1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.