For a plane domain, the complement, homology, primitive, logarithm, conjugate, conformal, homotopy, and contractibility conditions are equivalent

By [L1], conditions 1 and 2 are equivalent. By [L2], conditions 2, 4, and 6 are equivalent. By [L4], conditions 2 and 8 are equivalent. By [L9], conditions 4 and 5 are equivalent.

Condition 2 implies condition 7 by [L3]. Conversely, assume condition 7. Fix $p\in \mathbb{C}\setminus \mathrm{\Omega }$ and a closed rectifiable contour $\gamma :[a,b]\to \mathrm{\Omega }$. Applying condition 7 repeatedly to the nowhere-zero holomorphic function $z\mapsto z-p$ produces, for every $m\ge 1$, a nowhere-zero holomorphic function $g_m$ on $\mathrm{\Omega }$ with $g_m^{2^m}=z-p$. By [L10], the closed contour $g_m\circ \gamma$ admits a continuous logarithm ${\lambda }_m$; write ${\theta }_m=\mathrm{Im}{\lambda }_m$. Then $2^m{\lambda }_m$ is a continuous logarithm of $\gamma -p$ along $\gamma$, so [L11] gives $n(\gamma ,p)=2^m\frac{{\theta }_m(b)-{\theta }_m(a)}{2\pi }\mathrm{.}$ Applying [L11] again to $g_m\circ \gamma$ shows that $({\theta }_m(b)-{\theta }_m(a))\mathrm{/}(2\pi )$ is an integer. Therefore $n(\gamma ,p)\in 2^m\mathbb{Z}(m\ge 1)\mathrm{.}$ The only integer divisible by every power of $2$ is $0$, so $n(\gamma ,p)=0$. By [L11], this is equivalent to ${\int }_{\gamma }\frac{dz}{z-p}=0.$ Since $p$ and $\gamma$ were arbitrary and $z\mapsto 1\mathrm{/}(z-p)$ is continuous on $\mathrm{\Omega }$, [L9] makes $1\mathrm{/}(z-p)$ admit a primitive on $\mathrm{\Omega }$. Thus condition 2 holds by [L2], and conditions 2 and 7 are equivalent.

Assume condition 2. By [L5], condition 9 follows. Any conformal equivalence is in particular a homeomorphism, so condition 9 implies condition 10. Then [L6] gives condition 11, [L7] gives condition 3, and [L8] returns to condition 2. Thus [L5, L6, L7, L8] $2\Rightarrow 9\Rightarrow 10\Rightarrow 11\Rightarrow 3\Rightarrow 2.$

Steps 1.1, 1.2, and 1.3 connect every listed clause to condition 2, so all eleven conditions are equivalent.