I'm a physicist so forgive me for any sloppy jargon. I found that the homotopy group of maps from the 2-torus $T^2$ to the 2-sphere $S^2$ is $\mathbb{Z}$. Could someone please give me an explicit example of a map that has non-zero degree?
I know the question has been asked before here but, as a physicist, I don't find any of the answers useful. I would like an explicit smooth map from some torus coordinates to some sphere coordinates.
Sure. Let's say that the torus is the square $H = [-1, 1] \times [-1, 1]$ with its two vertical sides identified, and two horizontal sides identified, and the unit sphere is the unit disk $K$ in the plane with its boundary circle identified to a point.
I'm going to describe a map $F$ from $H$ to $K$ that respects these identifications. Fortunately for me, it turns out that $K$ is a subset of $H$.
Let $$ F(p) = \begin{cases} p & p \in K \\ \frac{p}{\|p\|} & \text{otherwise} \end{cases} $$
Clearly all points of the boundary of $H$ are mapped to the boundary of $K$, and in the quotient, these all become, say, the North pole, so everything's nice and continuous.
Visually, you can think of it like this: take your donut and draw a circle around the center hole, and then a 'perpendicular' circle around the 'handle' of the donut. Those two are the images of the sides of $H$ under the identification. The point where they cross is the image of all four corners under the identification.
If you shrink the center circle to a point, you get something that looks like a donut whose center didn't quite get punched all the way through before frying. If the then shrink the "handle" curve to a point, you get ... something that's topologically a sphere. You can think of the map from points on the donut to points on this 'sphere' as a kind of visual representation of the map $F$ I defined.
In coordinates, $(u, v)$ (I'm using these in place of $\phi_1, \phi_2$) will map to $(t, p)$ (in place of theta and phi). For me, $t$ runs from $-\pi$ to $\pi$ (longitude) and $p$ runs from $0$ to $\pi$ (latitude, offset by 90 degrees, so $p = 0$ at the south pole and $\pi$ at the north pole. Clear so far?
Then $$ t(u, v) = \begin{cases} 0 & (u, v) = (0, 0) \\ \operatorname{atan2}(u, v) & \text{otherwise} \end{cases} $$ where atan2 is the computer-scientist's version of arctan. Roughly, atan2(x, y) = arctan(x/y), except that it's well defined for all $x, y$.
What about $p(u, v)$? That's $$ p(u, v) = \begin{cases} \pi & \sqrt{u^2 + v^2} \ge \pi \\ \sqrt{u^2 + v^2} & \text{otherwise} \end{cases} $$
This function is probably not smooth, but it IS continuous, which is all that's required for homotopy computations. It's also, in your coordinates, exactly the map I described with the identification of the square: $p$ is the distance from the center of the square (clamped at $\pi$), and $t$ is the angular coordinate of a point in the square in planar polar coordinates.
If you really need the function to be smooth, I suppose you could convolve with some appropriate kernel, but...life's too short for me to go through that, esp. since it wasn't part of the question you posed.
