Let $M$ be an $n$-dimensional manifold. I am proving the statement "If $f:M\to\mathbb{R}^n$ is smooth and $M$ is compact then $f$ cannot have full rank." In other words its differential $DF:T_pM\to T_{f(p)}\mathbb{R}^n$ cannot be injective for some $p\in M$. Is my proof below sound?

Suppose $f$ has full rank. Then the inverse function theorem says that $f:M\to\mathbb{R}^n$ is a local diffeomorphism. This guarantees the existence of a collection of open sets $\{U_{\alpha}\}$ in $M$ such that $f\vert_{U_{\alpha}}:U_{\alpha}\to f\vert_{U_{\alpha}}(U_{\alpha})$ is a diffeomorphism. By continuity $f(M)$ is compact in $\mathbb{R}^n$ but at the same time it is also the union $\cup_{\alpha}f\vert_{U_{\alpha}}(U_{\alpha})$ with each $f\vert_{U_{\alpha}}(U_{\alpha})$ being open. This appears to be a contradiction.

  • $\begingroup$ Check the statement you're trying to prove. $DF_p$ can be injective for some $p$; what you've shown is that there exists at least one $p$ such that $DF_p$ is not injective. $\endgroup$ – Phillip Andreae Feb 4 '17 at 14:52
  • $\begingroup$ @PhillipAndreae Thanks for pointing that out! It still seems like the proof proceeds without modification though. $\endgroup$ – user375366 Feb 5 '17 at 0:25

The general idea of your argument should work, but you need to patch a couple holes:

  • Does your "collection" really need to be a cover?
  • What is the contradiction in an open set being compact?

Here's how I would prove this, avoiding any need for $U_\alpha$:

Assume $f$ is a local diffeomorphism $M \to \mathbb R^n$. Then by the inverse function theorem, $f(M)$ is open; and by continuity and compactness, $f(M)$ is compact. Compact subsets in $\mathbb R^n$ are closed, so $f(M)$ is both open and closed; and $\mathbb R^n$ is connected, so this implies $f(M) = \mathbb R^n$. But $\mathbb R^n$ is not compact, a contradiction.

  • $\begingroup$ Doesn't this method implicitly rely on the existence of $\{U_{\alpha}\}$ in order to conclude that $f(M)$ is open? The inverse function theorem doesn't seem to be strong enough to automatically conclude that $f(M)$ is open without first having to write it as a union of open sets. $\endgroup$ – user375366 Feb 5 '17 at 1:07
  • $\begingroup$ @user375366: that's one way to make it go through, sure; but if the only thing you're doing with the cover is taking its union then I think referring to it explicitly makes things seem more complicated than they really are. For arbitrary $y = f(x) \in f(M)$ the inverse function theorem tells us there is a neighbourhood of $y$ contained in $f(M)$, so $f(M)$ contains a ball around each of its points and is thus open. $\endgroup$ – Anthony Carapetis Feb 5 '17 at 1:13

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