Coproducts of join-semilattices

Question

I am trying to describe what is the coproduct of two objects in the category of join-semilattices (with a top element). Does anyone have an idea or a reference to read ?

Answer 1

The category of join-semilattices is a variety of algebras, i.e. the category of models of an equational theory, in the sense of universal algebra.

In such a category, we compute the coproduct of two algebras $A$ and $B$ by taking the free algebra generated by $A$ and $B$ - roughly speaking, its domain is the set of all terms formed from the elements of $A$ and $B$, modulo the equivalence relation generated by the equational axioms and the actual term evaluations in $A$ and $B$.

Let's start with the case of (unbounded) join semilattices. These have just one operation $\vee$, and equational axioms:

• $x \vee x = x$ (idempotence)
• $x \vee y = y \vee x$ (commutativity)
• $(x\vee y) \vee z = x \vee (y \vee z)$ (associativity)

Let $A$ and $B$ be join semilattices. Then a term formed from the elements of $A$ and $B$ is an element of $A$, an element of $B$, or a join of finitely many elements from $A$ and $B$. By associativity and commutativity, a term of the third type can be rewritten as the join of (1) finitely many elements of $A$ and (2) finitely many elements of $B$. But (1) and (2) can be evaluated in $A$ and $B$, respectively, so a general term is equivalent to one of the form $a\vee b$ with $a\in A$ and $b\in B$. So the coproduct is $$ A\sqcup B \sqcup \{a\vee b\mid a\in A,b\in B\}$$ (here $\sqcup$ denotes disjoint union) with the obvious join operation: If $a\in A$ and $b\in B$, their join is $a\vee b$. If $a,a'\in A$ and $b\in B$, then $(a\vee b) \vee a' = (a\vee a')\vee b$. If $a\in A$ and $b,b'\in B$, then $(a\vee b) \vee b' = a\vee(b\vee b')$. And if $a,a'\in A$ and $b,b'\in B$, then $(a\vee b)\vee (a' \vee b') = (a\vee a')\vee (b\vee b')$.

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Things get a bit simpler if we consider bounded join semilattices. These have an additional constant $\bot$ and an additional equational axiom:

• $\bot \vee x = x$ (identity)

The above discussion applies, but additionally, for any $a\in A$, $a$ is equivalent to the term $a\vee \bot_B$, and for any $b\in B$, $b$ is equivalent to the term $\bot_A \vee b$. So the coproduct is $$ \{a\vee b\mid a\in A,b\in B\}$$ with bottom element $\bot_A\vee \bot_B$ and the obvious component-wise join. Note that this is isomorphic to the product of $A$ and $B$! The category of bounded join semilattices has biproducts (finite products and coproducts coincide), just like the categories of abelian groups or vector spaces.

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But in the question, you asked about join semilattices with top elements. These have an additional constant $\top$ and an additional equational axiom:

• $\top \vee x = \top$

Now the above discussion applies, but we have $\top_A = \top_B = \top_A\vee b = a\vee \top_B$ for all $a\in A$ and $b\in B$. Let's write $A^*$ for $A\setminus \{\top_A\}$ and similarly for $B^*$. Then the coproduct is

$$ A^*\sqcup B^* \sqcup \{a\vee b\mid a\in A^*,b\in B^*\}\sqcup \{\top\}$$

with top element $\top$ and the obvious join operation.