Can nonisomorphic algebras generate isomorphic relatively free algebras?

Question

Suppose that we have two varieties of algebras $A$ and $B$, whose operators all have arities less than some regular cardinal, and such that every $B$-algebra (please correct me if this is not the usual notation) is also an $A$-algebra. It is known that the forgetful functor from the category of $B$-algebras to that of $A$-algebras has a left adjoint $F$, which takes every $A$-algebra $a$ to the relatively free $B$-algebra $Fa$ generated by $a$. My question is, if $a$ and $b$ are $A$-algebras such that $Fa$ is isomorphic (as a $B$-algebra) to $Fb$, are $a$ and $b$ necessarily isomorphic? If not in general, how about the special case where $A$-algebras are Boolean algebras and $B$-algebras are countably complete Boolean algebras?

Answer 1

Your left adjoint is not quite correct.

For example, take the case where $A$ is the variety of all groups, and $B$ is the variety of abelian groups. The forgetful functor from $\mathcal{A}b$ to $\mathcal{G}roup$ indeed has a left adjoint, but the left adjoint $\mathcal{F}$ takes a group $G$ to $G^{\rm ab}$, the abelianization of $G$. This easily yields examples of objects $a$ and $b$ such that $\mathcal{F}(a)\cong \mathcal{F}(b)$, even though $a\not\cong b$.

For the specific case of Boolean algebras I think the answer is still negative as stated: take a boolean algebra $a$ that is not [countably] complete and such that its [countable] completion has strictly larger cardinality than $a$ (I assume such things exist, but if this is not the case, then of course I'm falling flat on my case here). Let $b$ be the image of the [countable] completion under the forgetful functor. Since $a$ and $b$ have distinct cardinality, $a\not\cong b$, but $\mathcal{F}(b) = \mathcal{F}(\mathcal{U}(\mathcal{F}(a)))\cong \mathcal{F}(a)$, with the last equality using the unit and counits of the adjunction.

One way to construct $\mathcal{F}(a)$ if your varieties have a nice underlying set functor with an adjoint is to first consider $U(a)$, the underlying set of $a$, then take $\mathfrak{F}_B(U(a))$, the free $B$-algebra on the underlying set of $a$, and then let $\mathcal{F}(a) = \mathfrak{F}_B(U(a))/\Phi$, where $\Phi$ is the $B$-congruence generated by the relations corresponding to the $A$-operations on $a$. For instance, if $B$ is $\mathcal{G}roups$ and $A$ is $\mathcal{M}onoids$, given a monoid $M$, you first take $U(M)=\{[m]\mid m\in M\}$, the underlying set of $M$ (I'm using $[m]$ to denote the element $m$ in the set $U(M)$). Then you take the free group on $U(M)$, and you mod out by the the normal subgroup generated by the relations $[m][n]=[mn]$ (and since $[mn]=[nm]$, you end up with an abelian group).

Added: I need to think a bit more about the Boolean algebra case; you are correct that my assertion about $\mathcal{F}(\mathcal{U}(\mathcal{F}(a)))$ does not hold in general, and likely not in this case either. My gut feeling is that you can have $\mathcal{F}(a)\cong\mathcal{F}(b)$ even with $a\not\cong b$, though.

Added 2: Heh; reading the MO question you link to in the comments suggests that I am facing exactly what the long discussion between Joel David Hamkins and Todd Trimble was about, and I was making the same kind of assumptions Joel made to reach my conclusion that my construction would necessarily yield an example.