Problem 4.23

Prove that (a) any closed path is \(\mathbb{C}\)-contractible and (b) that any two closed paths are \(\mathbb{C}\)-homotopic.

(a)

Let \(\gamma\) be a closed path in \(\mathbb{C}\) parameterized by \(\gamma (t)_1\) for \(0\leq t \leq 1\) and let \(z\in\mathbb{C}\). Define \(h:[0,1]\rightarrow\mathbb{C}\) by \(h(t,s)=\gamma (t)(1-s)+sz\). Then \[h(t,0)=\gamma (t)\]
\[h(t,1)=z\]
\[h(0,s)=\gamma (0)(1-s)+sz = \gamma (1)(1-s)+sz = h(1,s)\] because \(\gamma\) is closed and \(0 \leq t \leq 1\) so \(\gamma (0)= \gamma (1)\). Thus, by definition of homotopy, \(\gamma \sim_\mathbb{C} z\). And since z is an arbitrary point in \(\mathbb{C}\), \(\gamma\) is \(\mathbb{C}\) contractible by definition.

(b)

Let \(\gamma_0\) and \(\gamma_1\) be closed piecewise smooth paths parameterized by \(\gamma_0 (t)\) and \(\gamma_1 (t)\) for \(0\leq t \leq 1\). Define \(h:[0,1]\rightarrow\mathbb{C}\) by \(h(t,s)=\gamma_0 (t)(1-s)+s\gamma_1 (t)\). Then \[h(t,0)=\gamma_0 (t)\]
\[h(t,1)=\gamma_1 (t)\]
\[h(0,s)=\gamma_0 (0)(1-s)+s\gamma_1 (0) = \gamma_0 (1)(1-s)+s\gamma_1 (1) = h(1,s)\] because \(\gamma_0\) and \(\gamma_1\) are closed and \(0 \leq t \leq 1\) so \(\gamma_0 (0)= \gamma_0 (1)\) and \(\gamma_1 (0)= \gamma_1 (1)\). Thus \(\gamma_0 \sim_\mathbb{C} \gamma_1\) by definition.

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