one-sided barr relator

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partowp 2026-05-06 15:54:12 +01:00
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@ -291,6 +291,7 @@
\newcommand{\emre}{\mathbf{L}} \newcommand{\emre}{\mathbf{L}}
\newcommand{\rto}{\mathrel{\tikz{\draw[-{Stealth}] (0,0) -- (0.4,0); \draw (0.17,0.07) -- (0.17,-0.07);}}} \newcommand{\rto}{\mathrel{\tikz{\draw[-{Stealth}] (0,0) -- (0.4,0); \draw (0.17,0.07) -- (0.17,-0.07);}}}
\newcommand{\powf}{\mathcal{P}} \newcommand{\powf}{\mathcal{P}}
\newcommand{\sappr}{\sqsupseteq}
\newcommand{\simeet}{% \newcommand{\simeet}{%
\mathbin{% \mathbin{%
@ -2209,6 +2210,38 @@ Barr relator is a generalization of the Egli-Milner relator, where the functor i
\todo{Finish.} \todo{Finish.}
\end{proof} \end{proof}
\begin{definition}[One-sided Barr relator]
A relator over a functor $F$ is a one-sided Barr relator, shown by $\overrightarrow{F}$, iff for a partial order $\appr$ over $F$, a relation $r\c X\rto Y$, and a span $(\pi_1\c A\to X,\pi_2\c A\to Y)$ that $r=\pi_2\comp\pi_1^\op$ we have:
\begin{gather*}
\overrightarrow{F}r=F\pi_2\comp\sappr\comp(F\pi_1)^\op
\end{gather*}
\end{definition}
\begin{prop}
For every functor $F\c\Set\to\Set$, the symmetrization of the one-sided Bar relator is equal with the Barr relator.
\end{prop}
\begin{proof}
Where there exist $\pi_1\c A\to X$ and $\pi_2\c A\to Y$ such that $r=\pi_2\comp\pi_1^\op$, we assume that $s \;\hat{\overrightarrow{F}}r\; t$, and we need to show that $s\;F\pi_2\comp(F\pi_1)^\op\;t$. Considering that $r^\op=\pi_1\comp\pi_2^\op$, we have:
\begin{align*}
s \;\hat{\overrightarrow{F}}r\; t&\\
&\iff s\;\overrightarrow{F}r\;t\qquad\&\qquad s \;(\overrightarrow{F}r^\op)^\op\; t\\
&\iff s\;F\pi_2\comp\sappr\comp(F\pi_1)^\op\;t \qquad\&\qquad s\;(F\pi_1\comp\sappr\comp(F\pi_2)^\op)^\op\;t\\
&\iff s\;F\pi_2\comp\sappr\comp(F\pi_1)^\op\;t \qquad\&\qquad s\;F\pi_2\comp\appr\comp(F\pi_1)^\op\;t
\end{align*}
Since $F\pi_1$ is a function, then $(F\pi_1)^\op$ is an injective map, so there exist exactly one $w\in FA$ such that $(F\pi_1)^\op(s)=w$, and:
\begin{gather*}
w\;F\pi_2\comp\sappr\; t \qquad\&\qquad w\;F\pi_2\comp\appr\; t
\end{gather*}
And similarly, since $F\pi_2$ is also a function we have exactly one $v\in FA$ such that $(F\pi_2^\op)(t)=v$, and:
\begin{align*}
&w\;\sappr\; v \qquad\&\qquad w\;\appr\; v\\
\iff&(F\pi_1)^\op(s)\;\sappr\; (F\pi_2^\op)(t) \qquad\&\qquad (F\pi_1)^\op(s)\;\appr\; (F\pi_2^\op)(t)\\
\iff&(F\pi_1)^\op(s)\;=\; (F\pi_2^\op)(t)\\
\iff&s\; F\pi_2\comp(F\pi_1)^\op\;t\\
\iff&s\;\bar{F}r\;t
\end{align*}\qed
\end{proof}
\begin{prop} \begin{prop}
Assuming that $\relar$ is a relator over $F\c\Set\to\Set$, and $\appr_{X}$ and $\appr_{Y}$ are posets over $FX$ and $FY$ respectively, then the relator that takes $r\c X\rto Y$ to $\appr_{X};\relar r;\appr_{Y}$ is a Barr relator. Assuming that $\relar$ is a relator over $F\c\Set\to\Set$, and $\appr_{X}$ and $\appr_{Y}$ are posets over $FX$ and $FY$ respectively, then the relator that takes $r\c X\rto Y$ to $\appr_{X};\relar r;\appr_{Y}$ is a Barr relator.
\end{prop} \end{prop}