Let $-1<\alpha<0$. Consider the domain $$\Omega=\{(x,y)| y>\alpha\wedge x^2+y^2>1\}$$ The purpose of this question is to present an argument that employs Van-Kampen's theorem, showing that $\Omega$ is simply connected, and then raise three questions.
Here is an attempt at a proof that $\Omega$ is simply connected. The figure attached below illustrates the notation.
Let $l_1$ denote the tangent to the circle at the point $(-\sqrt{1-\alpha^2},\alpha)$, and $l_2$ the tangent line to the circle at the point $(\sqrt{1-\alpha^2},\alpha)$. Choose a point $c=(\sin\theta,\cos\theta)$ and let $\tau_1$ denote the tangent line to the circle at the point $c$. The angle $\theta$ is chosen positive and sufficiently small as to guarantee that the lines $\tau_1,l_1$ are not parallel. Let $A$ denote their point of intersection. Put $$\Omega_1=\Omega\cap \{(x,y)| x<\sin\theta\}$$ $\Omega_1$ is a star-domain with respect to the point $A$. Similarly, we can choose a point $d=(-\sin\theta,\cos\theta)$, form the tangent line $\tau_2$ to the circle at $d$, and find its intersection point $B$ with the line $l_2$, and define $$\Omega_2=\Omega\cap \{(x,y)| x>-\sin\theta\}$$ There is full symmetry between the two sides, and $\Omega_2$ is also a star-domain, with respect to the point $B$. As star-domains, both $\Omega_1,\Omega_2$ are simply connected. Moreover, their intersection is simply $$\Omega_1\cap\Omega_2=\{(x,y)\in\Omega: |x|<\sin\theta\}$$ which is path-connected. By Van-Kampen's theorem, the union $\Omega_1\cup \Omega_2$ is also simply connected.
Question 1. Can anyone suggest an alternative argument, that does not use Van-Kampen's theorem?
It is visually obvious that there is no loop going around the origin contained in $\Omega$. It rigorously follows from the fact that $\Omega$ is simply connected.
Question 2. Is there a much simpler proof that there is no loop around the origin contained in $\Omega$?
Question 3. The argument above fails if $\alpha=-1$, but it is pretty clear -- visually -- that in that case the corresponding domain is simply connected as well. Does anyone know a proof?
With a little visualization, combined with a little trigonometry and calculus, you can write down a formula for a homeomorphism between your space and the obviously simply connected space $\mathbb R \times (1,\infty)$.
Let me indicate the visualization; I'll leave it for you to write down all the necessary formulas.
Let $A,B$ denote the two points where the circle intersects the line $y=\alpha$.
For each $r \in (1,\infty)$ consider the circle $C_r$ that contains the points $A$, $B$ and $P_r = (0,r)$. As $r$ varies, $C_r \cap \Omega$ is an open arc of the circle $C_r$, and these open arcs sweep out the whole of your region $\Omega$. So the trick now is to think of those open arcs as parameterized by the $x$-axis.
Here's a few more details. Let $\theta_B(r) \in (-\pi/2,0)$ be the angle of the point $B$ from the center of $C_r$, and let $\theta_A(r) \in (\pi,3\pi/2)$ be the angle of the point $A$ from the center of $C_r$.
Define a function $f = (f_1,f_2) : \Omega \to \mathbb R \times (1,\infty)$ as follows. First, $f_2(C_r \cap \Omega)=r$. Second, for each $(x,y) \in C_r \cap \Omega$, let $\theta(x,y) \in (\theta_B(r),\theta_A(r))$ be the angle of the point $(x,y)$ from the center of the circle $C_r$, and then define $$f_1(x,y) = \tan\left(\frac{\pi}{4}\frac{\theta(x,y)}{\theta_B(r)}-\frac{\pi}{4}\right) $$ In other words, we first scale the angle $\theta(x,y)$ so that it fits into the interval $(-\pi/2,\pi/2)$, and then we simply take the tangent of that angle (I hope I got that scaling right...).
After writing out all the formulas, you should be able to see with ordinary tools of calculus that this function $f$ is a continuous bijection with continuous inverse.