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Let $f: U\subseteq \mathbb{R}^3 \to \mathbb{R}^2$ be continuously differentiable, let $a\in U$, and suppose that ${\partial f \over {\partial (x_2, x_3)}}(a)$ is non-singular (as a $2 \times 2 $ matrix). Prove that there are open subsets $V$ and $W$ of $\mathbb{R}^3$ with $a\in W$, and a $C^1$-diffeomorphism $h: V \to W$, such that $f \circ h(x) = (x_2, x_3)$ for all $x\in V$.

Hint: let $F(x) = (x_1, f(x))$ and use the inverse function theorem

Does that fact that ${\partial f \over {\partial (x_2, x_3)}}(a)$ is non-singular somehow imply that $f$ is a $C^1$-diffeomorphism$? If so how?

Why do $V$ and $W$ have to be open? I'm lost.

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The hint gives you almost everything you need to know, including the openness requirement.

The key is to look at the Jacobian of $F(x)$. First, note which block of $\frac{\partial F}{\partial \mathbf{x}}$ does $\frac{\partial f}{\partial (x_2,x_3)}$ occupies. Now, note that $\frac{\partial F_1}{\partial x_1} = 1$; this should give you a complete picture of the determinant of the jacobian of $F$, and lead you to a straightforward application of the inverse function theorem. The rest should be bookkeeping.


Consider the jacobian of $f$:

$$\frac{\partial f}{\partial x} = \begin{pmatrix} \frac{\partial f_1}{\partial x_1} & \color{red}{\frac{\partial f_1}{\partial x_2}} & \color{red}{\frac{\partial f_1}{\partial x_3}} \\ \frac{\partial f_2}{\partial x_1} & \color{red}{\frac{\partial f_2}{\partial x_2}} & \color{red}{\frac{\partial f_2}{\partial x_3}} \end{pmatrix}.$$

The red submatrix, by assumption, is nonsingular.

Now, consider the jacobian of $F(x) \stackrel{\textrm{def}}{=} \begin{pmatrix} x_1 \\ f(x) \end{pmatrix}$:

$$J_F(x) \stackrel{\textrm{def}}{=}\frac{\partial F}{\partial x} = \begin{pmatrix} 1 & 0 & 0 \\ \frac{\partial f_1}{\partial x_1} & \color{red}{\frac{\partial f_1}{\partial x_2}} & \color{red}{\frac{\partial f_1}{\partial x_3}} \\ \frac{\partial f_2}{\partial x_1} & \color{red}{\frac{\partial f_2}{\partial x_2}} & \color{red}{\frac{\partial f_2}{\partial x_3}} \end{pmatrix}.\tag{1}$$

The determinant of this is quite easily computed as

$$\begin{align*}\det \frac{\partial F}{\partial x} &= 1\cdot \det \begin{pmatrix} \color{red}{\frac{\partial f_1}{\partial x_2}} & \color{red}{\frac{\partial f_1}{\partial x_3}} \\ \color{red}{\frac{\partial f_2}{\partial x_2}} & \color{red}{\frac{\partial f_2}{\partial x_3}} \end{pmatrix} + 0\cdot \det \begin{pmatrix}\frac{\partial f_1}{\partial x_1} & \color{red}{\frac{\partial f_1}{\partial x_3}} \\ \frac{\partial f_2}{\partial x_1} & \color{red}{\frac{\partial f_2}{\partial x_3}} \end{pmatrix} + 0 \cdot \det \begin{pmatrix}\frac{\partial f_1}{\partial x_1} & \color{red}{\frac{\partial f_1}{\partial x_2}} \\ \frac{\partial f_2}{\partial x_1} & \color{red}{\frac{\partial f_2}{\partial x_2}} \end{pmatrix} \\ &= \det \begin{pmatrix} \color{red}{\frac{\partial f_1}{\partial x_2}} & \color{red}{\frac{\partial f_1}{\partial x_3}} \\ \color{red}{\frac{\partial f_2}{\partial x_2}} & \color{red}{\frac{\partial f_2}{\partial x_3}} \end{pmatrix} \\ &\neq 0. \end{align*}$$

So now we apply the inverse function theorem to $F$. Namely, since $F$ is nonsingular in the open subset $U$, $F$ has an inverse near $p$; more technically, that there is an inverse function $F^{-1}$ in a neighborhood of $p$.

We also know the the form of the jacobian of this function:

$$J_{F^{-1}}(F(p)) = \left[J_F(p)\right]^{-1},$$

i.e., the inverse of (1) evaluated at $p$.

Now, for $F$ sends $W$ to $V$, and $F^{-1}$ sends $V$ to $W$. Let $y \in V$ (note: in your post you say $x \in V$, but this is confusing because we're using $x_1,x_2,x_3$ as variables in the domain, not the codomain). Then, $$F\circ F^{-1}(y) = y.$$ Or, writing this out explicitly, we have for $g_1,g_2,g_3 : \mathbb{R}^3 \to \mathbb{R}$,

$$F^{-1} \stackrel{\textrm{def}}{=} \begin{pmatrix} g_1 \\ g_2 \\ g_3\end{pmatrix}\\ F^{-1}(y) = \begin{pmatrix} g_1(y_1,y_2,y_3) \\ g_2(y_1,y_2,y_3) \\ g_3(y_1,y_2,y_3)\end{pmatrix} \\ F(F^{-1}(y)) = \begin{pmatrix} g_1 \\ f_1(g_1,g_2,g_3) \\ f_2(g_1,g_2,g_3) \end{pmatrix} = \begin{pmatrix} y_1 \\ y_2 \\ y_3\end{pmatrix}\\ \implies \begin{align*} g_1(y_1,y_2,y_3) &= y_1, \\ f_1(g(y)) &= y_2,\\ f_2(g(y)) &= y_3.\end{align*} $$


The short version is that the non-singular submatrix allows us to define an invertible extension to $f$ by way of the inverse function theorem. This guarantees the existince of an inverse map which happens to contain exactly the property that we need. Furthermore, the inverse function theorem gives us the exact form of the jacobian of this inverse function, guaranteeing that the desired mapping is at least $C^1$ differentiable.

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  • $\begingroup$ Could you please explain further? I dont understand :/ $\endgroup$ – JohanLiebert Feb 26 '14 at 21:17
  • $\begingroup$ @JohanLiebert Done. $\endgroup$ – Emily Feb 26 '14 at 21:55

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