# A $C^\infty$ map $M^n \to \mathbb R^n$ must have singular points if $M$ is compact

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.

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$\mathbb{R}^n$ 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.

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