Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $M,N$ be smooth manifolds where the dimension of $M$ is less than or equal to the dimension of $N$. Suppose that $F: M \rightarrow N$ is an injective immersion and $F(M)$ is an embedded submanifold, is it the case that $F$ is a smooth embedding?

This question came up when I was studying for my differential topology final with some friends. One of my friends is convinced that this is true, I feel strongly that it's not true. But neither of us could prove or disprove it in a satisfactory manner.

share|cite|improve this question
Do you allow manifolds with boundary? If so the map from $[0,1)$ wrapping around a circle once is an injective immersion, image is an embedded submanifold, but the map isn't a smooth embedding. – Jason DeVito Jun 5 '11 at 20:23
Bijective immersion is a diffeomorphism (by the inverse function theorem). Your $F$ is a bijective immersion if we see it as a map $M\to F(M)$. Therefore $F:M\to N$ is an embedding. And I hope what I wrote is true :) – user8268 Jun 5 '11 at 20:32
@Jason, I've considered that example actually and we decided to restrict the question to manifolds without boundary. – JSchlather Jun 5 '11 at 20:37
@User8268, It's not entirely clear to me where you're using the fact that F(M) is an embedded submanifold. And how you know that the smooth structure F(M) inherits from N lines up with the smooth structure of M. – JSchlather Jun 5 '11 at 20:38
@Jacob Schlather: I use this definition: $F:M\to N$ is an embedding iff $F(M)$ is an embedded submanifold and $F:M\to F(M)$ is a diffeomorphism. Here is a possible problem: by an embedded submanifold, do you mean a smooth embedded submanifold, or just topological emb.submfd? – user8268 Jun 5 '11 at 21:04
up vote 2 down vote accepted

user8268 is right. Let me just elaborate a little bit on his argument.

When we talk about the map $F\colon M\to N$, we really mean the composition of maps $M \xrightarrow{G} F(M) \xrightarrow{\iota} N,$ where $G\colon M \to F(M)$ is just the map $F$ with its range restricted, and $\iota\colon F(M) \to N$ is inclusion.

Now, we claim that $G$ is a diffeomorphism and $\iota$ is a smooth embedding, so that $F = \iota \circ G$ is a smooth embedding.

  • Since $F\colon M\to N$ is an injective immersion, and since $G$ is continuous, we have that $G\colon M\to F(M)$ is a bijective immersion, and so $G$ is a diffeomorphism.

  • Since $F(M)$ is embedded, we have that $\iota\colon F(M) \to N$ is a smooth embedding.

Some remarks:

(1) In the case of the Figure 8, what fails is that $\iota$ is not a smooth embedding, but instead an injective immersion. Note, though, that $G$ is still a diffeomorphism in this case because that is how we define both the smooth structure and topology of the Figure 8.

(2) In the first bullet point, we do need the fact that $G\colon M \to F(M)$ is continuous. Otherwise, we cannot immediately conclude that $G$ is smooth.

For instance, if the Figure 8 were given some weird topology -- neither the topology inherited from $M$ by declaring $G$ a diffeomorphism, nor the topology inherited from $N$ as a subspace, but instead something really weird -- then the map $G$ may fail to be continuous.

So why is $G$ continuous in our case? Well, since we were given that $F\colon M\to N$ is continuous, and since $F(M)$ carries the subspace topology (by virtue of being embedded), we can conclude that the restricted map $G$ is in fact continuous.

share|cite|improve this answer
So if we have an injective immersion $f: M \rightarrow N$ that is not a smooth embedding we can conclude that $F(M)$ is not an embedded submanifold of $N$? – JSchlather Jun 5 '11 at 22:44
Yes. That's equivalent to the question you just asked. Suppose $F\colon M \to N$ is an injective immersion. We just proved that if $F(M)$ is an embedded submanifold, then $F$ is a smooth embedding. The contrapositive is that if $F$ is not a smooth embedding, then $F(M)$ is an not embedded submanifold. – Jesse Madnick Jun 5 '11 at 23:35
To clarify: the conclusion is that $F(M)$ is not an embedded submanifold (or even an immersed submanifold) when given the subspace topology. However, $F(M)$ may be an immersed submanifold when given other topologies. – Jesse Madnick Jun 5 '11 at 23:45

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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