# What are products and coproducts of bounded lattices?

I understand the universal property that products and coproducts must satisfy, but I'm having trouble finding a concrete construction in the category Lat of bounded lattices with meet/join top/bottom preserving functions as arrows.

For example, does Lat have products and coproducts? What lattice would constitute the co/product of simple lattices like:

$$0\rightarrow a \rightarrow 1 \quad\text{and}\quad 0\rightarrow b \rightarrow 1$$

and what about more complicated lattices?

I suspect that one construct may be something like $\{ \langle a, b\rangle : a \in L_1, b\in L_2\}$ where we define $\langle a_1, b_1\rangle \preceq \langle a_2, b_2\rangle$ in $L_1\times L_2$ if and only if $a_1 \leq a_2$ and $b_1 \leq b_2$. The new top and bottom elements would then be $\langle 1, 1\rangle$ and $\langle 0,0\rangle$. If this is the coproduct, the inclusions might be $a\mapsto \langle a, 1\rangle$, for example. If this is the product, the projections might be $\langle a, b\rangle \mapsto a$ as in ordinary set projection.

## 1 Answer

Bounded lattices can be described by an equational theory, so they are examples of models of a Lawvere theory. Thus the category of bounded lattices is complete and cocomplete. Furthermore, the underlying set of the product is the product of the underlying sets. Basically, it will be a pair of bounded lattices with operations defined component-wise. In fact, this will be more or less the situation for all limits and as well as sifted colimits. So the product has elements $\{(0,0),(1,1),(a,0),(a,b),(a,1),(0,b),(1,b)\}$ where e.g. $(x_1,y_1)\land(x_2,y_2)=(x_1\land x_2,y_1\land y_2)$.

Coproducts are not sifted colimits so they are a bit more involved to describe. As a start, we can look at free bounded lattices generated from a set. The coproduct of two free bounded lattices is the free bounded lattice generated from the disjoint union of the generating sets. Now every bounded lattice is a (reflexive) coequalizer of a free bounded lattice. So coproducts of arbitrary bounded lattices can be derived by viewing them as (reflexive) coequalizers of the corresponding free bounded lattices.

Your examples are free bounded lattices generated by singleton sets, so the coproduct is going to be the free bounded lattice generated by a two element set. In particular, it will have elements $\{0,1,a,b,a\land b,a\lor b\}$.

Learning some basic general results from Lawvere/algebraic theories/universal algebra can provide a nice pile of results quickly with broad applicability (almost all of abstract algebra). Of course, the things that make lattices different from ring different from groups won't be captured this way, but it is good to separate the things that are generally true from the aspects that make a particular algebraic object distinct.

• Thanks! This is helpful. Does this same reasoning allow you to determine the structure of pullbacks and pushouts of lattices? – user326210 Jul 2 '17 at 2:04
• @user326210 Yes. Once you know some category is a category of models for a (given) Lawvere theory, you can explicitly construct all of the limits, colimits, and free objects. Of course it won't be in the most helpful representation, e.g. free vector spaces will be expressed as quotients of a term algebra rather than, say, Cartesian products of the base field, though you can of course prove that these representations are isomorphic. A lot of algebra is about finding representations that are handy for calculation. – Derek Elkins Jul 2 '17 at 2:47
• I'm not sure I understand. Could you show me how to do an explicit construction of a pullback or pushout of lattices for a simple case? – user326210 Jul 3 '17 at 2:17
• @user326210 Limits are straightforward. Just take the limit of the underlying sets and equip it with the (restricted) (product of) the original operation(s). The pullback of $f:A\to C$ along $g:B\to C$ of two lattice homomorphisms is the set $\{(a,b)\in A\times B\mid f(a)=g(b)\}$ equipped with the original lattice operations applied component-wise and restricted. Because $f$ and $g$ are lattice homomorphisms if $f(a_i)=g(b_i)$ then $f(a_1\land a_2)=g(b_1\land b_2)$ and similarly for $\lor$, so the component-wise operations will be closed on the subset the pullback picks out. – Derek Elkins Jul 3 '17 at 3:06