Spivak, Calculus on Manifolds, Problem 2-37 (b) 2-37

(a) Let $f:\mathbb{R^2}\to\mathbb{R}$ be a continuously differentiable function. Show that $f$ is $not\mbox{ 1-1}$. $Hint:$  If, for example, $D_1f(x,y)\neq0$ for all $(x,y)$ in some open set $A$, then consider $g:A\to\mathbb{R^2}$ defined by $g(x,y) = (f(x,y),y)$.
(b) Generalize this result to the case of a continuously differentiable function $f:\mathbb{R^n}\to\mathbb{R^m}$ with $m<n$.

I want to solve (b) analogously to (a), so here I first present my solution to (a).
If $D_1f(x,y) = 0$ for any $(x,y)\in\mathbb{R^2}$, then $f(x_1, y) = f(x_2,y)$ for any $x_1, x_2\in\mathbb{R}$; hence, $f$ is not $\mbox{1-1}$. Otherwise, there is $(x,y)$ such that $D_1f(x,y)\neq 0$. Then $g:\mathbb{R^2}\to\mathbb{R^2}$ defined as in the hint is continuously differentiable and
$$\det g'(x,y) = \det\left(\begin{smallmatrix}D_1f(x,y)&D_2f(x,y)\\0&1\end{smallmatrix}\right) = D_1f(x,y)\neq0,$$
so it satisfies the conditions for the inverse function theorem. Thus, there are open $V\ni (x,y)$ and $W\ni g(x,y)$ such that $g:V\to W$ is $\mbox{1-1}$.  For some unequal $(u,y_1),(u,y_2)\in W$ we get unequal $(x_1,y_1),(x_2,y_2)\in V$ such that $f(x_1,y_1) = f(x_2,y_2) = u$; hence, $f$ is not $\mbox{1-1}$.
I have three ideas on how to interpret the $D_1f$ from (a) in (b):

*

*as $D_1f^1$,

*as $D_1f$,

*as $\det\left(\begin{smallmatrix}D_1f^1(x)&\dots&D_mf^1(x)\\\vdots&\ddots&\vdots \\D_1f^m(x)&\dots&D_mf^m(x)  \end{smallmatrix}\right)$;

but neither of them seem to work.

*

*By assuming $D_1f^1(x)\neq 0$ for some $x$ and defining $g:\mathbb{R^n\to\mathbb{R^n}}$ by $g(x^1,...,x^n) = (f^1(x), x^2,...,x^n)$ I can prove that $f^1$ is not $\mbox{1-1}$. Similarily, any $f^i$ is not $\mbox{1-1}$, but it doesn't show that $f$ is not $\mbox{1-1}$.


*By  assuming $D_1f(x)\neq 0$ for some $x$ and defining $g:\mathbb{R^n\to\mathbb{R^n}}$ by $g(x^1,...,x^n) = (f(x), x^{m+1},...,x^n)$ I get
$$\det g'(x) = \det\left(\begin{smallmatrix}D_1f^1(x)&\dots&D_nf^1(x)\\\vdots&\ddots&\vdots \\D_1f^m(x)&\dots&D_nf^m(x)\\O_{m\times m}&&E_{(n-m)\times (n-m)}  \end{smallmatrix}\right)  = \det\left(\begin{smallmatrix}D_1f^1(x)&\dots&D_mf^1(x)\\\vdots&\ddots&\vdots \\D_1f^m(x)&\dots&D_mf^m(x)  \end{smallmatrix}\right);$$
but I cannot show that it's not equal to $0$ at some point $x$ to use the theorem.


*I want to define $g$ as in 2. but I cannot show why $f$ is not $\mbox{1-1}$, when
$$\det\left(\begin{smallmatrix}D_1f^1(x)&\dots&D_mf^1(x)\\\vdots&\ddots&\vdots \\D_1f^m(x)&\dots&D_mf^m(x)  \end{smallmatrix}\right) = 0$$ for all $x\in\mathbb{R^n}$.
The third one seems the most promising to me but I don't how to proceed. Do you have any tips?
 A: The appropriate way to think about the generalization is as follows: If the rank of $Df(x^1,\dots,x^n)$ is $m$ on some open set, then you can renumber your variables and write them as $x=(x^1,\dots,x^m)$, $y=(x^{m+1},\dots,x^n)$ to make the identical proof work.
However, you don't quite have the tools to finish. You will need the constant rank theorem (see Rudin or Spivak's A Comprehensive Introduction to Differential Geometry, volume 1, Theorem 9 on p. 42) to proceed. If the rank is everywhere $<m$, find a point $a\in\Bbb R^n$ where it's a maximum (say rank $r<m$), and proceed apply the rank theorem on a neighborhood of $a$.
A: I used the idea number 3 (of those you mentioned) using the problem 2.15 (from Spivak's Calculus on Manifolds) and mathematical induction to prove the lemma below. It applies to functions that are not constant, since it is obvious that a constant function is not 1-1.
Lema for 2.37.b: Let a function $ f:\mathbb{R}^n \to\mathbb{R}^n $, which is not a constant function, be a continuously differentiable function, then there exists $ x_0\in \mathbb{R}^n$, such that $ \det f'(x_0) \neq 0 $.
PROOF: We will prove the theorem using mathematical induction due to $n$. For $n=1$ a function takes the form $f:\mathbb R\to\mathbb R$ and $det f'(x)=f'(x)$, so using the mean value theorem it is easy to prove the theorem. Let us assume that the theorem is true for some $n \in \mathbb N$. We will show that it is also true for $n+1$.
Therefore, let$f:\mathbb R^{n+1}\to\mathbb R^{n+1}$ satisfy the assumptions of the lemma. Since $f$ is not a constant function, therefore there exist $\bar{x}_1, \bar{x}_2 \in\mathbb R^{n+1}$ and $k\in\{1,2,\dots,n+1\}$, such that $f^k(\bar x_1) \neq f^k(\bar x_2)$ (obviously then $\bar x_1 \neq \bar x_2$). Let us further define the sequence $\left\{u_m\right\} \subset \mathbb R^{n+1}$ as follows:
$$u_0 = \bar x_1$$
$$u_1 = \left(\bar x_2^1, \bar x_1^2,\dots,\bar x_1^{n+1}\right)$$
$$u_2 = \left(\bar x_2^1, \bar x_2^2,\bar x_1^3\dots,\bar x_1^{n+1}\right)$$
$$\dots$$
$$u_{n+1} = \bar x_2$$
Then there exists $l\in\{1,2,\dots,n+1\}$, such that
$$f^k(u_{l-1})\neq f^k(u_l) \quad\quad (1)$$
Assume without loss of generality that $k< n+1$ and $l< n+1$. If we now define the function $g:\mathbb R^n \to\mathbb R^n$ as follows:
$$g(x^1,\dots,x^n)=\left(f^1(x^1,\dots,x^n,\bar x_1^{n+1}),\dots,f^n(x^1,\dots,x^n,\bar x_1^{n+1})\right)$$
then the function  $g$ is not constant, since - given (1) - we have
$g\left(u_{l-1}^1,\dots,u_{l-1}^n\right) \neq g\left(u_l^1,\dots,u_l^n\right)$.
Let us further notice that
$$D_jg^i(x^1,\dots,x^n)=D_jf^i(x^1,\dots,x^n,\bar x_1^{n+1}) \quad\quad (2)$$
for $1\leq i,j\leq n$, therefore $g$ has continuous partial derivatives $D_jg^i$, which given Theorem 2.8 means that $g$ has a continuous derivative. Thus the function $g$ satisfies the assumptions of the lemma, which, given the inductive assumption, means that there exists $x_0\in\mathbb R^n$, such that
$$\det g'(x_0)\neq 0\quad\quad(3)$$.
Let in the rest of the proof  $\bar x_0 = (x_0^1, x_0^2,\dots, x_0^n,\bar x_1^{n+1})$. We will show that $D\left(\det f'(\bar x_0)\right) \neq 0$, which will imply that the function $\det f':\mathbb R^{n+1}\to \mathbb R$ is not a constant function and is sufficient to complete the proof. According to Problem 2.15 we have
$$D\left(\det f'(\bar x_0)\right)(h) = \sum_{i=1}^{n+1} \det\begin{bmatrix}Df^1(\bar x_0)  \\
\dots \\
h_i\\
\dots\\
Df^{n+1}(\bar x_0) \end{bmatrix}$$
where $h = (h_1, \dots, h_{n+1})$ and $h_i = (h_i^1,\dots,h_i^{n+1})$ and $Df^i(\bar x_0) = \left(D_1f^i(\bar x_0),\dots,D_{n+1}f^i(\bar x_0)\right) $ for $ 1\leq i\leq n+1$. Let’s denote
$$ A_i(h) = \det\begin{bmatrix}Df^1(\bar x_0)  \\
\dots \\
h_i\\
\dots\\
Df^{n+1}(\bar x_0) \end{bmatrix},$$
then ofcourse:
$$D\left(\det f'(\bar x_0)\right)(h) = \sum_{i=1}^{n+1} A_i(h) \quad\quad (4)$$
If we now insert into equation (4)  $h = (h_1,\dots,h_{n+1})$, where $h_i$ are defined as follow:
$$h_i = \left\{\begin{matrix} Df^{i+1}(\bar x_0), & \textrm{for }i < n+1\\
(0,\dots,0,1), & \textrm{for }i = n+1\end{matrix}\right. $$
then for $i<n+1$ we receive
$$A_i(h) = \det\begin{bmatrix} Df^1(\bar x_0)\\
\dots\\
 Df^{i+1}(\bar x_0)\\
Df^{i+1}(\bar x_0)\\
\dots\\
Df^{n+1}(\bar x_0) \end{bmatrix} = 0\quad\quad (5)$$
because the matrix has 2 identical rows, while for $i=n+1$ we have
$$A_i(h)=A_{n+1}(h)=\det\begin{bmatrix} D_1f^1(\bar x_0) & \dots & D_{n}f^1(\bar x_0) & D_{n+1}f^1(\bar x_0)\\
  \vdots & &\vdots & \vdots\\
D_1f^n(\bar x_0) & \dots &  D_{n}f^n(\bar x_0)& D_{n+1}f^n(\bar x_0)\\
0 & \dots & 0 & 1\end{bmatrix}= $$
$$=(-1)^{2(n + 1)}\cdot 1 \cdot\det\begin{bmatrix} D_1f^1(\bar x_0) & \dots & D_{n}f^1(\bar x_0)\\
  \vdots & &\vdots\\
D_1f^n(\bar x_0) & \dots &  D_{n}f^n(\bar x_0)\end{bmatrix}=$$
$$\stackrel{(2)}{=}  \det\begin{bmatrix} D_1g^1(x_0) & \dots & D_{n}g^1(x_0)\\
  \vdots & &\vdots\\
D_1g^n(x_0) & \dots &  D_{n}g^n(x_0)\end{bmatrix}\stackrel{(3)}{\neq} 0$$
Finally, given the above, (4) and (5) yield
$$ D\left(\det f'(\bar x_0)\right)(h) = \sum_{i=1}^{n+1} A_i(h) = A_{n+1}(h) \neq 0$$,
which in turn means that the differential $D\left(\det f'(\bar x_0)\right)$ is not a constant function equal to $0$, thus the function $\det f'$ is not a constant function, that is, there are $a_1, a_2 \in R^{n+1}$, such that $ \det f'(a_1)\neq\det f'(a_2)$, and this finishes the proof.
