4.23
4.23
a Prove that any closed path is \(\mathbb{C}\)-contractable.
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Assume \(\gamma\) is a closed path in \(\mathbb{C}\) which is paramaterized by \(\gamma(t)_1\) for some \(0 \leq t \leq 1\). Let \(z \in \mathbb{C}\). Define \(h:=[0,1]\rightarrow\mathbb{C}\) by \(h(t,s)=\gamma(t)(1-s)+sz\). So: \[\begin{aligned}
h(t,0)&=\gamma(t)\\
h(t,1)&=z\\
h(0,s)&=\gamma(0)(1-s)+sz\\
&=\gamma(1)(1-s)+sz\\
&=h(1,s)\end{aligned}\] Because \(\gamma\) is closed and \(0 \leq t \leq 1\), we know \(\gamma(0)=\gamma(1)\). Thus by definition of homotopy, we know \(\gamma \sim_{\mathbb{C}z}\). Now since \(z\) is arbitrary, we know for any \(z \in \mathbb{C}\) that \(\gamma\) is \(\mathbb{C}\) contractible by the definition of contractibility.
b Prove that any two closed paths are \(\mathbb{C}\)-homotopic.
Let \(\gamma_0\) and \(\gamma_1\) be closed, piecewise smooth paths. Parameterize these paths by \(t\) where \(0 \leq t \leq 1\), so we have \(\gamma_0t\) and \(\gamma_1t\) respectively.
Define \(h:=[0,1] \rightarrow \mathbb{C}\) by \(h(t,s)=\gamma_0(t)(1-s)+s\gamma_1t\). So: \[\begin{aligned}
h(t,0)&=\gamma_0(t)\\
h(t,1)&=\gamma_1(t)\\
h(0,s)&=\gamma_0(0)(1-s)+s\gamma_1(0)\\
&=\gamma_0(1)(1-s)+s\gamma_1(1)\\
&=h(1,s)\end{aligned}\] Since \(\gamma_0\) and \(\gamma_1\) are closed and \(t\) was parameterized from \(0 \leq t \leq 1\), we know \(\gamma_0(0)=\gamma_0(1)\) and \(\gamma_1(0)=\gamma_1(1)\). Therefore \(\gamma_0\sim_{\mathbb{C}}\gamma_1\) by definition of contractibility.