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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.

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    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.2011-06-05
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    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 :)2011-06-05
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    @Jason, I've considered that example actually and we decided to restrict the question to manifolds without boundary.2011-06-05
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    @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.2011-06-05
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    @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?2011-06-05
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    @User I do mean a smooth embedded submanifold.2011-06-05
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    @JS: then I don't see any problem. $F$ being injective immersion implies $F:M\to F(M)$ is a bijective immersion, hence a diffeomorphism.2011-06-05
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    @User, Are you implicitly using the fact that F(M) is an embedded submanifold. Because it's not the case that an injective immersion must be a topological embedding, as shown by the figure eight space, etc.2011-06-05
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    @JS: I use it quite explicitly. To say that $F:M\to F(M)$ is a bijective immersion I need that $F(M)$ is an immersed submanifold. Then, when we know that $F:M\to F(M)$ is a diffeomorphism, I use the fact that $F(M)$ is an embedded submanifold to conclude that $F:M\to N$ is an embedding.2011-06-05

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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.

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    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 $N$?2011-06-05
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    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.2011-06-05
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    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.2011-06-05