# 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.

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 $t\in I$.    $\,\,\square$