Let $\mathbb{T}$ be a finitary algebraic theory. For each $\mathbb{T}$-algebra $A$, let $Q (A)$ be the join semilattice of finitely generated congruences on $A$. There is an evident pushforward operation, yielding a functor $Q$ from the category of $\mathbb{T}$-algebras to the category of join semilattices.
Let $f : A \to B$ be a homomorphism of finitely presented $\mathbb{T}$-algebras. Say $f : A \to B$ is a weak $Q$-equivalence if $Q (f) : Q (A) \to Q (B)$ is an isomorphism of join semilattices. It is easy to find examples of weak $Q$-equivalences that are not isomorphisms. (For example, this happens when $\mathbb{T}$ is the theory of commutative rings: take $A$ and $B$ to be finite fields.)
Say $f : A \to B$ is a strong $Q$-equivalence if it satisfies the following condition:
- For every finitely presented $\mathbb{T}$-algebra $C$, the homomorphism $f \ast \mathrm{id}_C : A \ast C \to B \ast C$ is a weak $Q$-equivalence, where $\ast$ denotes the coproduct in the category of $\mathbb{T}$-algebras.
Question. If $f : A \to B$ is a strong $Q$-equivalence, then must $f : A \to B$ be an isomorphism?
For some $\mathbb{T}$, this is true: for example, if $\mathbb{T}$ is the theory of boolean algebras, then even weak $Q$-equivalences are isomorphisms. (Use Stone duality.)
In general, we can guarantee that strong $Q$-equivalences are epimorphisms (in the sense of category theory). Also, every surjective weak $Q$-equivalence is an isomorphism. So, if $\mathbb{T}$ is an algebraic theory such that every epimorphism of $\mathbb{T}$-algebras is a surjective homomorphism (e.g. the theory of left $R$-modules for some ring $R$), then strong $Q$-equivalences are isomorphisms. But what happens otherwise – for instance, if $\mathbb{T}$ is the theory of commutative rings?
