Construction of a special homotopy on $S^3$ Here $S^3$ denotes the real 4-dimensional unit sphere. Denote by $A: S^3 \rightarrow S^3$ the antipodal map $x \mapsto -x$ and $I$ the identity map. Since the antipodal map has degree $(-1)^{3+1}=1$, we see that $A$ is homotopic to $I$. We can construct the homotopy map as follows: 
Regard a point $x \in S^3$ as a 2-tuple $(z_1, z_2)$ of complex numbers satisfying $|z_1|^2 + |z_2|^2 =1$,  and define the homotopy map by $F: (z_1, z_2, t) \mapsto (z_{1}e^{i\pi t},z_{2}e^{i\pi t} )$, we see that $F=I$ when $t=0$, $F=A$ when $t=1$ and $F$ is continuous. Moreover, we can let $F$ be of the following form: $F: (z_1, z_2, t) \mapsto (z_{1}e^{i f_1(t)},z_{2}e^{i f_2(t)} )$, where $f_i(0)=0, f_i(1)=\pi$ are continuous real functions, $i=1, 2$.
However, they are all the possible forms of $F$ I have found out. It seems that $F$ must be of the form  $(z_1, z_2, t) \mapsto (F_1 (z_{1}, t),F_2 (z_{2}, t) )$, just like "two suitable rotations acting independently".
My question is as follows:
Is it true that homotopy between $A$ and $I$ must be of the form  $F: (z_1, z_2, t) \mapsto (F_1 (z_{1}, t),F_2 (z_{2}, t) )$? For example, can we construct a $F$ with more variables changing together $F: (z_1, z_2, t) \mapsto (F_1 (z_{1}, z_2,  t),F_2 (z_1, z_{2}, t) )$?
Please give a construction if there exists some $F$ in a different form. Or give a proof if there is no other form.
Any hint would be greatly appreciated!
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EDIT: Since the antipodal map $A: S^n \rightarrow S^n$ has degree $(-1)^{n+1}$, we see that $A$ is not homotopic to $I$ when $n$ is even. So if your construction on $S^3$ can be generalized to $S^4$ or $S^{2n}$ for some $n$, then perhaps there are some mistakes in your construction.
 A: No, consider any homotopy $H:[0,1]\times S^3\rightarrow S^3$ beginning and starting with the identity and let $\tilde F=H\circ F$.
Explicit maps:
Let $f:R^3\rightarrow R^3$ be any map that is compactly supported. Let $H(t,x)=x+t(t-1)f(x)$. This is a non-trivial homotopy from the identity to the identity on $R^3$, that doesn't do anything outside a compact set. By stereographic projection this defines also a homotopy on $S^3$ from the identity to the identity.
Another way of producing non-trivial homotopies is using the flows of vector fields on $S^3$ 
A: It is not true. Let $q \in S^3$. There are at least three obvious ways to interpret $q$:


*

*$q = (x_1,x_2,x_3,x_4)$ with $x_i \in \mathbb R$.

*$q = (z_1,z_2)$ with $z_i \in \mathbb C$.

*$q \in \mathbb R^4$.
These interpretations yield three questions:


*

*Is it true that homotopy between $A$ and $I$ must be of the form  $F: (x_1, x_2,x_3,x_4,t) \mapsto (F_1 (x_1, t),F_2(x_2, t),F_3 (x_3 t), F_4 (x_4, t))$? 

*Is it true that homotopy between $A$ and $I$ must be of the form  $F: (z_1, z_2,t) \mapsto (F_1 (z_1, t),F_2(z_2, t))$?

*Is it true that homotopy between $A$ and $I$ must be of the form  $F: (q,t) \mapsto F (q, t)$?
It is obvious that the answer to 3. is "yes", but there is no reason to believe that 2. is true. If it were, why shouldn't also 1. be true?
However, let us argue a little more formally. Identify $\mathbb R^4$ with the quaternions $\mathbb H$. Define $\phi: I \to \mathbb H, \phi(t) = (1-2t) + 4t(1-t)(i+j+k)$. We clearly have $\phi(t) \ne 0$ for all $t$, thus $\psi(t) = \phi(t)/\lVert \phi(t)\rVert \in S^3$ is well-defined and $\psi(0) =1, \psi(1) = -1$. Now
let $F(q,t) = q \cdot \psi(t)$ (quaternion multiplication). This is a homotopy from $I$ to $A$. The real part of $F(x_1 + x_2 i + x_3 j + x_4 k,t)$ is
$$\dfrac{(1-2t)x_1 - 4t(1-t)(x_2+x_3+x_4)}{\lVert \phi(t)\rVert}$$
thus the "complex" part  (= real part + $i$-part) does certainly not have the form $F_1(z_1,t) = F_1(x_1,x_2,t)$.
