Implicit Function Theorem: a counter-example

Question

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?

Answer 1

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.)

Answer 2

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

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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).

Answer 3

Let $F:\mathbb{R}^{n-p}\times\mathbb{R}^p\to\mathbb{R}^{n-p}\times\mathbb{R}^p$ be a function such that $F(x_1,x_2)=(x_1,f(x_1,x_2))$.
$F$ is a function of class $C^1$ in an open set containing $a=(a_1,a_2)$.
Since $$F'(a_1,a_2)=\left( \begin{array}{c:c} I_{n-p} & O_{n-p,p} \\ \hdashline \frac{\partial(f^1,\dots,f^p)}{\partial(x^1,\dots,x^{n-p})}(a_1,a_2) & \frac{\partial(f^1,\dots,f^p)}{\partial(x^{n-p+1},\dots,x^n)}(a_1,a_2) \\ \end{array} \right)=\left( \begin{array}{c:c} I_{n-p} & O_{n-p,p} \\ \hdashline \frac{\partial(f^1,\dots,f^p)}{\partial(x^1,\dots,x^{n-p})}(a_1,a_2) & M\\ \end{array} \right)$$, $\det F'(a_1,a_2)=\det M\neq 0$.
By the inverse function theorem, there are an open set $A\subset\mathbb{R}^{n-p}\times\mathbb{R}^p$ containing $F(a_1,a_2)=(a_1,0)$ and an open set $B\times C\subset\mathbb{R}^{n-p}\times\mathbb{R}^p$ containing $(a_1,a_2)$ such that $F:B\times C\to A$ has the inverse function $h:A\to B\times C$ of class $C^1$.
Since $F(x_1,x_2)=(x_1,f(x_1,x_2))$, $h$ is of the form $h(x_1,x_2)=(x_1,k(x_1,x_2))$, where $k$ is a function of class $C^1$.
Let $\pi:\mathbb{R}^{n-p}\times\mathbb{R}^p\to\mathbb{R}^p$ be defined by $\pi(x_1,x_2)=x_2$.
Since $(x_1,x_2)\stackrel{F}{\mapsto}(x_1,f(x_1,x_2))\stackrel{\pi}{\mapsto}f(x_1,x_2)$, $\pi\circ F=f$.
For $(x_1,x_2)\in A$, $$(f\circ h)(x_1,x_2)=((\pi\circ F)\circ h)(x_1,x_2)=(\pi\circ(F\circ h))(x_1,x_2)=\pi(x_1,x_2)=x_2$$ holds.

So, I think we can correct the statement of Theorem 2-13 as follows:

Theorem 2.13: Let $f: \mathbb{R}^n \to \mathbb{R}^p$ be continuously differentiable in an open set containing $a=(a^1,a^2)$, 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 $(a^1,0)$ and a differentiable function $h: A \to \mathbb{R}^n$ with differentiable inverse such that

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