# Smooth mapping $v \colon [0,1] \to S^{n-1}$

I have a smooth mapping $v \colon [0,1] \to S^{n-1}$ such that for any $u \in S^{n-1}$ exists $t \in [0,1]: v(t)\cdot u = 0$ and $n \geq 3$. So a have an assumption that such a mapping $v(\cdot)$ doesn't exist.

I tried to consider a function $f(t,s) = v(t)\cdot v(s)$, then I selected a function $s(t)$: $v(t)\cdot v(s(t)) = 0$ for any $t$ and I tried to show that $s(t)$ can be choosen to be contunious. If it is true, then a mapping $s(t)$ has a fixed point and then $|v(t)|^2=0$ for some $t$. Maybe contunious $s(t)$ doesn't exists at all.

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What does this mean: "So a have an assumption that such a mapping $v(\cdot)$ doesn't exist"? –  Paul Nov 24 '11 at 11:15
I think that function $v(\cdot)$ with specified properties doesn't exist. –  Nimza Nov 24 '11 at 11:26
So your question is this: Prove that there does not exist a smooth mapping $v:[0,1]\rightarrow\mathbb{S}^{n-1}$ where $n\geq 3$ with the following property: for any $u\in\mathbb{S}^{n-1}$ there exists $t\in[0,1]$ such that $v(t)\cdot u=0$. Moreover, $v(t)\cdot u=0$ means the dot product in $\mathbb{R}^n$ by considering $\mathbb{S}^{n-1}\subset\mathbb{R}^n$. Am I correct? –  Paul Nov 24 '11 at 11:30
You are correct :) –  Nimza Nov 24 '11 at 11:37
For $n=3$, any smooth parametrisation of the equator is such a mapping. –  PseudoNeo Nov 24 '11 at 11:57