rename: order -> ordered sets

this was already done before, but I missed these

Author: ScriptRaccoon

database/data/001_categories/005_order-theory.sql

@@ -31,7 +31,7 @@ VALUES
 ),
 (
 	'FinOrd',
-	'category of finite orders',
+	'category of finite ordered sets',
 	'$\mathbf{FinOrd}$',
 	'finite (totally) ordered sets',
 	'order-preserving maps',

database/data/004_property-assignments/FinOrd.sql

@@ -45,13 +45,13 @@ VALUES
 	'FinOrd',
 	'mono-regular',
 	TRUE,
-	'Let $i : A \to B$ be a monomorphism of finite orders. If $A$ is empty, then $i$ is clearly regular, so assume it is not. The map $i$ is injective (see below), hence order-reflecting. Define maps $u,v  : B \to A$ by $u(b) := \max \{a \in A : i(a) \leq b \}$ and $v(b) := \min \{a \in A : b \leq i(a) \}$. These are order-preserving and satisfy $u \circ i = v \circ i$, both sides are $\mathrm{id}_A$. Conversely, if $b \in B$ satisfies $u(b) = v(b) =: a$, then $i(a) \leq b$ and $b \leq i(a)$, hence $b = i(a)$. This shows that $i$ is the equalizer of $u,v$.'
+	'Let $i : A \to B$ be a monomorphism of finite ordered sets. If $A$ is empty, then $i$ is clearly regular, so assume it is not. The map $i$ is injective (see below), hence order-reflecting. Define maps $u,v  : B \to A$ by $u(b) := \max \{a \in A : i(a) \leq b \}$ and $v(b) := \min \{a \in A : b \leq i(a) \}$. These are order-preserving and satisfy $u \circ i = v \circ i$, both sides are $\mathrm{id}_A$. Conversely, if $b \in B$ satisfies $u(b) = v(b) =: a$, then $i(a) \leq b$ and $b \leq i(a)$, hence $b = i(a)$. This shows that $i$ is the equalizer of $u,v$.'
 ),
 (
 	'FinOrd',
 	'epi-regular',
 	TRUE,
-	'Let $f : A \to B$ be an epimorphism of finite orders. It is surjective (see below). Define $u,v : B \to A$ by $u(b) := \min(f^{-1}(b))$ and $v(b) := \max(f^{-1}(b))$. One can easily check that $u,v$ are order-preserving maps with $f \circ u = f \circ v$ (both sides are $\mathrm{id}_B$). Let $h : A \to T$ be an order-preserving map with $h \circ u = h \circ v$. Then $h(a)$ only depends on $b := f(a)$: We have $u(b) \leq a \leq v(b)$, hence $h(u(b)) \leq h(a) \leq h(v(b)) = h(u(b))$. Therefore, there is a unique map $\tilde{h} : B \to T$ with $\tilde{h}(f(a)) = h(a)$, and one easily checks that it is order-preserving. This shows that $f$ is the coequalizer of $u,v$.'
+	'Let $f : A \to B$ be an epimorphism of finite ordered sets. It is surjective (see below). Define $u,v : B \to A$ by $u(b) := \min(f^{-1}(b))$ and $v(b) := \max(f^{-1}(b))$. One can easily check that $u,v$ are order-preserving maps with $f \circ u = f \circ v$ (both sides are $\mathrm{id}_B$). Let $h : A \to T$ be an order-preserving map with $h \circ u = h \circ v$. Then $h(a)$ only depends on $b := f(a)$: We have $u(b) \leq a \leq v(b)$, hence $h(u(b)) \leq h(a) \leq h(v(b)) = h(u(b))$. Therefore, there is a unique map $\tilde{h} : B \to T$ with $\tilde{h}(f(a)) = h(a)$, and one easily checks that it is order-preserving. This shows that $f$ is the coequalizer of $u,v$.'
 ),
 (
 	'FinOrd',