# Why isn't there a continuously differentiable injection into a lower dimensional space?

How to show that a continuously differentiable function $f:\mathbb{R}^{n}\to \mathbb{R}^m$ can't be a 1-1 when $n>m$? This is an exercise in Spivak's "Calculus on manifolds".

I can solve the problem in the case $m=1$. To see this, note that the result is obvious if the first partial derivative $D_1 f(x)=0$ for all $x$ as then $f$ will be independent of the first variable. Otherwise there exists $a\in \mathbb{R}^n$ s.t. $D_1 f(a)\not=0$. Put $g:A\to \mathbb{R}^n, g(x)=(f(x),x_2, \ldots, x_n)$ (with $x=(x_1,\ldots, x_n)$). Now the Jacobian $$g'(x) = \left[ \begin{array}{cc} D_1 f(x) & 0 \\ 0 & I_{n-1} \end{array} \right]$$ so that $\text{det}\, g'(a)=D_1 f(a)\not=0$. By the Inverse Function Theorem we have an open set $B\subseteq A$ s.t. $g:B\to g(B)$ is bijective with a differentiable inverse. In particular, $g(B)$ is open.

Pick any $g(b)=(f(b),b_2,\ldots, b_n)\in g(B)$. Since $g(B)$ is open there exists $\varepsilon > 0$ such that $(f(b), b_2, \ldots, b_n+\varepsilon) \in g(B)$. Thus we can find $b'\in B$ s.t. $g(b')=(f(b), b_2, \ldots, b_n+\varepsilon)$. By the definition of $g$, $f(b')=f(b)$ with $b'$ and $b$ differing in the last coordinate. Thus $f$ isn't injective.

However, I cannot generalize this argument to higher dimenssions.

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Let $k\leq m$ be the maximum rank of $f'(x)$ when $x$ runs through $\mathbb R^n$ and $a\in \mathbb R^n$ be a point where the maximum is attained: $rank(f'(a))=k$.
We will have $rank(f'(u))=k$ for all $u$ in some open neighbourhood $U$ of $a$ [obtained by choosing a non-zero $k\times k$ minor of $f'(a)$].
The constant rank theorem will then assure us that locally near $a$ the map $f$ looks like $(x_1,...,x_n) \mapsto (x_1,...,x_k,0,...,0)$ and is thus not injective since $n\gt k$ (because $n\gt m\geq k$)

Edit
Here is a 13-line proof of the constant rank theorem, written by a high-quality contributor to this site: @Pierre-Yves Gaillard

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Thx for the answer I'm not really familiar with the constant rank theorem. The book doesn't mention it either so I wait if someone comes up with a more elementary solution. –  KPK Jun 17 '12 at 15:42
Dear KPK, I'll be very happy if someone else gives a solution more to your taste. But I suspect that it would amount to reproving the constant rank theorem in a special case. By the way, that's exactly what you did for $m=1$! Indeed (in your notation) $(f\circ g^{-1})(f(x),x_2,...,x_n)=f(x)$ can be rewritten as $(f\circ g^{-1})(y_1,y_2,...,y_n)=y_1$. In other words, up to the coordinate change effected by your ingeniously constructed $g$, the map $f$ is simply projection on the first factor. –  Georges Elencwajg Jun 17 '12 at 20:32
Perhaps that's the idea of the whole exercise. In the book this problem comes right after the Inverse Function Theorem. There's an a)-part about the case $n=2$, $m=1$ (construction of $g$ is given as a hint) and b)-part asks the general result. I wonder what the precise statetements of the constant rank theorem might be and whether it would be difficult to prove in this special case? –  KPK Jun 18 '12 at 7:55
Dear KPK, I have added (in an Edit to my answer) a link to an incredibly short proof of the constant rank theorem. I think it would be a worthwhile investment to read the proof or at least to remember the precise statement. –  Georges Elencwajg Jun 18 '12 at 8:32
Thank you for the link as well as for all the other help. I red the proof altough I'm not 100 percent sure if I understood all the details on first reading; have to look that again some other day:) Anyway, I believe I can now at least answer to my original question. Somehow the idea of arguing using the rank had escaped me. –  KPK Jun 18 '12 at 21:40

Here's another answer which probably uses machinery you don't have, but has the advantage of working in the topological category.

First, recall the Borsuk-Ulam theorem:

If $g:S^n\rightarrow \mathbb{R}^n$ is continuous, then there is a point $p \in S^n$ such that $g(p) = g(-p)$.

In particular, there are no continuous injective maps from $S^n$ to $\mathbb{R}^n$.

Now, suppose $f:\mathbb{R}^n\rightarrow \mathbb{R}^m$ is continuous and injective. Let $i:\mathbb{R}^m\rightarrow \mathbb{R}^{n-1}$ be the canonical inclusion.

Then $i\circ f:\mathbb{R}^n\rightarrow \mathbb{R}^{n-1}$ is continuous. Now, restricting $i\circ f$ to $S^{n-1}\subseteq \mathbb{R}^n$ gives a map $i\circ f|_{S^{n-1}}:S^{n-1}\rightarrow \mathbb{R}^{n-1}$.

By Borsuk-Ulam, this new map cannot be injective. Since $i$ is injective, this means $f$ cannot be injective either.

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Borsuk Ulam is true for $n\ge 2$ right? –  La Belle Noiseuse Jun 21 '12 at 16:07
@Mex: It's true for $n\geq 0$. –  Jason DeVito Jun 21 '12 at 16:52
could you help me to solve this one with out borsuk ulam? show $C^{\infty}$ function from $\mathbb{R}^2$ to $\mathbb{R}$ can not be injective? –  La Belle Noiseuse Jun 21 '12 at 17:32
@Mex: In the $C^\infty$ case, I'd suggest Georges's proof above mine. –  Jason DeVito Jun 21 '12 at 19:50

I don't have a full solution for the general case, myself, but I can boil it down to a simpler problem which I feel may have a more elementary solution.

Firstly, I believe you miscalculated the top row of $g'$. In the $m=1$ case, it comes out to:

$$g'(x) = \left[ \begin{array}{c|ccc} D_1 f(x) & D_2 f(x) & \dots & D_n f(x) \\ \hline 0 & & I_{n-1} & \\ \end{array} \right]$$

The determinant still comes out to be $\det(g'(x)) = D_1 f(x)$.

In the general case, $g'$ becomes

$$g'(x) = \left[ \begin{array}{ccc|ccc} D_1 f^1(x) & \dots & D_m f^1(x) & D_{m+1} f^1(x) & \dots & D_n f^1(x) \\ \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\ D_1 f^m(x) & \dots & D_m f^m(x) & D_{m+1} f^m(x) & \dots & D_n f^m(x) \\ \hline & 0 & & & I_{n-m} & \\ \end{array} \right]$$

The determinant then becomes

$$\det(g'(x)) = \left| \begin{array}{ccc} D_1 f^1(x) & \dots & D_m f^1(x) \\ \vdots & \ddots & \vdots \\ D_1 f^m(x) & \dots & D_m f^m(x) \\ \end{array} \right|$$

As before, this determinant cannot be nonzero anywhere, or else you can apply the inverse function theorem again, to show that $f$ is not 1-1. Therefore it remains to be shown that if this determinant is zero everywhere, $f$ cannot be 1-1. It seems intuitively clear to me that that $(f^1, \dots, f^m)$ should fail to be 1-1, and therefore so will $f$, but I'm having difficulties showing it using the earlier results from Spivak.

To be clear, I claim that the general case boils down to: if $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$ is continuously differentiable and satisfies $\det(f'(x)) = 0$ for all $x$, then $f$ is not 1-1.

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If we are able to show that $f^1,...,f^m$ all fail to be 1-1, what is the argument we can use to show that $f$ is not 1-1? These component functions are continuously differentiable maps from $\Bbb R^n$ to $\Bbb R$, so we can apply the OP's work above. –  Open Season Oct 2 '14 at 21:05
Not sure. Maybe consider the "level curves" (level surfaces?) of the f^i, in some not ill-chosen point, and then show that they intersect in more than one point? (Of course, this is just a re-statement of your question.) –  aleino Oct 4 '14 at 4:50
Hmm okay. There are definitely counterexamples for $m > n$. I'll think about it. –  Open Season Oct 4 '14 at 15:14