Example 11.3:

Let $L$ be the line $\{(0, 0, x_3)/ x_3 \in \mathbb{R}\}$ in $\mathbb{R}^3$ and $C$ be the circle

$$(x_1-1)^2 + x_2^2 = 1/4,\quad x_3 = 0.$$

We show that the torus is a deformation retract of the space $X = \mathbb{R}^3 - (L\cup C)$. The idea is simple but some details ought to be examined. Let us begin with the punctured half plane

$$H_0^{\prime} = \{(x_1, 0, x_3)/ x_1 > 0\} - \{(1, 0, 0)\}$$

which clearly deformation retracts to the circle $C_0$ given by

$$C_0: (x_1-1)^2 + x_3^2 = 1/4,\quad x_2 = 0.$$

The homotopy $F:H_0^{\prime}\times [0, 1]\longrightarrow H_0^{\prime}$ is simply given by the convex combination:

$$F({\bf x}, t) = (1-t){\bf x} + t\Big( {\bf e}_1 + \frac{{\bf x} - {\bf e}_1}{\Vert{\bf x} - {\bf e}_1 \Vert}\Big),\quad {\bf e}_1 = (1, 0, 0).$$

The idea is to rotate the picture about the $x_3$-axis. It is expedient to use spherical polar coordinates given by

$$x_1 = \rho\cos\theta\sin\phi,\; x_2 = \rho\sin\theta\sin\phi,\; x_3 = \rho\cos\phi,\quad 0 < \phi < \pi,\;\theta \in \mathbb{R}.$$

Let $H_{\theta}^{\prime}$ be the half plane bounded by the $x_3$-axis making angle $\theta$ with $H_0^{\prime}$ and $R_{\theta}$ denote the rotation about the $x_3$-axis mapping $H_{\theta}^{\prime}$ onto $H_0^{\prime}$ namely,

$$R_{\theta}(\rho\cos\theta\sin\phi, \rho\sin\theta\sin\phi,\rho\cos\phi) = (\rho\sin\phi, 0, \rho\cos\phi)$$

The homotopy we are looking for is then the map $G:X\times [0,1]\longrightarrow X$ given by

$$G({\bf x}, t) = R_{\theta}^{-1}\circ F(R_{\theta}({\bf x}), t).\eqno(11.8)$$

It is easy to see using the properties of rotations, that

(i)

$G$ is well defined, that is the image of $G$ avoids the circle $C$

(ii)

Satisfies the requisite boundary conditions at $t = 0$ and $t = 1$.

However, the continuity of $G$ is not automatic since the $\theta$ appearing in the definition of $G$ depends also on ${\bf x}$ and we know that $\theta$ cannot be defined as a continuous function of ${\bf x}$ on $X$. One can either write a formula (which is easy) and see that $\theta$ occurs in (11.8) only as $\cos\theta$ and $\sin\theta$ which are continuous functions on $X$ or better still use the property of quotients. We leave the amusing details to the reader.