It is well-known that composing meromorphic functions on $\mathbb C$ does not necessarily result in a meromorphic function (e.g., $\exp\circ\frac1x$, which has an essential singularity at $x=0$.)
Question: What is the "closure" of the meromorphic functions under composition? That is, what is the minimal extension field $K$ of the field $\mathcal M(\mathbb C)$ such that for each $f,g\in K\setminus\mathbb C$, we have $f\circ g\in K$.
My naïve guess would be the field of holomorphic functions defined on $\mathbb C\setminus I$, for some zero-dimensional complex analytic subset $I$. Such a field will be certainly closed under composition, but I am not sure how to prove/disprove it is minimal.
Here, I view meromorphic functions as entire functions $f\colon\mathbb C_\infty\setminus\{\infty\}\to\mathbb C_\infty$, where $\mathbb C_\infty:=\mathbb C\cup\{\infty\}$ is the Riemann sphere. Then, we can define the composition of entire functions $f\colon\mathbb C_\infty\setminus I\to\mathbb C_\infty$ and $g\colon\mathbb C_\infty\setminus J\to\mathbb C_\infty$ with $f\circ g\colon \mathbb C_\infty\setminus(J\cup g^{-1}(I))\to\mathbb C_\infty:z\mapsto f(g(z))$.