Consider two doubly-connected open subsets $A$ and $A'$ of the Riemann sphere. We assume these two domains to be of same modulus (the moduli space being one real parameter), i.e. we assume that there exists a holomorphic bijection $\phi:A\rightarrow A'$. Note that the map $\phi$ is then unique up to precomposition by the automorphisms of the annulus $A$, loosely speaking the choice of whether we switch the two boundary components, plus a rotation.

**Question**

Is it possible to factor $\phi$ as a composition of holomorphic maps $\phi=f_1\circ g_1 \circ \cdots \circ f_n \circ g_n$, where the maps $f_i$ and $g_i$ are defined on simply-connected domains?

If yes, can we find such a factorization with $n=1$?

**Motivation: case $\phi=f_1$.**

To fix ideas, suppose the annuli $A$ and $A'$ separate $0$ and $\infty$, and that the map $\phi$ sends outer boundary to outer boundary. Assume that the map $\phi$ can be extended to the inside of $A$, i.e. $\phi$ is the restriction to $A$ of a holomorphic function defined on a simply-connected domain containing $0$ and bounded by the outer boundary of $A$. I claim that this gives no information on the outer boundary of the domain $A'$. However, once we fix the outer boundary of $A'$, its inner boundary has to live in a (small) family of three real parameters. Inded, fix the outer boundary of $A'$ to what you wish, and see by Riemann uniformization that the possible maps $\phi$ form a three real parameters family.

Thinking in terms of degrees of freedom, it seems it could be possible, for any couple of annular domains $A$ and $A'$, to write $\phi$ as a composition $f\circ g$, where $f$ extends to the inside, and $g$ extends to the outside: either one of these maps $f$ and $g$ allows complete freedom on either the exterior or the interior boundary of its image.

**Note: uniqueness if $n=1$.**

In the $n=1$ case, the factorization, if it exists is unique up to Moebius transformation.

Indeed, suppose $\phi = f\circ g = \tilde f \circ \tilde g$. Let us consider the holomorphic function $\tilde f^{-1} \circ f = \tilde g \circ g^{-1}$, a priori a holomorphic bijection $g(A)\rightarrow \tilde g(A)$. However, $\tilde f^{-1} \circ f$ extends inside $g(A)$ i.e. it is actually defined on a simply-connected domain bounded by the outer boundary of $g(A)$. Whereas $\tilde g \circ g^{-1}$ extends outside of $g(A)$. The function $\tilde f^{-1} \circ f$ can thus be extended to the whole sphere, and hence is a Moebius transformation.