Stein simply connected slit STATEMENT: Prove that the complex plane slit along the union of the rays $\cup_{k=1}^n\left\{A_k+iy: y\leq 0\right\}$ is simply connected.
This is question 19 in chapter 8 of Stein's Complex Analysis text.
QUESTION: I don't understand what it's asking for. Could someone please provide a picture or rewording of the statement to make it clearer. Note that I don't want an answer, rather I just want help with parsing what is being said in the question. Thanks.
QUESTION 2: I am unsure of how to proceed with this problem. I assume that given two points and two curves connecting those points that I could shift it into the upper half plane which is connected, but I can't seem to find a homotopy that deforms one curve into another without conflict. The only resolution I can think of is that since the upper half plane is connected we might be able to assume that there exists a homotopy that is completedly contained in the space between the two curves, inclusive. Any suggestions.
 A: Each of the sets in the union is a ray that is a horizontal shift of the closed negative $y$-axis. All of these rays are to be removed from the plane, making a sequence of slits in the plane, like removing the downward-pointing teeth of a comb. You are being asked to show that what remains is a simply connected subset of the plane.
A:                                 
Let $A_1, A_2, \dots, A_n$ be real numbers and$$\Omega = \mathbb{C} \setminus \bigcup_{k=1}^n \{A_k + iy : y \le 0\}.$$We will show that $\Omega$ is simply connected. Set$$\Gamma = \{z \in \mathbb{C}: \text{Im}\,z = y = 1\}.$$Then $\Gamma$ is just a line and thus simply connected. To show that $\Omega$ is simply connected, we will show that $\Gamma$ is a retract of $\Omega$. We define a function $H$ as follows:$$H: \Omega \times [0, 1] \to \Omega,\text{ }H(z, t) = (x, (1 - t)y + t) = z + it(1 - y).$$To check that $H$ is well-defined, we take $z = A_k + iy$, where $y > 0$, and check that $H(z, t) \in \Omega$ for all $t \in [0, 1]$. We have$$H(z, t) = A_k + i((1 - t)y + t)$$and$$(1 - t)y + t >0$$for all $t \in [0, 1]$. Thus, $H$ is well-defined. We see that $H$ is also continuous and $$H(z, 0) = z \text{ for all }z \in \Omega,\text{ }H(z, 1) = x + i \text{ for all }z \in \Omega.$$Thus, $H$ is a deformation retract of $\Omega$ onto $\Gamma$. Thus,$$\pi_1(\Omega, *) \simeq \pi_1(\Gamma, *) \simeq \{0\}.$$Therefore, $\Omega$ is simply connected.
