Let $f:S^1\rightarrow \mathbb{R}^3$ be a smooth map with $df_x \neq 0$, for all $x \in S^1$. Show that there is a plane $C$ through the origin of $\mathbb{R}^3$ such that $p\circ f:S^1\rightarrow C$ has $d(p\circ f)_x \neq 0$, for all $x \in S^1$, where $p$ denotes orthogonal projection of $\mathbb{R}^3$ onto $C$.
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1Is this homework? What do you know and what have you tried? – Ted Shifrin Jul 15 '13 at 23:43
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Surely there exists a better title... – Mariano Suárez-Álvarez Jul 15 '13 at 23:46
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How could this be homework whenever it is summer? – dunkindonuts Jul 16 '13 at 00:02
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1@dunkindonuts It's winter in the Southern hemisphere... – 40 votes Jul 16 '13 at 00:10
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Consider the function $g$ that sends any $x\in S^1$ to the unit tangent vector to the curve $f$ at $x$ (up to notational issues, $g(x)=df_x/\Vert df_x\Vert$). This is a smooth map from $S^1$ to $S^2$. So there is a vector $v\in S^2$ such that neither $v$ nor $-v$ is in the image of $g$. Choose the plane $C$ to be perpendicular to such a $v$, so that your projection $p$ is in the direction of $v$.
Andreas Blass
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Hence my reason for asking the background of the OP; he was too busy being snide to bother answering. Your "So there is ..." is quite a leap. One needs either Sard's Theorem or a good exercise on how the image of a smooth map from $\mathbb R^k$ to $\mathbb R^\ell$ with $k<\ell$ has measure $0$. – Ted Shifrin Jul 16 '13 at 02:30
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@TedShifrin I was, indeed, making optimistic assumptions about the OP's background, not so much because I really believed those assumptions but because they made possible a rather short answer. I figured that, if the OP comes back and says he doesn't follow this short answer, then I or someone else can amplify it. – Andreas Blass Jul 16 '13 at 02:51