SFN:Math-test

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

$

Definition 1  (Definition of set-based categories)

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:

Lemma 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