- Prove that the complex plane slit along the union of the rays \(\cup _ { k = 1 } ^ { n } \{ A _ { k } + i y : y \leq 0 \}\) is simply connected.
[Hint: Given a curve, first “raise” it so that it is completely contained in the upper half-plane.]
Solution
Let \[ \Omega=\mathbb C\setminus\bigcup_{k=1}^n\{A_k+iy:y\le0\}. \] After deleting repetitions, assume the real numbers \(A_k\) are distinct. We prove that every closed curve in \(\Omega\) is null-homotopic.
Let \(\gamma:[0,1]\to\Omega\) be closed. Its compact image has positive distance from the closed deleted set, so there is \(\varepsilon>0\) such that the horizontal \(\varepsilon\)-neighborhood of \(\gamma([0,1])\) still avoids every slit whenever a homotopy below crosses the real line away from the \(A_k\)’s. First perturb \(\gamma\) slightly, if necessary, so that whenever \(\Re\gamma(t)=A_k\) one has \(\Im\gamma(t)>0\); compactness then gives a number \(h>0\) such that \[ \Im\gamma(t)+h>0\qquad(0\le t\le1). \] Consider the vertical translation homotopy \[ H(t,s)=\gamma(t)+ish, \qquad 0\le s\le1. \] If \(H(t,s)\) lay on the \(k\)th deleted ray, then \(\Re\gamma(t)=A_k\) and \(\Im\gamma(t)+sh\le0\). But at every \(t\) with \(\Re\gamma(t)=A_k\) we arranged \(\Im\gamma(t)>0\), a contradiction. Thus \(H\) is a homotopy in \(\Omega\) from \(\gamma\) to the closed curve \(\gamma+h i\), which lies entirely in the upper half-plane.
The upper half-plane is convex. Hence \(\gamma+hi\) contracts there, by the straight-line homotopy to any fixed point of the upper half-plane. Since the upper half-plane is contained in \(\Omega\), this is also a contraction in \(\Omega\). Therefore every closed curve in \(\Omega\) is null-homotopic, and \(\Omega\) is simply connected.