Free-lattice functors weakly preserve epi-pullbacks

Abstract

Suppose $p(x,y,z)$ and $q(x,y,z)$ are terms. If there is a common "ancestor" term $s(z_{1},z_{2},z_{3},z_{4})$ specializing to $p$ and $q$ through identifying some variables \begin{align*} p(x,y,z) & \approx s(x,y,z,z)\\ q(x,y,z) & \approx s(x,x,y,z), \end{align*} then the equation \[ p(x,x,z)\approx q(x,z,z) \] is trivially obtained by syntactic unification of $s(x,y,z,z)$ with $s(x,x,y,z).$ In this note we show that for lattice terms, and more generally for terms of lattice-ordered algebras, the converse is true, too. Given terms $p,q,$ and an equation \begin{equation} p(u_{1},\ldots,u_{m})\approx q(v_{1},\ldots,v_{n})\label{eq:p_eq_q} \end{equation} where $\{u_{1},\ldots,u_{m}\}=\{v_{1},\ldots,v_{n}\},$ there is always an "ancestor term" $s(z_{1},\ldots,z_{r})$ such that $p(x_{1},\ldots,x_{m})$ and $q(y_{1},\ldots,y_{n})$ arise as substitution instances of $s,$ whose unification results in the original equation. In category theoretic terms the above proposition, when restricted to lattices, has a much more concise formulation: Free-lattice functors weakly preserves pullbacks of epis.