2
$\begingroup$

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!)

$\endgroup$
4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.