Just wondering if this a valid construction to build a torus. Sorry in advance if it's completely wrong.
Consider four maps $f_n:\Bbb R^2 \to \Bbb R^2$ for $n=1,2,3,4$ with:
$$ f_1(x,y)=(-e^{-x},-e^{-y})$$
$$ f_2(x,y)=(e^x,-e^{-y}) $$
$$ f_3(x,y)=(e^x,e^y) $$
$$ f_4(x,y)=(-e^{-x},e^y) $$
The subscripts represent the quadrant that each map "acts on." So $f_3$ for example, maps points from the third quadrant to $[0,1]^2.$ So I'll say that $f_3$ "acts on" points in the third quadrant. (including points on the axes).
Then translate all the image spaces for $f_n$ to $[0,1]^2.$ The image space for $f_3$ already sits in $[0,1]^2$ so that stays where it is. What I mean by "image space" is where the points are mapped to. Note that each quadrant is accounted for by the maps so it's essentially the same as $\Bbb R^2$ except for the fact that the origins are at the corners of the unit square instead of just being at $(0,0).$
Thinking of these four maps acting on their respective quadrants separately, and then combining them all into $[0,1]^2,$ one can see that the corners of this unit square actually represent the images of the original origin(s) of $\Bbb R^2.$ To see this for $f_3,$ we have that $(0,0)\mapsto(e^0,e^0)$ which is equivalent to saying $(0,0)\mapsto(1,1).$ So $(1,1)$ is the image of the origin using $f_3$ and $(1,1)$ is the upper right corner of the unit square.
Identify the sides of the square to make a torus so that all the four corners of the square (the separate images of the origin of $\Bbb R^2$) actually become one point.
This does look like a valid, if somewhat unorthodox, construction of the torus.
If I understand what you are doing, it looks like the act of "translating the image spaces for $f_n$ to $[0,1]^2$" restricts to a 4-to-1 map from the set $\mathbb R^2 - \{\text{($x$-axis)} \cup \text{($y$-axis)}\}$ to the set $(0,1)^2$. The standard quotient map from $[0,1]^2$ to the torus is one-to-one on $(0,1)^2$. Composing these maps, one obtains a quotient map from $\mathbb R^2$ to the torus whose restriction to the set $\mathbb R^2 - \{\text{($x$-axis)} \cup \text{($y$-axis)}\}$ is 4-to-1 onto the portion of the torus covered by $(0,1)^2$.