Section 54: Problem 5 Solution

Working problems is a crucial part of learning mathematics. No one can learn topology merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.

James R. Munkres

Consider the covering map $p\times p:\mathbb{R}\times\mathbb{R}\rightarrow S^{1}\times S^{1}$ of Example 4 of §53. Consider the path $f(t)=(\cos2\pi t,\sin2\pi t)\times(\cos4\pi t,\sin4\pi t)$ in $S^{1}\times S^{1}$ . Sketch what $f$ looks like when $S^{1}\times S^{1}$ is identified with the doughnut surface $D$ . Find a lifting $\tilde{f}$ of $f$ to $\mathbb{R}\times\mathbb{R}$ , and sketch it.

Let $p'\times p':\mathbb{R}^{2}\rightarrow D$ be the composition of $p\times p$ and $h:S^{1}\times S^{1}\rightarrow D$ , the homeomorphism described in the example, which is $h(a,b,c,d)=((1+\frac{1}{3}a)c,(1+\frac{1}{3}a)d,\frac{1}{3}b),$ then $\begin{eqnarray*} p'\times p'(x,y) & = & h\circ(p\times p)(x,y)=\\ & = & h(\cos2\pi x,\sin2\pi x,\cos2\pi y,\sin2\pi y)=\\ & = & ((1+\frac{1}{3}\cos2\pi x)\cos2\pi y,(1+\frac{1}{3}\cos2\pi x)\sin2\pi y,\frac{1}{3}\sin2\pi x). \end{eqnarray*}$ The path $f$ in $D$ is the path $\begin{eqnarray*} f'(t) & = & h\circ f(t)=\\ & = & h(\cos2\pi t,\sin2\pi t,\cos4\pi t,\sin4\pi t)=\\ & = & ((1+\frac{1}{3}\cos2\pi t)\cos4\pi t,(1+\frac{1}{3}\cos2\pi t)\sin4\pi t,\frac{1}{3}\sin2\pi t), \end{eqnarray*}$ see Figure 1↓. Its lifting is

sketched below.

figure lifting_of_a_path_in_torus_to_R2.png

Figure 1 A path in the torus and its lifting.

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Section 54: Problem 5 Solution

Section 54: Problem 5 Solution