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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.

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    $\begingroup$ 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). $\endgroup$ – Daniel Fischer Apr 14 '17 at 15:40
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    $\begingroup$ 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. $\endgroup$ – Daniel Fischer Apr 14 '17 at 15:47
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Lets try method 2.

To simplify matters, we can make use of the following:

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.

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