Simply Connected Plane Domains: the Grand Equivalence — Examples

The complement $\widehat{\mathbb{C}}\setminus \mathbb{D}=\{z:\mathrm{\mid }z\mathrm{\mid }\ge 1\}\cup \{\mathrm{\infty }\}$ is connected: the exterior $\{z:\mathrm{\mid }z\mathrm{\mid }>1\}$ is path-connected, and $\mathrm{\infty }$ is its point at infinity.

By [L1], condition 1 from step 1.1 forces every other clause on the grand-equivalence list. So $\mathbb{D}$ satisfies them all.

The spherical complement of $\mathbb{C}$ is the singleton $\{\mathrm{\infty }\}$, hence connected.

By [L1], the connected complement from step 1.1 forces all remaining clauses. In particular, every cycle in $\mathbb{C}$ is null-homologous, every entire function has a primitive, and $\mathbb{C}$ is contractible.

The spherical complement of $\mathrm{\Omega }$ is $\widehat{\mathbb{C}}\setminus \mathrm{\Omega }=(-\mathrm{\infty },0]\cup \{\mathrm{\infty }\},$ which is connected: it is the closure in $\widehat{\mathbb{C}}$ of one arc from $0$ to $\mathrm{\infty }$.

Therefore [L1] makes $\mathrm{\Omega }$ simply connected.

Regard $\mathrm{\Omega }$ as a convex subset of ${\mathbb{R}}^2$. Then [L1] gives trivial fundamental group.

By [L2], clause 3 from step 1.1 implies that $\mathrm{\Omega }$ is simply connected.

Fact [L1] makes the given domain $\mathrm{\Omega }$ homologically simply connected.

By [L2], step 1.1 implies that $\mathrm{\Omega }$ is simply connected.

The set $\mathrm{\Omega }$ is open and connected by construction. Its spherical complement is connected: outside the two discs and corridor one can move continuously around the exterior, and the slits cut out by the corridor attach to the same unbounded exterior region instead of creating a hole. Therefore [L1] makes $\mathrm{\Omega }$ simply connected.

The domain is not star-shaped. Set $u_{\pm }=(\pm 2,\frac{3}{4}),{\mathrm{\ell }}_{\pm }=(\pm 2,-\frac{3}{4}),$ which all lie in $\mathrm{\Omega }$. Let $z_0=(x_0,y_0)\in \mathrm{\Omega }$. If $x_0\le 0$ and $y_0\ge 0$, then along the segment from $z_0$ to $u_{+}$ the point with $x$-coordinate $1$ has $y$-coordinate at least $\frac{3}{8}>\frac{1}{4}$, so it lies above the corridor and also outside the right unit disc because its distance to $(2,0)$ exceeds $1$. Hence that segment leaves $\mathrm{\Omega }$. If $x_0\le 0$ and $y_0<0$, the same argument with ${\mathrm{\ell }}_{+}$ gives a point at $x=1$ with $y\le -\frac{3}{8}<-\frac{1}{4}$, again outside $\mathrm{\Omega }$. By symmetry, if $x_0\ge 0$ then one of the segments from $z_0$ to $u_{-}$ or ${\mathrm{\ell }}_{-}$ leaves $\mathrm{\Omega }$. Thus no point of $\mathrm{\Omega }$ sees all of $\mathrm{\Omega }$ by straight segments, so $\mathrm{\Omega }$ is not star-shaped.

Steps 1.1 and 1.2 give the required witness: simply connected need not imply star-shaped.

The complement of $\mathrm{\Omega }$ in $\mathbb{C}$ is the singleton $\{0\}$, hence connected. But the spherical complement is $\widehat{\mathbb{C}}\setminus \mathrm{\Omega }=\{0,\mathrm{\infty }\},$ which is disconnected.

Assuming the Axiom of Choice, [L1] makes the disconnected spherical complement from step 1.1 enough to conclude that $\mathrm{\Omega }$ is not simply connected. Fact [L2] records the same failure independently as the nontrivial group ${\pi }_1({\mathbb{C}}^{\times })\cong \mathbb{Z}$.

Therefore connected complement in $\mathbb{C}$ does not imply simple connectivity.

The annulus $\mathrm{\Omega }$ is open and connected, and the unit circle $\gamma$ lies in $\mathrm{\Omega }$. The spherical complement of $\mathrm{\Omega }$ has two pieces: the closed inner disc $\{\mathrm{\mid }z\mathrm{\mid }\le \frac{1}{2}\}$ and the exterior region $\{\mathrm{\mid }z\mathrm{\mid }\ge 2\}\cup \{\mathrm{\infty }\}$. So it is disconnected.

By [L1], step 1.1 already shows that $\mathrm{\Omega }$ is not simply connected. Fact [L2] gives the familiar loop witness: the unit circle still winds once around the omitted origin.

Therefore connectedness of the domain does not force simple connectivity.

The punctured disc is open and connected, and the loop $\gamma$ lies in it. Its spherical complement is the union of the singleton $\{0\}$ and the outer region $\{\mathrm{\mid }z\mathrm{\mid }\ge 1\}\cup \{\mathrm{\infty }\}$, so it is disconnected.

By [L1], step 1.1 shows that the punctured disc is not simply connected. Fact [L2] gives the same geometric reason numerically: the loop $\gamma$ still winds once around the omitted point $0$.

This refutes the claim.

Apply [L1] to $\mathrm{\Omega }={\mathbb{C}}^{\times }$. Its complement in $\mathbb{C}$ is the connected set $\{0\}$, but the domain is not simply connected.

So the displayed implication fails. The missing point is exactly that plane simple connectivity depends on the complement in $\widehat{\mathbb{C}}$, not merely in $\mathbb{C}$.

By [L1], the slit plane is simply connected.

It is not convex: the points $-1+i$ and $-1-i$ lie in the slit plane, but the straight segment between them is the vertical line segment $\{-1+it:-1\le t\le 1\}$, which contains the removed point $-1\in (-\mathrm{\infty },0]$. Thus the segment is not contained in the domain.

Therefore simple connectivity does not imply convexity.

Apply [L1]. It gives a plane domain that is simply connected and simultaneously not star-shaped.

Hence the universal claim is false.