# 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.