (of their respective spaces $M$ and $\mathbb R^{n_p}$), then I think we have an analogous 'local smooth submanifold' property for an immersed submanifold $P$:

Even though $P$ itself is not a smooth regular/embedded submanifold, Let $M$ be a manifold such that $P$ is an immersed submanifold of $M$. For each $p \in P$, there exists a subset $W_p$ of $P$ that contains $p$ such that $W_p$ is a smooth regular/embedded submanifold of $M$ (I'm not too sure about the with or without dimension here for $M$ or $W_p$.).

Question: Is this property correct?

( Edit: I removed another question asking for a property where $W_p$ is an immersed submanifold of $M$ because $W_p$ is an immersed submanifold since $W_p$ is a regular/an embedded submanifold of $M$. )

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    Every immersion $F:N \to M$ (whether it is injective or not) is locally on the domain isomorphic to a slice: for each $n \in N$ there are charts around $n$ and $F(n)$ with the property that the map $F$ is, in these charts, equal to the map $(a_1,\dots,a_n) \mapsto (a_1,\dots,a_n,0,\dots,0)$. See e.g. chapter 1 of Warner's excellent book "Foundations of differentiable manifolds and Lie groups." – Stephen Jul 31 '19 at 13:16
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    @Stephen What? I know immersions are equivalent to local embeddings. My questions are about immersed submanifolds and not (directly) about immersions themselves. –  Jul 31 '19 at 13:17
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    Well, the point is that such a slice is diffeomorphic as a subset to Euclidean space (and in fact, something stronger is true: the pair $(M,F(N))$ is locally isomorphic to the pair $(\mathbf{R}^m,\mathbf{R}^n)$). I doubt you will find a cleaner proof. (In particular, $S_q$ in your post above can always be chosen in such a way that $F(S_q)$ is open in $F(N)$.) – Stephen Jul 31 '19 at 13:19
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    @Stephen Ugh...thanks I think...(Trying to understand) Do you mean your answer to the title (i'll make this into "question 0" in an edit now) is "yes", your answer to question 1 is "N/A. Here's what might be the cleanest proof", and your answer to question 2 is "I can find a property stronger than the one in title namely 'the pair...to the pair...' " ? –  Jul 31 '19 at 13:24
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    I think that's right, if I have understood your question correctly. – Stephen Jul 31 '19 at 13:25
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    Thanks @Stephen (even if I haven't understood your comment yet)! You can answer if you want –  Jul 31 '19 at 13:25
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    @Stephen FYI I edited to a better question but hopefully not a vastly different question. You can still answer, I believe. –  Aug 01 '19 at 03:26

1 Answers1

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Yes. Proof:

  1. By definition of $P$'s being an immersed smooth submanifold, there exist smooth manifolds $N$ and $M$ and a smooth map $F: N \to M$ such that $F$ is an immersion with image $F(N)=P$. (I'm not sure if $N$ and $M$ necessarily have dimensions.)

  2. Let $p \in P = F(N)$. Then there exists $q \in N$ such that $F(q)=p$.

  3. For $q$ in (2), with immersions as equivalent to local embeddings, there exists, by (1), a neighborhood $S_q$ of $q$ in $N$ such that $F|_{S_q}: S_q \to M$ is an embedding.

  4. image$(F|_{S_q}) = F|_{S_q}(S_q)=F(S_q)$, is a smooth regular/embedded submanifold of $M$, by (3).

  5. Choose $W_p = F(S_q)$, by (4).

QED

Note : I believe (3) is the key step.

  • Regular smooth submanifold = Embedded smooth submanifold.
  • Immersions are local embeddings.
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