# Map of smooth manifolds

Let $M$ and $N$ be smooth, connected $n$-dimensional manifolds. Let $M$ be compact and non-empty.

Show that every embedding $f: M \to N$ is a diffeomorphism.

So because $f$ is a embedding we have that $f(M) \subset N$ is a submanifold and especially that $f: M \to f(M)$ is a diffeomorphism. My idea would be to show that $f(M) = N$ and that's somehow related to $M$ being compact but I don't really know how.

• Since $M$ is compact, what do you know about $f(M)$? – Daniel Fischer Jun 30 '15 at 20:53
• $f(M)$ is also compact – Cosmare Jun 30 '15 at 20:54
• Right. And as a compact subset of a Hausdorff space, it is ...? – Daniel Fischer Jun 30 '15 at 20:57
• Yes. And an embedding of a manifold into another manifold of the same dimension is what kind of mapping? – Daniel Fischer Jun 30 '15 at 21:00
• It's good to know that, but here we can get by with the plain old inverse function theorem. The differential (derivative) is by assumption invertible at all points of $M$, so $f$ is a local diffeomorphism (hence open). Now, $f(M)$ is an open and closed subset of $N$, ... – Daniel Fischer Jun 30 '15 at 21:50