A proper homologically simply connected plane domain has a bounded univalent competitor

Because $\mathrm{\Omega }⊊\mathbb{C}$, choose $a\in \mathbb{C}\setminus \mathrm{\Omega }$. The function $z\mapsto z-a$ is holomorphic and nowhere zero on $\mathrm{\Omega }$, so [L1] gives a holomorphic $q:\mathrm{\Omega }\to \mathbb{C}$ with $q(z{)}^2=z-a$.

If $q(z_1)=q(z_2)$ then $z_1-a=q(z_1{)}^2=q(z_2{)}^2=z_2-a$, so $z_1=z_2$; thus $q$ is injective. Also $0\mathrm{\notin }q(\mathrm{\Omega })$, and $q(\mathrm{\Omega })\cap (-q(\mathrm{\Omega }))=\mathrm{\varnothing }$ because $q(z_1)=-q(z_2)$ would again force $z_1=z_2$, hence $q(z_1)=0$, impossible.

Put $w_0:=q(z_0)$. Since $q$ is nonconstant, [L3] makes $q(\mathrm{\Omega })$ open, so choose $\rho >0$ with $D(w_0,\rho )\subseteq q(\mathrm{\Omega })$. Step 2.1 gives $q(\mathrm{\Omega })\cap (-q(\mathrm{\Omega }))=\mathrm{\varnothing }$, hence $D(-w_0,\rho )=-D(w_0,\rho )\subseteq -q(\mathrm{\Omega })$ is disjoint from $q(\mathrm{\Omega })$. Therefore $\mathrm{\mid }q(z)+w_0\mathrm{\mid }\ge \rho$ for every $z\in \mathrm{\Omega }$.

Define $h(z):=\frac{\rho }{2(q(z)+w_0)}\mathrm{.}$ Step 3.1 gives $\mathrm{\mid }h(z)\mathrm{\mid }\le 1\mathrm{/}2$, so $h(\mathrm{\Omega })\subseteq \mathbb{D}$. The reciprocal affine map is injective away from $-w_0$, and step 2.1 makes $q$ injective, so $h$ is holomorphic and injective on $\mathrm{\Omega }$.

Let $b=h(z_0)\in \mathbb{D}$. By [L2], $g:={\phi }_b\circ h$ is holomorphic and injective from $\mathrm{\Omega }$ into $\mathbb{D}$, and $g(z_0)=0$. Differentiating $q^2=z-a$ gives $q^{\mathrm{\prime }}(z_0)=1\mathrm{/}(2w_0)\mathrm{\ne }0$, so $g^{\mathrm{\prime }}(z_0)\mathrm{\ne }0$. Multiplying by the unimodular constant $\overline{g^{\mathrm{\prime }}(z_0)}\mathrm{/}\mathrm{\mid }{g}^{\mathrm{\prime }}(z_0)\mathrm{\mid }$ makes the derivative at $z_0$ positive. The resulting map lies in $\mathcal{F}(\mathrm{\Omega },z_0)$.