Issues related to finding an algebra in S (P 𝒦) which is the domain of an epimorphism onto 𝑨

[williamdemeo]

I'm combining the following into one issue because they all have roughly the same response, which is that finding an F in S (P K) that is the domain of an epimorphism onto an arbitrary algebra in Mod (Th K) proved exceedingly difficult. It was the hardest part of the proof. The suggestions below all seem to point out that (in theory) it's easy to see the F's they suggest are in S (P K). However, proving this formally is hard.

1. Review 2 (point 1). The second half of the paper (starting with § on relatively free algebra) is confusing. The reviewer was hoping for the following argument:

   • Define 𝔽[X] as 𝑻(X)/≈, where x ≈ y iff given any homomorphism f into an element of K, f x = f y (in other words, x ≈ y iff (x,y) ∈ Th 𝒦). Then, if 𝑨 is in Mod (Th 𝒦), the surjective morphism 𝑻(A) → 𝑨 factors through 𝑻(X) → 𝔽[X], so it remains to show that 𝔽[X] ∈ S (P 𝒦) (then, 𝑨 ∈ HSP 𝒦).
   • 𝔽[X] is easily shown to be a sd prod of all algebras in 𝒦. However, because of size issues, this product may not exist. Fortunately, it's also a subproduct of the algebras in { 𝑻(X)/Θ }, because any hom factors as an epimorphism followed by a monomorphism, so that x ≈ y iff for any epimorphism f into an element of S 𝒦, f x = f y.

2. Review 2 (point 12).

   Let 𝑨 ∈ 𝒦⁺; it suffices to find an algebra 𝑭 ∈ S (P 𝒦) such that 𝑨 is a homomorphic image of 𝑭, as this will show that 𝑨 ∈ H (S (P 𝒦)) = 𝒦.

   Why can't we conclude the proof here by choosing 𝑭 = 𝔽[A], since we already showed that 𝑨 is an homomorphic image of 𝔽[A] before §6 and that 𝔽[A] ∈ S (P 𝒦) (in the § on relatively free algebra in theory)?

3. Review 2 (point 14).

   Let 𝑪 be the product of all algebras in S 𝒦, so that 𝑪 ∈ P (S 𝒦).

   Can we comment on the reason we chose a different route from the mathematical proof that we describe informally? (see also: comment on the universe hierarchy).

[JacquesCarette]

Sounds like all of this should be done in referee comments indeed - and maybe future work for a journal version?

[williamdemeo]

Actually, I labeled them incorrectly. All three points were made by the same referee. (The "weak accept" ...I guess s/he really didn't like our proof very much :roll_eyes: )

[williamdemeo]

Sounds like all of this should be done in referee comments indeed - and maybe future work for a journal version?

Yes, good idea.

[williamdemeo]

done