# Simple Connectedness of bounded region in $\mathbb{C}$ split by line

Note: This is one of the problems from Complex Analysis by Stein and Shakarchi.

Let $\Omega \subset \mathbb{C}$ be a bounded region (open, connected), and let $L \subset \mathbb{C}$ be a line that intersects $\Omega$, you may assume that $\Omega \cap L = I$ is an interval.

The line $L$ splits the complex plane in two halves, and since $\Omega$ is open, $\Omega$ will have a nonempty intersection with each half. Let's call these intersections $\Omega_l$ and $\Omega_r$. In particular, $\Omega$ will be the disjoint union $\Omega_l \cup I \cup \Omega_r$.

To prove: If $\Omega_l$ and $\Omega_r$ are simply connected, then so is $\Omega$.

Intuitively, this seems rather obvious: if $\Omega_l$ and $\Omega_r$ "have no holes", then $\Omega$ can't have any holes either. However, proving this rigorously has me stumped.

There seem to be a few possible approaches here:

1. Use the following theorem/lemma:

A bounded region is simply connected if and only if its complement is connected.

This gives us that $\Omega^c \cup I \cup \Omega_l$ and $\Omega^c \cup I \cup \Omega_r$ are both connected, and that it suffices to prove that $\Omega^c$ is connected. I don't see how any of these help, though.

2. Use the definition of simple connectedness:

An open set is simply connected if any two curves with the same initial and final points inside the set are homotopic.

We could take two curves inside $\Omega$ and try to deform one into the other. Inside $\Omega_l$ and $\Omega_r$ we can deform curves easily, but when they intersect $I$ I can't figure out how to do it.

3. Use the winding number:

A bounded region $\Omega$ is simply connected if and only if $W_\gamma(z) = 0$ for any closed curve $\gamma$ in $\Omega$ and any point $z \notin \Omega$.

Ideally, we would take any curve $\gamma$ in $\Omega$ and somehow split this into two curves $\gamma_l, \gamma_r$ in $\Omega_l$ and $\Omega_r$ respectively, then write the winding number $W_\gamma(z)$ in terms of the winding numbers of these new curves. But, once again, I can't figure out how to handle this when $\gamma$ and $I$ intersect.

Any hints, suggestions or solutions are appreciated.

• I suggest method 3. You can restrict to polygonal paths, with no segment parallel to $I$, to have an easy to handle set of possible intersections with $I$. And you can assume that a segment of the path either crosses $I$ or doesn't intersect $I$. Use segments on $I$ to get two closed curves in $\Omega_l \cup I$ and $\Omega_r \cup I$ (if $\gamma$ isn't contained in one part already). – Daniel Fischer Apr 14 '17 at 15:40
• Actually, method 2 works pretty much the same, and if you have enough topology to throw at it, you can just mumble Seifert/van Kampen. – Daniel Fischer Apr 14 '17 at 15:47

If $\Omega\subseteq \Bbb C$ is open, then any curve $\gamma$ in $\Omega$ is homotopic to a polyline.
In fact, we may even impose that the segments of the polyline run only parallel to any two directions of our choice. In particular, we can avoid segments parallel to $L$. Thus the worst that can happen for a closed polyline starting at $a\in \Omega_l$, say, is: It intersects $L$ finitely many times, and between any two consecutive intersections it runs all in $\Omega_l$ or all in $\Omega_r$.
Claim. Any closed polyline in $\Omega$ (with segments not parrallel to $L$) that intersects $L$ is homotopic to such a polyline with less intersection points with $L$.
Proof. Let $a_0a_1\ldots a_n$ (where $a_0=a_n$) be the closed polyline. We may assume wlog. that $a_0\notin$L$. We may assume that every intersection point is in fact a node of the polyline. Case A: If there is an intersection point$a_i$such that$a_{i-1}$and$a_{i+1}$are on the same side of$L$(both in$\Omega_l$, say), we can do the following: Pick$r>0$with$B_r(a_i)\subseteq \Omega$. Pick points$p$within$[a_{i-1}a_i]$and$q$within$a_ia_{i+1}]$. Clearly$pa_iq$is homotopic to$pq$, and hence$a_0\ldots a_n$is homotopic to$a_0\ldots a_{i-1}pqa_{i+1}\ldots a_n$, which has one intersection less, as promised. Case B: If the just treated case A does not occur, the polyline must switch sides at each intersection point. In particular, there must be at least two intersection points. If$a_i$and$a_j$are consecutive such intersection points, we can do as follows: The polyline stays on one side of$L$(wlog. within$\Omega_l$) between$a_i$and$a_j$(in particular,$j>i+1$). For each$z\in I$, there is some$r=r(z)>0$such that$B_r(z)\subseteq \Omega$. Since the closed line segment$[a_i,a_j]\subset I$is compact, it is already covered by finitely many of such open disks$B_r(z)$with$z\in I$. Conclude from this that for some$\epsilon>0$, we have$U:=U_\epsilon([a_i,a_j]):=\bigcup_{z\in[a_0,a_n]}B_\epsilon(z)\subseteq \Omega$. Now we can describe another polyline from$a_i$to$a_j$that runs inside$\Omega_l$(except for the endpoints): Let$L'$be the line parallel to$L$at some distance$\delta<\epsilon$on the$\Omega_l$side. If$\delta$is small enough,$L'$intersects$[a_ia_{i+1}]$in a point$p\in U$and$[a_{j-1}a_j]$in a point$q\in U$. Then the polyline$a_ipqa_j$is homotopic (within$\Omega$) to$a_ia_{i+1}\ldots a_j$because$pa_{i+1}\ldots a_{j-1}q$is homotopic (within$\Omega_l$) to$pq$. By the simple shape of$U$,$a_ipqa_j$is homotopic (in$\Omega$) in a straightforward manner to its reflection at$L$. So far we did not change the number of intersection points. But now$a_i$(and$a_j$) are points where the polyline does not change sides, hence case A applies and we can reduce the number of intersection points, again.$\square$As a corollary, every closed line in$\Omega$(which we may assume to start/end outside$I$) is homotopic to a line living completely on one side of$L$. As$\Omega_l$and$\Omega_r$are simply connected, the closed line is ultimately homotopic to a single point. In other words,$\Omega\$ is simply connected.