Finn Lawler allegory

Recall the notion of a unitary pre-tabular allegory. Bicategories of relations are equivalent, but this has yet to be shown rigorously and published, as far as I’m aware.

This PDF file gives an explicit argument, which I put at bicategory of relations until Mike Shulman suggested the neater proof that’s there now. But I’ll keep this here anyway.

The following lemma is used at allegory.

Lemma

If $f \colon X \to A$ and $g \colon X \to B$ are maps in a unitary pre-tabular allegory, then $\langle f, g \rangle = \pi_1^\circ f \cap \pi_2^\circ g$

Proof

Let $r = \pi_1^\circ f \cap \pi_2^\circ g$. Then $\pi_1 r = f$ and $\pi_2 r = g$, by the modular law and the fact that projections tabulate top elements. So $r = \langle f, g \rangle$ if and only if it is a map.

$$\begin{aligned} r r^\circ & = (\pi_1^\circ f \cap \pi_2^\circ g)(f^\circ \pi_1 \cap g^\circ \pi_2^\circ) \\ & \leq \pi_1^\circ f f^\circ \pi_1 \cap \pi_2^\circ g g^\circ \pi_2 & \text{(distrib.)} \\ & \leq \pi_1^\circ \pi_1 \cap \pi_2^\circ \pi_2 \\ & = 1 & \text{(proj'ns tabulate)} \end{aligned}$$

For the unit inequality, we have

$$\begin{aligned} r^\circ r & = (f^\circ \pi_1 \cap g^\circ \pi_2^\circ) (\pi_1^\circ f \cap \pi_2^\circ g) \\ & = (f^\circ \pi_1 \cap g^\circ \pi_2^\circ) \pi_1^\circ f \cap (f^\circ \pi_1 \cap g^\circ \pi_2^\circ) \pi_2^\circ g\\ & = (f^\circ \cap g^\circ \pi_2^\circ \pi_1) f \cap (f^\circ \pi_1 \pi_2^\circ \cap g^\circ) g & \text{(mod.)} \\ & = f^\circ f \cap g^\circ g & \text{(dist., proj. tab. top)}\\ & \geq 1 \cap 1 \\ & = 1 \end{aligned}$$

The second equality follows from distributivity $(s \cap t) r \leq s r \cap t r$, which is an equality because $r r^\circ \leq 1$.