Find homotopy in $S^2$ $\subset$ $R^3$ between $f_0(x,y,z)=(z,-x,-y)$ and $f_1(x,y,z)=(x,y,z)$ Find homotopy in $S^2$ $\subset$ $R^3$ between $f_0(x,y,z)=(z,-x,-y)$ and $f_1(x,y,z)=(x,y,z)$
My ideas:
Use quaternions or complex numbers to write the formula
 A: Note that $t(z,-x,-y)=(x,y,z)$ equation over $S^2$ has no solution $t\in\mathbb{R}$, because
$$x=tz$$
$$y=-tx=-t^2z$$
$$z=-ty=t^3z$$
$$x=tz=-t^2y=t^3x$$
$$y=-tx=t^3y$$
and since at least one of $x,y,z$ has to be nonzero then this implies that $t=0$ which cannot happen.
In particular $(z,-x,-y)\in S^2$ and $(x,y,z)\in S^2$ are always linearly independent. Define
$$F(t,x,y,z)=t(x,y,z)+(1-t)(z,-x,-y)$$
This already is a homotopy, except that in $\mathbb{R}^3$, not necessarily in $S^2$. So all we have to do is normalize it, if possible, i.e. if it is nonzero everywhere. But since $(z,-x,-y)$ and $(x,y,z)$ are linearly independent then
$$F(t,x,y,z)\neq 0$$
for any $t$ and $(x,y,z)\in S^2$. Therefore
$$\lVert F(t,x,y,z)\rVert\neq 0$$
and so the homotopy you are looking for is:
$$H(t,x,y,z):=\frac{1}{\lVert F(t,x,y,z)\rVert}\cdot F(t,x,y,z)$$
Note that this "normalized linear homotopy" approach works well whenever you have two continuous maps $f_0,f_1:S^n\to S^n$ such that $f_0(v)$ and $f_1(v)$ are linearly independent for all $v\in S^n$.
