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Problem 2-37 on p. 39 of Spivak's Calculus on Manifolds asks

  • Let $f:\mathbb{R}^2 \to \mathbb{R}$ be a continuously differentiable function. Show that $f$ is not 1-1. (Hint: If, for example, $D_1f(x,y) \neq 0$ for all $(x,y)$ in some open set $A$, consider $g: A \to \mathbb{R}^2$ defined by $g(x,y)=(f(x,y),y)$.)
  • Generalize this result to the case of a continuously differentiable function $f:\mathbb{R}^n \to \mathbb{R}^m$ with $m<n$.

This previous question has the same question I do: how does he want us to do the second part? One response there suggests just proving the constant-rank theorem. I can't imagine that is what he has in mind, since he states and proves that theorem in the next section.

This is a suggested solution to the second part, but I think it is erroneous. (It assumes that if vector-valued $f$ is 1-1, then at some $x$ there is some component $f^i$ that is both 1-1 near $x$ and has nonzero gradient at $x$. I don't see why this would be true a priori. I'm trying to think of a counterexample when $n \leq m$.)

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Eric Auld
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    If you do go down the constant rank route, this may help http://math.stackexchange.com/a/901671/27978. ā€“ copper.hat Jan 26 '15 at 17:41
  • By the Open Domain Principle, each continuous injective map $f:Uā†’\Bbb R^n$ defined on an open subset $U\subset \Bbb R^ n$ is an open topological embedding. But this seems to be much powerful tool than you use. ā€“ Alex Ravsky Jan 27 '15 at 01:06

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