Immersive Injections whose images are Embedded Submanifolds 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. 
 A: 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.
