Pullback of Boolean algebras

Question

Given two Boolean algebra maps $f : B \to A$ and $f' : B' \to A$. Does there exists a nice characterisation of the pull-back of $f$ and $f'$? (I am looking for a reference (or an explanation).)

Answer 1

To expand on Max's comment, all categories of models of an (essentially) algebraic theory are complete (and cocomplete). This covers anything that can be defined in universal algebra. So the limits (and colimits) exist. Furthermore, since the underlying set functor has a left adjoint in these cases, the underlying set of a limit of algebraic objects is the limit of the undelying sets. For your specific case, the underlying set of the pullback of Boolean algebras is the pullback of the underlying sets which is the set Max described.

We can extend this to some colimits, specifically (filtered/)sifted colimits. For this, it's beneficial to take the categorical view of universal algebra given by Lawvere theories. That is, an algebra for an (essentially) algebraic theory is a finite (limit-/)product-preserving functor from a small category with finite (limits) products into $\mathbf{Set}$. Since $\mathbf{Set}$ is complete and cocomplete, a general result about limits and colimits in functor categories states that they are computed point-wise. What we don't necessarily have is that the limit/colimit of a diagram of finite (limit-/)product-preserving functors is finite (limit-/)product-preserving. This is easy to show for limits as limits commute and by the same stroke it is easy to show for colimits that commute with finite (limits/)products which are exactly (filtered) sifted colimits. Since the underlying set of an algebra is the image of an object with respect to the functor modeling the algebra, this gives another way of showing that the underlying set of a limit of algebras is the limit of the underlying sets of the algebras. We further get that the underlying set of a (filtered/)sifted colimit of algebras is the (filtered/)sifted colimit of the underlying sets of the algebras. This doesn't hold for arbitrary colimits such as coproducts which are neither filtered nor sifted. Sure enough, the coproduct of, say, a group is not built on the disjoint union of the underyling sets of the components. Showing cocompleteness of the categories of models is quite a bit trickier than showing completeness or cocompleteness with respect to (filtered) sifted colimits, but it is doable generally.

The general categorical analysis gives some very handy take-aways applicable in a large variety of situations. First, categories of algebraic objects (rings, groups, lattices, etc.) are complete and cocomplete. Second, limits and (filtered/)sifted colimits of algebras are built on limits and (filtered/)sifted colimits of their underlying sets. Third, the underlying set functor is a right adjoint and reflects isomorphisms.