Hatcher 0.1

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


“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


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