Existence of continuous family of uniformising parameters

Question

I asked this question on MSE a while ago but didn't receive any useful answers.

Suppose I have a $1$-parameter family continuous maps $f_t: \mathbb{S}^2\rightarrow \mathbb{C}P^1$ from a topological $2$-sphere to the Riemann sphere which is a local homeomorphism away from isolated points (for example, imagine a continuous family of rational functions $\mathbb{C}P^1 \longrightarrow \mathbb{C}P^1$ and then "forget" the complex structure on the domain). Suppose that $f_t$ gives us a complex structure on $\mathbb{S}^2$ pulls back a complex structure on $\mathbb{S}^2$ for each $t$. With this new complex structure on $\mathbb{S}^2$, there exists a uniformising parameter $z_t: \mathbb{C}P^1 \longrightarrow \mathbb{S}^2$ (up to Mobius transformations) such that $f_t\circ z_t$ is holomorphic for each $t$.

Question: Does there exist a continuous choice of $z_t$ so that $f_t\circ z_t: \mathbb{C}P^1 \longrightarrow \mathbb{C}P^1$ is a continuous family of holomorphic maps?

Some comments

If $f_t$ is a homeomorphism for all $t$, then clearly this is true because $f_t$ itself gives the parameter $z_t$.

I am not familiar with the proof of the uniformisation theorem well enough to answer this. An equivalent question would be: If you have a continuous family of atlases $\mathcal{A}_t$ on $\mathbb{S}^2$, then can you choose a continuous family of biholomorphisms $z_t : \mathbb{C}P^1 \longrightarrow (\mathbb{S}^2,\mathcal{A}_t)$? (although, I am not exactly sure how to define a "continuous family of atlases" since the domains and codomains of the coordinate charts may change.)

Answer 1

That your $f_t$ are local homeomorphisms away from isolated points is not sufficient for the conclusion you want. Your $f_t$ must be at least topologically holomorphic. (A continuous map is called topologically holomorphic if it is open and discrete).

Now one needs some stronger restrictions on $z\mapsto f(z)$ (how do you pull back the conformal structure via just local homeomorphism?)

If $z\mapsto f_t(z)$ safisfy a mild regularity condition, namely that they are quasiregular (with respect to some fixed conformal structure on $S^2$), then the proof can proceed as follows: Let $\mu_t=(f_t)_{\overline{z}}/(f_t)_z$ be the Beltrami coefficient and suppose that $t\mapsto \mu_t$ is continuous (as a function with values in $L^\infty$). Then to find homeomorphisms $\phi_t$ such that $f_t\circ\phi^{-1}_t$ are holomorphic, one has to solve the Beltrami equation $$\phi_{\overline z}=\mu_t\phi_z,$$ and there is a theorem which guarantees the existence of a unique normalized homeomorphic solution of Beltrami equation which depends continuously on $t$, see

Ahlfors, Lectures on quasiconformal mappings, Ch. V, C.

This involves stronger restrictions on both $z\mapsto f_t(z)$ and $t\mapsto f_t(z)$. Perhaps they can be somewhat relaxed but I do not think that the conclusion of continuity can be obtained under your restrictions.