SFN:Math-test
$ \newcommand\cc{\mathcal{C}} \newcommand\set{\mathsf{Set}} \newcommand\sbcat{\mathit{Category}} \newcommand{\lcat}[1][S]{\mathbb{L}(#1)} \newcommand{\rcat}[1][S]{\mathbb{R}(#1)} \newcommand{\cc}{\mathcal{C}} \newcommand{\cd}{\mathcal{D}} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\de}{\delta} \newcommand{\De}{\Delta} \newcommand{\la}{\lambda} \newcommand{\La}{\Lambda} \newcommand{\set}{\mathsf{Set}} \newcommand{\sbcat}{\mathcal{S}-\text{Category}} \newcommand{\preord}{\mathcal{P}} \newcommand{\scat}{\mathsf{Cat_{o}}} \newcommand{\vrt}[1]{\boldsymbol{\mathfrak{v}}{#1}} \newcommand{\Grp}{\mathsf{Grp}} \newcommand{\abgrp}{\mathsf{Ab}} \def\inc#1#2{{\jmath}_{{#1}}^{{#2}}} \newcommand{\cod}{\operatorname{cod}} \newcommand{\dom}{\operatorname{dom}} \newcommand{\e}{\epsilon} \newcommand{\im}{\operatorname{im}} \newcommand{\dom}{\operatorname{dom}} \newcommand{\vmap}[1][{}]{{#1}_{\mathfrak{A}}}
$
Definition 1 (Definition of set-based categories)
say that $\cc$ is set-based ($\sbcat$ for short) with respect to $U$ if the pair $(\cc,U)$ satisfy the following:
(Sb:a) $U$ is an embedding.
(Sb:b) $U(\im f)=\im U(f)$ for all $f\in\cc$.
(Sb:c) The functor $U$ has the following property: for $c,c'\in\vrt \cc$ and $x\in U(c)\cap U(c')$ there is $d\in\vrt\cc$ such that
\[ d\subseteq c,\quad d\subseteq c'\quad\text{and}\quad x\in U(d). \]Notice that the condition (Sb:b) implies that the functor $U$ preserves image--factorizations:
Lemma 1
(Sb:d) $U(f)^{\circ}=U(f^{\circ})$ for all $f\in\cc$.
If (Sb:b) (or (Sb:b)$^{*}$) hold, then \begin{equation}\label{eq:5}
U(f\mid c)=U(f)