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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.

user100101212
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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 $\mathbb{R}^n \to \mathbb{R}^m$.)

anomaly
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  • Could you elaborate on how we show that $f(M)$ is open? – user100101212 Feb 29 '20 at 22:46
  • The restriction of $f$ to sufficiently small coordinate charts has open image. – anomaly Feb 29 '20 at 22:49
  • Are we using that $M$ is homeomorphic to its image (by injectivity) and hence open? I am not understanding the coordinate charts statement. Are we restricting to a neighborhood $p \in U \subset M$ which has open image, how does that tell us $f(M)$ is open? – user100101212 Feb 29 '20 at 22:56
  • $M$ and $N$ are locally homeomorphic to $\mathbb{R}^n$, so invariance of domain applies to $f\vert U$ for $U$ a small coordinate chart. But the image of $f$ is the union of the image of all the $f\vert U$ restrictions. – anomaly Mar 01 '20 at 01:25