View source for SFN:Math-test

$
\newcommand\cc{\mathcal{C}}
\newcommand\set{\mathsf{Set}}
\newcommand\sbcat{\mathit{Category}}

$
{{defn|name=Definition 1|note=Definition of set-based categories|label=df:sbcat
|Let $\cc$ be a small category and let $U:\cc\to\set$ be a functor. We
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:

{{remark|name=Lemma 1|label=lem:1|Let $\cc$ be a category with images satisfying the condition (Sb:a).  Then it satisfies (Sb:b) if and only if

(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:4}
      U\left(f^{\circ}\inc{\im f}{\cod f}\right)=U(f)^{\circ}\inc{\im
        U(f)}{\cod U(f)} 
\end{equation}
for all $f\in\cc$. Moreover, we have
\begin{equation}
      \label{eq:5}
      U(f|c)=U(f)|U(c)\quad\text{and}\quad 
      U\left(f(c)\right)=U(f)\left(U(c)\right)
\end{equation}
for any $f\in\cc$ and $c\subseteq\dom f$.
}}