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This seems obvious, but I'm having trouble carrying through the details.

Suppose there is a smooth function $f$ with zero derivative on a manifold $M$ with $n$ connected components. Why is $f$ constant on each connected component?

Detailed answers are very much appreciated. Thanks!

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    @MarianoSuárez-Alvarez I know what a manifold is, and how to define functions on them. I'm a little fuzzy on derivative taking, that is all.2012-01-07

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Let $M$ be an $m$-manifold. We'll concentrate on a connected component of $M$, say $U$. Pick $p\in U$, let $V_p$ be a neighborhood of $p$ in $U$ admitting a local Euclidean chart, and let $\phi: D\subset\mathbb{R}^m\rightarrow V_p$ be a coordinate chart. If $f: M\rightarrow \mathbb{R}$ is a differentiable function on $M$, this really means that $f\circ \phi: \mathbb{R}^m\rightarrow \mathbb{R}$ is a differentiable function (in fact, this is the definition of a differentiable function on $M$). Now prove that $f\circ \phi$ is constant using standard calculus. So $f$ is constant on $V_p$. From the fact that $U$ is connected, conclude by standard topological arguments that $f$ is constant on $U$.

I would suggest picking up a book on differential geometry and topology.

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    @WNY Do not worry. I am familiar with multivariable analysis, just not so much with differential geometry.2012-01-07