The following theorem is stated in Spivak's "Calculus on Manifolds" as a follow-up on the Implicit Function Theorem:

Theorem 2.13: Let $f: \mathbb{R}^n \to \mathbb{R}^p$ be continuously differentiable in an open set containing $a$, where $p \le n$. if $f(a) = 0$ and the $p \times n$ matrix $(D_jf_i(a))$ has rank $p$, then there is an open set $A \subset \mathbb{R}^n$ containing $a$ and a differentiable function $h: A \to \mathbb{R}^n$ with differentiable inverse such that

$$f \circ h (x^1, \dots, x^n) = (x^{n-p+1}, \dots, x^n).$$

I don't see how this can be true. For a simple counter-example, let $f(x) = \sin(x)$ with $n=p=1$. Since $f'(2\pi)=1$, the theorem should hold at $a = 2\pi$, and since $a \in A$ we get for $x = a = 2\pi$:

$$\sin(h(a)) = a = 2\pi,$$

which cannot be true for any $h$. Where is the mistake?

  • It seems that the direction of $h$ got mixed up here (or there should be $f\circ h^{-1}$ in a neighbourhood of $0$). Then your example $h$ would be $h\colon (-1,1)\to (\pi,3\pi)$, $x\mapsto \arcsin(x)+2\pi$ so that $\sin(h(x))=x$ – Hagen von Eitzen Oct 2 '12 at 19:16
  • could you please explain by a geometric view of the statement? – Wow Jan 23 '13 at 2:06
up vote 5 down vote accepted

You are correct, the Theorem as stated is false. You get the correct statement by replacing $h$ in the equation by $h^{-1}$ (and you also really want $h(a) = 0$). Then it is a consequence of the Implicit Function Theorem. (In fact, it is a more general version of the Inverse Function Theorem.)

  • could you please explain by a geometric view of the statement? – Wow Jan 23 '13 at 2:05
  • 1
    However, the error is trivial. @TaxiDriver There's one: If $f$ is smooth enough and $Df(a)$ is surjective, then there's a local right inverse $g$ of $f$ such that $f\circ g=\operatorname{id}$ locally. – Frank Science Jul 7 '13 at 5:06

You're right, the statement of the theorem is incorrect.


As the proof of the theorem shows, Spivak is talking about the function $h$ that was constructed in the proof of the Implicit Function Theorem (Theorem 2-12). So, a correct version of Theorem 2-13 can look like this:

Theorem 2-13. Let $f: \mathbb{R}^n \to \mathbb{R}^p$ be continuously differentiable in an open set containing $a$, where $p \leq n$. If $f(a) = 0$ and the $p \times n$ matrix $(D_j f^i(a))$ has rank $p$, then there is an open set $A \subset \mathbb{R}^n$ containing $a$, an open set $B \subset \mathbb{R}^n$ and a differentiable function $h : B \to A$ with differentiable inverse such that $$f \circ h(x^1,\dots,x^n) = (x^{n-p+1},\dots,x^n).$$

Also, note that $h$ and $h^{-1}$ are in fact continuously differentiable (and are $C^\infty$ if $f$ is).

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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