Injective Map between Manifolds is Surjective

Question

Suppose $f : M \rightarrow N$ is a smooth map of compact connected manifolds of the same dimension which is an injection. How do I show that $f$ must be a surjection? Any help/hints would be appreciated.

Answer 1

Here's an outline of a proof. Since $N$ is connected, it's enough to show that the image $f(M)$ is both open and closed in it. Closedness is easy: $f(M)$ is compact. To prove openness, use invariance of domain: For any open $U\subset \mathbb{R}^n$, any injective continuous map $g:U \to \mathbb{R}^n$ is a homeomorphism onto its image $V = g(U)$, and $V$ is open in $\mathbb{R}^n$. (Smoothness isn't required for that result, and in fact the result you mention holds in just the continuous category. Also note that $g$ is a map between subspaces of $\mathbb{R}^n$; invariance of domain doesn't hold for arbitrary maps $\mathbb{R}^n \to \mathbb{R}^m$.)