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How to show that there are 2 points of $S^n$ where a smooth map $g: S^n \to\mathbb R$ has tangent map equal to zero?

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    Use compactness of $S^n$.2012-02-19

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$S^n$ is compact, so on a purely topological level, $g$ (as a continuous function on it) must have a maximum and a minimum. Now if $g$ is also smooth, these must be critical values, by the same argument as in elementary calculus. Since the range of $g$ is 1-dimensional, anywhere $dg_x$ does not have full rank it must be zero.

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    (cont'd) What this means for you is that the first move is to observe: $S^n$ is a compact topological space, so any real-valued function $g$ attains a maximum and a minimum, say at $p,q\in S^n$. *Now put local coordinates around $p$.* You now have a function $\mathbb{R}^n\rightarrow \mathbb{R}$ *that has a maximum at $p$*. (More precisely, at $p$'s image in the local coordinates.) The problem has turned into the exercise in multivariable calculus that Leandro mentioned: prove a function $\mathbb{R}^n\rightarrow \mathbb{R}$ has vanishing derivative at a maximum point. Then do the same for $q$.2012-02-21