Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Can anyone give me an hint:

If $M$ is a compact manifold of dimension $n$ and $f:M\rightarrow \mathbb{R}^n$ is $C^{\infty}$, then $f$ can not be everywhere nonsingular.

Suppose $f$ is everywhere non-singular. Then $df_m:T_m(M)\rightarrow \mathbb{R}^n_{f(m)}$ would be an isomorphism, right? But I do not understand where is the contradiction.

share|cite|improve this question
$\mathbb{R}^n$ is a manifold of dimension $n$ and the identity map is smooth. – Neal Jun 28 '12 at 21:37
Mex: I edited the ${\mathbb C}^\infty$ to $C^\infty$; the notation ${\mathbb C}$ means complex numbers. – KCd Jun 28 '12 at 21:47
Consider the canonical coordinate functions on $\mathbb{R}^n$, pulled back to $M$. $M$ is compact, so these functions must attain their maxima and minima. Therefore... – Zhen Lin Jun 28 '12 at 22:24
Just project onto the first coordinate. You then get a smooth function $M\to \mathbb{R}$. It attains its max, so now conclude something about the differential. – Matt Jun 28 '12 at 23:23
Alternately, if $f$ has no singular points, then the image of $M$ is open (by the inverse function theorem). But by compactness it is closed... – Qiaochu Yuan Jun 29 '12 at 1:58

As has been discussed in the comments, there are several approaches to this problem. Here is one of them in some detail.

Let $f : M \to \mathbb{R}^n$ be smooth, then $f = (f_1, \dots, f_n)$ where $f_i : M \to \mathbb{R}$ is the $i^{\text{th}}$ coordinate function. Let $p \in M$ and let $(U, (x^1, \dots, x^n))$ be a chart centred at $p$. Then in local coordinates, the derivative of $f$ at $p$ has standard matrix

$$\left[\begin{array}{ccc} \frac{\partial f_1}{\partial x^1}(p) & \dots & \frac{\partial f_1}{\partial x^n}(p)\\ \vdots & & \vdots\\ \frac{\partial f_n}{\partial x^1}(p) & \dots & \frac{\partial f_n}{\partial x^n}(p)\end{array}\right].$$

Note that $f_1 : M \to \mathbb{R}$ is smooth and $M$ is compact, so $f_1$ attains a maximum. If we denote by $p$ a point where $f_1$ attains its maximum, and use the coordinate system as above, we have

$$\frac{\partial f_1}{\partial x^1}(p) = \dots = \frac{\partial f_1}{\partial x^n}(p) = 0.$$

But then the standard matrix of the differential of $f$ at $p$ has a row of zeroes, and is therefore singular.

Note, we could have done this for any of the functions $f_i$. The approach would also work if $p$ were a point where one of the functions attained a minimum.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.