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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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    Yes, it works. Thank you.2011-11-24

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