Study Plan + Hatcher Section 0 solutions

So going into the summer, I had a pretty concise game plan of how I wanted things to work. I wanted (a) to exempt all my qualifying exams so that I’d have fewer have to’s on my agenda, and (b) to spend all the time I’d save through completing (a) split between (i) studies to help me with my required coursework next semester and (ii) studies to help me towards being able to research things I want to research.

Worst case scenario, of course, was that I didn’t exempt qualifying exams and had to incorporate preparation for those into the mix. As it stood, though, I had a clear idea of what my summer was going to consist of, academia-wise. I planned on taking the first week (i.e., last week) to casually do some independent research without a rigid schedule in place, which would turn into a regimented plan – beginning today and lasting until the beginning of fall – consisting of one hour of studying per day in each of the following fields:

  • Clifford Analysis.
  • Differential Topology/Geometry.
  • Algebraic Topology.
  • Abstract Algebra.

Among the four fields listed above, the first two were for me and me alone, whereas the last two were to help me for next semester.

It’s now 4:30am Tuesday. I didn’t spend Monday adhering to that regimen and, for all intents and purposes, I probably won’t today. There’s a long list of excuses I could give about how I had to go to the ER, about how my infant son is teething and angry, etc. etc., but the story is simply this: I made a regimen that I don’t seem to be able to stick to. Bing, bang, boom.

I’m going to do my best to get something constructive out of this summer, but I’m facing the sad reality that it probably won’t be what I’d planned. Boo hoo.

To aid my attempts, I’m going to try to put some of my solutions here as a way of forcing myself to stay dedicated. Somewhat. For now, here are some solutions for the Chapter 0 material from Hatcher.

Hatcher, Algebraic Topology, Chapter 0
1. 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.

Proof. Let T be the torus and let X=T\setminus\{\text{pt}\} be the space in question. By considering the square gluing diagram of the torus T sans a point, a diagrammatic representation for X can be given as shown in Figure 1 below.

Hatcher 0.1
Figure 1

Note that the gray part of Figure 1 represents the fact that the torus T is filled in and that one can visualize the deformation retract in question by grabbing hold of the hole in X and stretching it out so that the diagram in Figure 1 is hollow (i.e., not filled in, i.e. all white, etc.). Thus, the deformation retract in question amounts to the projection of the interior of a square (the square representing the gluing diagram of T) onto its boundary, the formula for which can be attained by presupposing that the gluing diagram for T is placed in \mathbb{R}^2 as J\times J where J=[-1,1]. Basic arithmetic shows that, for t\in I, the family f_t:I\times I\to\mathbb{R}^2 given by

f_t(x,y)=(1-t)(x,y)+t\left(\dfrac{(x,y)}{\max\{|x|,|y|\}}\right)

“does the trick.” Indeed, note that f_0(x,y)=(x,y), that f_1(x,y)=(x,y)/\max\{|x|,|y|\} is an element of \partial(J\times J), and that f_t(x,y)|_{\partial(J\times J)}=(x,y) due to the fact that \max\{|x|,|y|\}=1 on \partial(J\times J). Finally, note that continuity of the family f_t is given due to the fact that f_t is the composition of continuous functions of x,y,t for all t\in I. \,\,\square

2. Construct an explicit deformation retraction of \mathbb{R}^n\setminus\{0\} onto S^{n-1}.
Proof. This problem is essentially a problem from Calculus III. Note that for a vector \mathbf{x} in \mathbb{R}^n\setminus\{0\}, the normalized vector \mathbf{x}/\|\mathbf{x}\| lies on S^{n-1}. It suffices, then, to do the normalization process in a way that’s continuous for a time parameter t\in I, and one way to accomplish this is to define a family f_t:\mathbb{R}^n\setminus\{0\}\to S^{n-1} so that, for each \mathbf{x}=(x_1,x_2,\ldots,x_n) in the domain,

f_t(x_1,x_2,\ldots,x_n)=\left(\dfrac{x_1}{t\|\mathbf{x}\|+(1-t)},\cdots,\dfrac{x_n}{t\|\mathbf{x}\|+(1-t)}\right).

As noted in problem 1, the function f_t is continuous for each t\in I. Moreover, f_0(\mathbf{x})=\mathbf{x}, f_1(\mathbf{x})=\mathbf{x}/\|\mathbf{x}\|, and f_t(\mathbf{x})|_{S^{n-1}}=\mathbf{x} due to the fact that \|\mathbf{x}\|=1 for all \mathbf{x}\in S^{n-1}. \,\,\square

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