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Spivak's Proof Not Clear. Theorem 2.13

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Hello MHB.

I am having trouble understanding the proof of theorem 2.13 given in Spivak's Calculus on Manifolds.

Theorem 2.12 (Implicit Function Theorem)
Suppose $f:\mathbb R^n\times \mathbb R^m \to \mathbb R^m$ is a continuously differentiable function in an open set containing $(a,b)$ and $f(a,b)=0$. Let $M$ be the $m\times m$ matrix $$(D_{n+j}f^i(a,b))~~~ 1\leq i,j\leq m.$$
If $\det M\neq 0$, there is an open set $A\subseteq \mathbb R^n$ containing $a$ and an open set $B\subseteq \mathbb R^m$ containing $b$, with the following property: for each $x\in A$ there is a unique $g(x)\in B$ such that $f(x,g(x))=0$.
The function $g$ is differentiable.

Proof: (Spivak) Define $F:\mathbb R^n \times \mathbb R^m\to\mathbb R^n \times \mathbb R^m$ by $F(x,y)=(x,f(x,y))$. Then $\det F'(a,b)=\det M\neq 0$.
By the inverse function theorem, there is an open set $W\subseteq \mathbb R^n\times\mathbb R^m$ containing $F(a,b)=(a,0)$ and an open set $V=A\times B\subseteq \mathbb R^n\times\mathbb R^m$ containing $(a,b)$, where $A$ and $B$ are open in $\mathbb R^n$ and $\mathbb R^m$ respectively, such that $F:A\times B\to W$ has a differentiable inverse $h:W\to A\times B$. It's easy to see that $h$ is of the form $h(x,y)=(x,k(x,y))$ for some $k:W\to B$. Since $h$ is differentiable, it follows that $k$ too is differentiable. Let $\pi:\mathbb R^n\times\mathbb R^m\to\mathbb R^m$ be defined as $\pi(x,y)=y$. Then $\pi\circ F=f$. Thus $f(x,k(x,y))=f\circ h(x,y)=\pi\circ(F\circ h)(x,y)=y$. So $f(x,k(x,0))=0$. Thus $g(x)=k(x,0)$ is the required function.


Here's the theorem I am having trouble with.

Theorem 2.13
Let $f:\mathbb R^n\to\mathbb R^p$ be continuously differentiable in an open set containing a point $c\in \mathbb R^n$, where $p\leq n$. If $f(c)=0$ and the $p\times n$ matrix $[D_jf_i(c)]$ has rank $p$, then there is an open set $C\subseteq \mathbb R^n$ which contains $c$ and a diffeomorphism $h:C\to \mathbb R^n$ such that $f\circ h(x_1,\ldots,x_n)=(x_{n-p+1},\ldots,x_n)$.

Proof: (Spivak) We can consider $f$ as a function $f:\mathbb R^{n-p}\times \mathbb R^p \to \mathbb R^p$. If $\det M\neq 0$, then $M$ is the $p\times p$ matrix $(D_{n-p+j}f_i(c))$, $1\leq i,j\leq p$, then we are precisely situation considered in the proof of the above theorem.
NOW, In that proof we showed that there is an $h$ such that $f\circ h(x_1,\ldots,x_n)=(x_{n-p+1},\ldots,x_n)$ and we are done.
(Now Spivak deals with the case when $\det M$ is not zero but I am ok with that part of gthe proof)

My problem with the proof is:

In the above proof $h$ is a diffeomorphism from $W$ (as in the proof of theorem 2.12) to $V$ (as in the proof of theorme 2.12 again). What we needed was to find a diffeomorphism $h$ from an open set $C$ containing $c$ to $\mathbb R^n$. So we need to make some adjustments. Let $\gamma:V\to\mathbb R^n$ be a diffeomorphism and define $h_1=\gamma \circ h$. Let $W$ play the role of $C$.
Now,
1. We are not sure if $C=W$ contains $c$. What $W$ contains for sure is $F(c)=(c_1,\ldots,c_{n-p},0,\ldots,0)$, where $F$ is defined in the proof of theorem 2.12.
2. Because of this new gamma we are not sure if $f\circ h_1$ satisfies $f\circ h_1(x_1,\ldots,x_n)=(x_{n-p+1},\ldots,x_n)$.
___

Can somebody please help me on this.
I know this is a long post and it must be a headache to go through all this so I thank you big time in advance for taking your time out and helping.
 
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