Derivative 0 => constant function

**EinStone** · October 7th 2010, 05:15 AM

Suppose that M, N are smooth manifolds and that M is connected. Let $f: M \rightarrow N$ be differentiable so that $d_x f = 0$ for all $x \in M$ .

Show that f must be constant.

Anyone got any hint? Why do we need M to be connected?

**Defunkt** · October 7th 2010, 05:52 AM

If we don't require M to be connected, then we can let $M=[0,1] \cup [2,3] ~, ~ N = [0,1]$ and $f(x) = \left\{ \begin{array}{lr} 1 & x \in [0,1]\\ 0 & x \in [2,3] \end{array} \right.$

**EinStone** · October 7th 2010, 09:18 AM

Ok, I see. Now how to start proving this?

**HallsofIvy** · October 8th 2010, 06:15 AM

Use proof by contradiction. Suppose that $f(x)\ne f(y)$ and show that this leads to a non-zero derivative.

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