There is a set of solid rings; is there also a set of epimorphisms $R→A$ for $R$ fixed?

Question

All rings here are commutative. A ring $R$ is called solid if the unique morphism $ℤ→R$ is an epimorphism. For instance, $ℤ/nℤ$ for $n∈ℕ$, or $ℚ$, or even $ℤ/nℤ × ℚ$. See this answer.

In the comments of this post was raised the question of whether the class of epimorphisms $R→A$, for a fixed ring $R$, is a set (modulo isomorphism). This is true for $R=ℤ$. Is it true for any commutative ring?

Intuitively, I would have said it's not true there is a set of solid rings because in Isbell's zigzag lemma, I would have thought we can make the criterion "go to infinity" somehow and construct arbitrarily large solid rings. So I find this surprising, and would like to know about the general case.

Answer 1

According to this MathOverflow thread and the sources that it links to, if $R\to A$ is a commutative ring epimorphism then the cardinality of $A$ is at most that of $R$, from which it easily follows that for a given commutative ring $R$ there is only a set of (isomorphism classes of) ring epimorphisms $R\to A$.