My example in fact concerns integral cohomology rings of $p$-groups. There are no abelian examples because $H^2(G;\mathbb{Z})\cong \mathrm{Hom}(G,U(1))$ determines $G$ uniquely.

Given that the integral cohomology groups are difficult to calculate, this is the sort of question that is extremely difficult to counterexample.

Of course, a codimension one real subspace in $A$ causes the complement to be disconnected. Codimension two real subspaces in some sense generalize codimension one complex subspaces.

Suppose that when I constructed my simply connected domain I made countably many arbitrary choices. I don't see how you can compute a uniformizing function for my domain in general. To be more explicit, I'm going to take as my domain the union of the lower half plane and discs of radius $1/3$ centered at some elements of $\mathbb{Z}$. I'm thinking of the case when my subset of $\mathbb{Z}$ is not recursively enumerable. Surely you will need at least Choice for countable sets.

Finding explicit maps that uniformize (if that is the word) even fairly simple shaped simply connected domains is incredibly difficult. So isn't this just the sort of problem that is bound to require the Axiom of Choice?

Finding a bound for a module presentation over $H^*(BU(n);\mathbb{F}_p)$ has been known a lot longer than Symonds' result giving a bound for a ring presentation. Symonds' result was a major advance.

Piotr: thanks for your message and correcting yourself with an edit.