= Solution
For a plane domain, the following are equivalent descriptions of a <simply connected domain>: every <based loop> contracts in the domain with its basepoint fixed; any two <paths> with the same endpoints are <homotopic> while those endpoints stay fixed; its complement in the <Riemann sphere> is <connected>; every <holomorphic function> has a single-valued primitive; and every nowhere-zero <holomorphic function> has a <holomorphic logarithm>. The complement must be taken in the extended plane, including infinity. An equivalent <winding number> formulation is that every closed curve has index zero about every point outside the domain.
Here is a direct equivalence proof for the first two topological formulations. If all <paths> with given endpoints are <homotopic> relative to those endpoints, apply this to a loop based at $a$ and the constant <path> at $a$, obtaining a contraction. Conversely, let $\gamma_0,\gamma_1$ join $a$ to $b$. The loop $\gamma_0*\gamma_1^{-1}$ contracts with basepoint fixed. Concatenating this contraction with $\gamma_1$ produces a <homotopy> from $\gamma_0*\gamma_1^{-1}*\gamma_1$ to $\gamma_1$, relative to $a,b$. The middle retracing excursion contracts by shortening its traversal in both directions, giving a <homotopy> from that first <path> to $\gamma_0$. Hence $\gamma_0$ and $\gamma_1$ are <homotopic> relative to their endpoints. Planar domains are <path-connected>, so no extra path-component condition is needed.
The analytic descriptions follow from <path independence> of the holomorphic integral. Local primitives exist on discs by integrating <power series>; a <homotopy> can be subdivided into pieces lying in such discs, and local <path> integrals cancel along its common edges. Thus integrals over <homotopic> <paths> agree, giving a global primitive. Applied to $h'/h$ for a nowhere-zero function, a primitive $H$ makes $he^{-H}$ constant and hence supplies a <holomorphic logarithm>. These observations are also the tools used in the mapping proof below.
The <Riemann mapping theorem> states that every nonempty <simply connected> proper domain $D\subsetneq\mathbb C$ admits a <conformal bijection> onto the <unit disc>. Given $z_0\in D$, there is exactly one such map with $f(z_0)=0$ and $f'(z_0)>0$. The exclusion of the whole plane is necessary: an <entire> map into the disc would be constant by <Liouville's theorem>.
An extremal proof has the following concrete steps. First construct an <injective> bounded holomorphic map. Choose $a\notin D$ and a <holomorphic square root> $h$ of $z-a$. It is <injective>, since equality of its values implies equality of their squares. Its image is disjoint from its negative: $h(z)=-h(w)$ would imply $z=w$ and then $h(z)=0$, impossible. If $b=h(z_0)$, choose $r>0$ with $B(b,r)\subset h(D)$; the disc $B(-b,r)$ is omitted. Therefore $1/(h+b)$ is <injective> and bounded. Scaling it and applying an <automorphism of the unit disk> gives a map into the disc taking $z_0$ to zero; a <rotation> makes its <derivative> positive.
Let $\mathcal U$ consist of all <injective> maps $D\to\mathbb D$ with this normalization. It is nonempty. The <derivative> at $z_0$ is bounded above by the <Cauchy integral formula> on a small disc compactly contained in $D$. Choose $f_n\in\mathcal U$ with <derivatives> tending to their positive supremum $L$. Boundedness, the <Cauchy estimates> and a diagonal <Arzelà-Ascoli theorem> argument give a subsequence converging locally uniformly to $f$; this is the bounded-family form of the <Montel theorem>. Then $f(z_0)=0$, $f'(z_0)=L>0$, and the <maximum modulus principle> puts its image in the open disc. The limit is <injective>: if distinct $z,w$ had $f(z)=f(w)$, apply <Hurwitz's theorem> near $w$ to $f_n(\cdot)-f_n(z)$, which has no zero there; the limit would vanish identically, contradicting $f'(z_0)>0$.
Finally suppose $a\in\mathbb D$ is omitted by $f$. It is nonzero since $f(z_0)=0$. Let $T_a(w)=(w-a)/(1-\overline a w)$ and choose a <holomorphic square root> $h$ of $T_a\circ f$. It is <injective> and takes values in the disc. Put $b=h(z_0)$, so $|b|=\sqrt{|a|}$, and normalize $T_b\circ h$ by a <rotation>. The new map's <derivative> magnitude at $z_0$ is
$$
\frac{(1-|a|^2)L}{2\sqrt{|a|}(1-|a|)}
=\frac{1+|a|}{2\sqrt{|a|}}L>L.
$$
This <square-root improvement of a normalized conformal map> contradicts extremality. Thus $f$ is onto. Two normalized <bijections> differ by a <disc automorphism> fixing zero; <Schwarz lemma> applied to it and its inverse makes it a <rotation>, and positivity of the normalized <derivatives> forces that <rotation> to be the identity. This completes the requested proof outline.
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