A homologically simply connected plane domain has connected spherical complement

Suppose, toward a contradiction, that $F:=\widehat{\mathbb{C}}\setminus \mathrm{\Omega }$ is disconnected. Then $F=A\bigsqcup B$ for disjoint nonempty closed subsets $A,B\subseteq F$. Since $\mathrm{\infty }\in F$, relabel so that $\mathrm{\infty }\in B$. Because the Riemann sphere is a metric space, the disjoint closed sets $A$ and $B$ have disjoint open neighbourhoods $U$ and $V$ in $\widehat{\mathbb{C}}$ with $A\subseteq U$ and $B\subseteq V$. The inclusion $\mathrm{\infty }\in V$ forces $U\subseteq \mathbb{C}$, so $A$ is compact in $\mathbb{C}$ and $U\cap (\mathbb{C}\setminus \mathrm{\Omega })=U\cap F=A$. Hence $U\setminus A\subseteq \mathrm{\Omega }$.

Apply [L2] to the compact set $A\subseteq U$, viewing $\mathbb{C}$ as ${\mathbb{R}}^2$. It gives a finite union $J$ of closed grid rectangles with $A\subseteq \mathrm{int}J\subseteq J\subseteq U$. Give each rectangle boundary its positive orientation, sum those boundary chains, and cancel every interior edge with its opposite by [L3]. Let $\mathrm{\Gamma }$ be the remaining chain. Then $\mathrm{\Gamma }$ is a cycle, and its trace is the frontier of $J$, so ${\mathrm{\Gamma }}^{\ast }\subseteq U\setminus A\subseteq \mathrm{\Omega }$.

Let $p\in \mathrm{int}J$ lie on no grid line. Exactly one grid cell $Q$ of $J$ contains $p$. Along the positively oriented boundary of $Q$, the continuous argument of $\zeta -p$ increases by $2\pi$, while along the boundary of every other grid cell it has increment $0$ because $p$ lies outside that cell. Therefore [L4] gives winding number $1$ for ${\mathrm{\partial }}^{+}Q$ about $p$ and $0$ for every other cell boundary, and additivity from [L3] yields $n(\mathrm{\Gamma },p)=1$.

Fix $a\in A$. Because $A\subseteq \mathrm{int}J$, choose a disc $D(a,r)\subseteq \mathrm{int}J$ and then choose $p\in D(a,r)$ on no grid line. The disc misses ${\mathrm{\Gamma }}^{\ast }$, so local constancy from [L5] and step 3.1 give $n(\mathrm{\Gamma },a)=n(\mathrm{\Gamma },p)=1$.

The point $a$ lies in $\mathbb{C}\setminus \mathrm{\Omega }$, while step 2.1 gives ${\mathrm{\Gamma }}^{\ast }\subseteq \mathrm{\Omega }$. Step 4.1 yields $n(\mathrm{\Gamma },a)=1\mathrm{\ne }0$, so [L1] says that $\mathrm{\Gamma }$ is not null-homologous in $\mathrm{\Omega }$, contradicting the homological simple connectivity of $\mathrm{\Omega }$. Therefore the assumption of step 1.1 was false, and $\widehat{\mathbb{C}}\setminus \mathrm{\Omega }$ is connected.