Can you think of a continuous transformation that maps every point on the (surface of the) unit sphere to a different point on the (surface of the) unit sphere?
More formally, we seek a continuous $f:S^2\rightarrow S^2$ such that $s\neq f(s)\neq -s \forall s\in S^2$.
(No simple rotation of the unit sphere will do, since the points on the axis of rotation are mapped to themselves.)
Is this even possible or is there some topological barrier to accomplishing this?
Background
Given some vector $u\in\mathbb{R}^3$ I want to simply derive a vector $v$ that is guaranteed to be non-parallel to $u$. At first glance this seems easy since even a random vector would suffice with probability 1 regardless of input vector $u$. Yet for all this freedom, I can't think of any continuous transformation that would work for all vectors.
$f(x) = -x$ maps every point to a different point.
However, every continuous map $f$ from $S^2$ to itself will have at least one point $x$ for which either $f(x) = x$ or $f(x) = -x$, and in particular $f(x)$ is parallel to $x$. This follows from the hairy ball theorem.