Hatcher Chapter $0$
Construct an explicit deformation retraction of the torus with one point deleted onto a graph consisting of two circles intersecting in a point, namely, longitude and meridian circles of the torus.
I read a solution online that starts with the unit square with a hole in the centre and then it gives the following deformation retraction:
But I don't know how they came up with this formula. I realize this is a major issue for me. I am completely lost when it comes to writing this explicitly.
Even now that I see the solution, I don't know why it makes sense. I see the 't' and I know they're using parametric equations but it seems like my own calculus training is failing me because I wasn't able to think of this when confronted with the question.
My doubt (as best as I can describe it): Where is the 'hole' represented in this formula? By my understanding, the torus is punctured somewhere (since one point is missing) in Hatcher's question. But this formula is just showing movement directly from the centre of the rectangle, no? Where does it account for the puncture?
Also, why are we dividing by $\textrm{max}\{|x|, |y|\}$?
Reference request: Is there some text I can use to brush up on my parametric equations skill for this kind of application? I feel like the books I used basically tell you how to represent a line and not much else.
The hole is at $(0, 0)$. Note that $(0, 0)$ can't be in the domain because $(0, 0) / \max(0, 0)$ is undefined.
This formula is moving directly away from $(0, 0)$. Notice that $f_t(x, y)$ is always a multiple $(x, y)$.
You divide by $\max\{|x|, |y|\}$ so that $f_1(x, y)$ lies on an edge of the square. (These edges will become circles in the quotient space.)