Lifting maps of Sphere to the Torus

Question

If on the square $S=\left[-\frac{1}{2},\frac{1}{2}\right]\times\left[-\frac{1}{2},\frac{1}{2}\right]$ we identify the points

1. $\left(-\frac{1}{2},y\right)\sim\left(\frac{1}{2},y\right)$ for every $y\in\left[-\frac{1}{2},\frac{1}{2}\right]$
2. $\left(x,-\frac{1}{2}\right)\sim\left(x,\frac{1}{2}\right)$ for every $x\in\left[-\frac{1}{2},\frac{1}{2}\right]$,

then we obtain the two-dimensional Torus $\mathbb{T}^2$. If, in addition, we identify:

3. $(x,y)\sim(-x,-y)$ for every $(x,y)\in S$,

then we obtain the two-dimensional Sphere $\mathbb{S}^2$. Let $q\colon\mathbb{T}^2\to\mathbb{S}^2$ be the respective projection.

If a map $f\colon\mathbb{R}^2\to\mathbb{R}^2$ that induces a map on the Torus satisfies $f(-x,-y)=-f(x,y)$ for every $(x,y)\in S$, then it also induces a map on the Sphere.

My question is: what are the self-homeomorphisms of the Sphere that can be lifted to self-homeomorphisms of the Torus? In other words, does there exist a sufficient condition that if a homeomorphism $f\colon \mathbb{S}^2\to\mathbb{S}^2$ satisfies it, then there exists a homeomorphism $g\colon \mathbb{T}^2\to\mathbb{T}^2$ satisfying $q\circ g = f\circ q$?

Answer 1

As was noted in the comments, this map $q$ is a branched double covering from the torus to the sphere. One convenient way of writing it down concretely is to regard $\mathbb{T}^2$ as the quotient of $\left[-\frac 12,\frac 12\right]$ described in the question and consider a slightly modified version of polar coordinates: $$ q(x,y)=\begin{pmatrix} |\sin(2\pi y)|\cos(2\pi x) \\ \sin(2\pi y)\sin(2\pi x) \\ \cos(2\pi y) \end{pmatrix} $$ This map has the property that $q(x,y)=q(x,y+1)=q(x+1,y)$, (i.e. it descends to a continuous map $\mathbb{T}^2\to\mathbb{S}^2$), and $q(x,y)=q(-x,-y)$ (i.e. it is equivalent to the quotient given in the problem).

Regarding $q$ as a map from $\mathbb{T}^2\to\mathbb{S}^2$, we have that $q$ is a quotient which maps the two circles $y=[0]$ and $y=[1/2]$ to the north and south pole respectively, and is a double covering everywhere else. I'll refer to these circles as $C_0,C_{1/2}\subset\mathbb{T}^2$ and to the two poles as $n,s\in S^2$.

If a homeomorphism $f:\mathbb{S}^2\to\mathbb{S}^2$ can be lifted up $q$ to a homeomorphism $g:\mathbb{T}^2\to\mathbb{T}^2$, then $g$ must be fiber preserving, and so $g$ must must either fix or swap the circles $C_0$, $C_{1/2}$, and thus $f$ must either fix or swap $n$ and $s$. Since the map $(x,y,z)\mapsto(x,y,-z)$ can be lifted to the map $([x],[y])\mapsto([x],[1/2-y])$, we can consider only the case $f(n)=n$, $f(s)=s$ without loss of generality.

If $f$ fixes $n$ and $s$, we can always lift the restriction $\bar f:\mathbb{S}^2\setminus\{n,s\}\to\mathbb{S}^2\setminus\{n,s\}$ to a homeomorphism $\bar{g}:\mathbb{T}^2\setminus(C_0\cup C_{1/2})\to\mathbb{T}^2\setminus(C_0\cup C_{1/2})$ and it only remains to determine if $\bar{g}$ extends to a homeomorphism $g:\mathbb{T}^2\to\mathbb{T}^2$. This will happen only if $f$ is "sufficiently well behaved" around $n$ and $s$. Below is one way to make this precise:

Let $\psi:\mathbb{S}^1\times(-1,1)\to\mathbb{S}^2\setminus\{n,s\}$ be the homeomorphism $\psi((x,y),z)=(\sqrt{1-z^2}\ x, \sqrt{1-z^2}\ y, z)$, let $r_y:\mathbb{S}^1\to\mathbb{S}^1$ be the reflection about the $y$-axis, and let $r_z:\mathbb{S}^2\to\mathbb{S}^2$ be the reflection about the $z$-axis. A homeomorphism $f:\mathbb{S}^2\to\mathbb{S}^2$ can be lifted up $q$ to a homeomorphism of $\mathbb{T}^2$ if and only if the following hold:

1. $f(\{n,s\})=\{n,s\}$. If $f(n)=s$ replace $f$ with $f\circ r_z$ in the subsequent criteria.
2. The map $\psi^{-1}\circ f\circ\psi$ extends to a homeomorphism $h:\mathbb{S}^1\times[-1,1]\to\mathbb{S}^1\times[-1,1]$. Note that by fixing the second argument $h$ must restrict to two homeomorphisms $h_{1},h_{-1}:\mathbb{S}^1\to\mathbb{S}^1$.
3. $h_1$ and $h_{-1}$ both commute with $r_y$.

There are many functions which fit these criteria (e.g. all maps which are the identity on a neighborhood of $n$ and $s$, a finite collection of rotations). If $f$ is differentiable/smooth at $n$ and $s$, you can probably phrase the above conditions in terms of those derivatives, at least in some cases.