# Implicit Function Theorem: a counter-example

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?

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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? – La Belle Noiseuse Jan 23 '13 at 2:06

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.)
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