This is my version of the programming exercise. I would upload it to the blog so you guys can try it out, but it is an executable file and it won’t upload for security reasons. I used VB in Visual Studio. Attached is a picture of the code and the GUI. Click on the code image, and then click on it again after it opens to see the code in readable size.
Suppose we have a compact \(K \subseteq \mathbb{C} \). We wish to show that the boundary of each path component of \(\mathbb{C} \setminus K \) is a subset of \(K\). Consider a path component \(P\), where \(\partial P \) is the collection of boundary points of \(P\). By definition, \( \forall q \in \partial P, q \notin P\). Thus \(q \in K \lor q\in L\), where \(L\) is a different path component of \(\mathbb{C} \setminus K\). Suppose \(q \in L\). We know that \( \mathbb{C} \setminus K\) is open, as its complement (\(K\)) is closed, as it is a compact subset of a metric space. Thus \(\exists \epsilon > 0 \ s.t. \ B\left(q, \epsilon\right) \subseteq \mathbb{C} \setminus K\). As \(q\) is a boundary point of \(P\), \( P \cap B\left(q, \epsilon\right) \neq \emptyset\). As \(B\left(q, \epsilon\right) \subseteq \mathbb{C} \setminus K\), there is a path between \(q\) and another point \(\hat q\), where \(\hat q \in B\left(q, \epsilon\right) \cap P\). This follows because \(B\left(q, \epsilon\right) \cap K \ = \emptyset\). As \(q\) and \(\hat q\) are path connected, they must be in the same path component by definition. This is a contradiction, so \(q \in K\). Thus, \(\forall q \in \partial P, q \in K\), so \(\partial P \subseteq K\).
Exercise 8.1:
Claim: The winding number of a concatenation of paths is equal to the sum of the winding numbers.
Proof: We could solve 8.2 first and have this as a corollary, but perhaps it is more illustrative to go by direct computation. We have
\[ \gamma (t) = \begin{cases} \gamma_1 (2t) & t \in [0, 1/2] \\ \gamma_2 (2t-1) & t \in [1/2,1] \end{cases} \]
Now, let \( g_1 \) and \( g_2 \) be lifts (based at 0) of \( \gamma_1 \) and \( \gamma_2 \). That is, \( \gamma_1 (t) = \exp (2 \pi i g_1 (t)) \) and \( \gamma_2 (t) = \exp (2 \pi i g_2 (t)) \) with \( g_1(0) = g_2(0) = 0 \). Consider
\[ g(t) = \begin{cases} g_1 (2t) & t \in [0, 1/2] \\ g_2 (2t-1) + g_1 (1) & t \in [1/2, 1] \end{cases} \]
Then, \( g\) is well-defined since \( g_1 (1) = g_2 (0) + g_1 (1) \) and is continuous by the gluing lemma of HW1. Also,
\[ \exp(2 \pi i g (t) ) = \begin{cases} \exp(2 \pi i g_1 (2t) ) & t \in [0, 1/2] \\ \exp(2 \pi i g_2 (2t -1) ) & t \in [1/2, 1] \end{cases} = \begin{cases} \gamma_1 (2t) & t \in [0, 1/2] \\ \gamma_2 (2t-1) & t \in [1/2,1] \end{cases} = \gamma (t) \]
where we have used that \( g_1 (1) \in \mathbb{Z} \). Thus, \( g \) is the unique lift of \( \gamma \) with base point 0.
Therefore, \( \text{wn}(\gamma) = g(1) = g_2 (1) + g_1(1) = \text{wn}(\gamma_1) + \text{wn}(\gamma_2). \)
Exercise 8.2:
Claim: \( \gamma_1 \gamma_2 \sim \gamma_1 * \gamma_2 \)
Proof: Let \( \Gamma : [0,1] \times [0,1] \to \mathbb{C} \setminus \left\{ 0 \right\} \) be given by
\[ \Gamma (t,s) = \begin{cases} \gamma_1 \left( \frac{2t}{1+s} \right) \gamma_2 \left( \frac{2st}{1+s} \right) & t \in [0, 1/2] \\ \gamma_1 \left( \frac{2st+1-s}{1+s} \right) \gamma_2 \left( \frac{2t – 1 +s}{1+s} \right) & t \in [1/2, 1] \end{cases} \]
\( \Gamma \) is well-defined since at 1/2 we get \( \gamma_1 \left( \frac{1}{1+s} \right) \gamma_2 \left( \frac{s}{1+s} \right) \) from both sides. It is continuous by the gluing lemma. It is always nonzero since \( \gamma_1 \) and \( \gamma_2 \) are nonzero. Also, \( \Gamma (0,s) = \gamma_1 (0) \gamma_2 (0) = 1 \) and \( \Gamma (1,s) = \gamma_1 (1) \gamma_2 (1) = 1 \). Finally, \( \Gamma(t, 0) = \gamma_1 * \gamma_2 (t) \) and \( \Gamma (t,1) = \gamma_1 (t) \gamma_2 (t) \).
So we have constructed a based homotopy of loops from \( \gamma_1 * \gamma_2 \) to \( \gamma_1 \gamma_2 \).
