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I am a little confused about if I am understanding the definition of an immersed submanifold correctly, and characterizing which immersed submanifolds are embedded submanifolds.

From Lee's Introduction to Smooth Manifolds, we have the following definition:

An immersed submanifold of a smooth manifold $N$ (with or without boundary) is a subset $S \subset N$ endowed with a topology (not necessarily the subspace topology) w.r.t. which it is a topological manifold (without boundary), and a smooth structure w.r.t. which the inclusion map $i: S \to N$ is a smooth immersion.

I am interested in the following Proposition 4.22 (also from Lee).

Suppose $M$ and $N$ are smooth manifolds with or without boundary, and $F: M \to N$ is an injective smooth immersion. If any of the following holds, then $F$ is a smooth embedding

  • $F$ is an open or closed map

  • $F$ is a proper map.

  • $M$ has empty boundary and $\text{dim}M = \text{dim} N$.

In the context of an immersed submanifold $S \subset N$, in above proposition, am I able to identify $S$ with $M$, and the inclusion map $i$ as our $F$?

Suppose $i$ or $S$ satisfies one of the properties above. Can I conclude $i$ is a smooth embedding, and therefore $S$ is an embedded submanifold?

Or does the above theorem actually show that $i$ is a smooth embedding, but not w.r.t. the subspace topology on $S$ inherited from $N$, so we cannot conclude (directly from this proposition) that $S$ is an embedded submanifold of $N$?

I am aware of Proposition 5.21 characterizing which immersed submanifolds are embedded submanifolds, but I would like to understand why I cannot apply the above Proposition 4.22.

(Apologies if the above questions don't make sense - I am still new to this!)

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1 Answer 1

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Since I seem to be responsible for your confusion, let me see if I can clarify.

In the context of an immersed submanifold $S \subset N$, in above proposition, am I able to identify $S$ with $M$, and the inclusion map $i$ as our $F$?

Yes.

Suppose $i$ or $S$ satisfies one of the properties above. Can I conclude $i$ is a smooth embedding, and therefore $S$ is an embedded submanifold?

Yes.

Or does the above theorem actually show that $i$ is a smooth embedding, but not w.r.t. the subspace topology on $S$ inherited from $N$, so we cannot conclude (directly from this proposition) that $S$ is an embedded submanifold of $N$?

If $i$ is a smooth embedding, then it is in particular a topological embedding, which means a homeomorphism onto its image (where the image has the subspace topology). Since $i$ is basically the identity map from $S$ to its image, this means $S$ itself has the subspace topology.

I think you actually understand what's going on quite well. You just need to get a little more practice with these concepts so you can start to feel more confident in your knowledge.

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