The condensed Whitehead conjecture for complete Boolean algebras

Let $A$ be an abelian group and let $\mathbb{B}$ be a complete Boolean algebra. Write $S(\mathbb{B})$ for the associated Stone space, and let $\Vdash_{\mathbb{B}} “A$ is not Whitehead$"$ mean that $\mathbb{B}$ forces that $A$ is not a Whitehead group. Let $\underline{A}$ and $\underline{\mathbb{Z}}$ denote the corresponding condensed abelian groups, and let $\underline{\mathrm{Ext}}^1_{\mathsf{CondAb}}$ denote the condensed Ext functor.

The condensed Whitehead conjecture. Suppose that $A$ is an abelian group, $\mathbb{B}$ is a complete Boolean algebra, and

$$\Vdash_{\mathbb{B}} “A \text{ is not Whitehead}".$$

Then

$$\underline{\mathrm{Ext}}^1_{\mathsf{CondAb}}(\underline{A},\underline{\mathbb{Z}})(S(\mathbb{B}))\neq 0.$$

The conjecture proposes that the technical strengthening of the failure of $A$ to be Whitehead used in the paper's theorem is unnecessary. It is refuted: the converse does not hold, as shown in the discussion following Corollary.