1

Let $M$ be a connected smooth manifold, $\dim M \ge 2$. Prove：any smooth map $f:M \rightarrow \mathbb{R}$ can't be one-to-one.

Martin Sleziak
  • 53,687
Edward Grammer
  • 67
  • Hint: Look at the differential of your map. – Jacky Chong Apr 14 '17 at 05:55
  • @Jacky: well, what special property does the differential of a one-to-one map have? – Georges Elencwajg Apr 14 '17 at 06:02
  • @GeorgesElencwajg Is it not true that if you have a smooth injective map then there exists a point in $M$ such that the differential is not zero? Then use inverse function theorem (implicit function theorem)? – Jacky Chong Apr 14 '17 at 06:20
  • Dear @Jacky: yes, it is true that there is a point where where the differential is non zero but that does not allow you to use the inverse function theory.See Pedro's answer for a correct proof. – Georges Elencwajg Apr 14 '17 at 06:28

3 Answers3

2

The rank of $f$ at every point is $\leq 1$. Consider the subset of points where the rank is exactly $1$. This is an open subset (since rank is upper semicontinuous or whatever). Certainly it can't be empty or else you'd have a constant map. So there's a non-empty open subset where $f$ has constant rank $1$. Then the constant rank theorem makes it clear that $f$ can't be injective.

Pedro
  • 6,518
2

No continuous map from $M$ to $\mathbb{R}$ can be injective.

As $M$ has dimension at least $2$ it contains a subspace $C$ homeomorphic to a circle. Now there is no continuous injective map $g$ from $C$ to $\mathbb{R}$. If there were, then there are $a$ and $b$ on $C$ with $g(a)<g(b)$. Then on each arc of $C$ with endpoints $a$ and $b$ there is a point mapped to $\frac12(g(a)+g(b))$ (intermediate value theorem).

Angina Seng
  • 158,341
0

Here is an argument independent of the smooth structure of $M$.

Let $M$ be $n$-dimensional ($n\geq 2$) and pick any $p\in M$. If $f:M\to\mathbb{R}$ is one-to-one, then $f$ restricted on a neighborhood of $p$, is also one-to-one, and we may assume that $f$ is now defined on $\mathbb{R}^n$ by a neighborhood chart around $p$. Now by continuity of $f$, there exists some $r>0$ such that $f(B(0,r))\subset (f(0)-1,f(0)+1)$, where $B(0,r)$ denotes the open ball centered at $0$ with radius $r>0$. Since $B(0,r)$ is connected, $f(B(0,r))$ is a bounded interval. Say the two boundary points of this interval is $a$ and $b$ with $a<b$ ($a\neq b$ since $f$ is one-to-one on $B(0,r)$). Then it is not hard to show that in fact $f(\overline{B(0,r)})=[a,b]$.

Now let $p,q\in \overline{B(0,r)}$ such that $f(p)=a$ and $f(q)=b$. Denote by $l$ the compact line segment connecting $p$ and $q$, then we have $f(l)=[a,b]$. This contradicts with the fact that $f$ is one-to-one on $\overline{B(0,r)}$.

shrinklemma
  • 1,755