Difference between revisions of "SFN:Math-test"
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preserves image--factorizations: | preserves image--factorizations: | ||
| − | {{ | + | {{dfn|name=Lemma 1|label=lem:1|statement=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 $f\in\cc$. | (Sb:d) $U(f)^{\circ}=U(f^{\circ})$ for all $f\in\cc$. | ||
Revision as of 05:50, 29 March 2014
$ \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: