remove |- from yaml seed files, put string values into one line for now

Author: ScriptRaccoon

databases/catdat/data/categories/Alg(R).yaml

@@ -56,10 +56,7 @@ unsatisfied_properties:
     reason: We may copy the proof for the <a href="/category/CAlg(R)">category of commutative algebras</a> (since the proof there did not use that $P$ is commutative). Alternatively, any regular quotient object classifier in $\Alg(R)$ would produce one in $\CAlg(R)$ by <a href="/lemma/subobject_classifiers_coreflection">this lemma</a> (dualized).
 
   - property_id: cocartesian cofiltered limits
-    reason: |-
-      Consider the ring $A = R[X]$ and the sequence of rings $B_n = R[Y]/(Y^{n+1})$ with projections $B_{n+1} \to B_n$, whose limit is $R[[Y]]$. Every element in the coproduct of rings $R[X] \sqcup R[[Y]]$ has a finite "free product" length. Now consider the elements
-      	$$w_n = (1 + XY) (1+XY^2) \cdots (1+X Y^n) \in A \sqcup B_n.$$
-      	Because of $w_n \equiv w_{n-1} \bmod Y^n$ these form an element $w \in \lim_n (A \sqcup B_n)$. Expanding $w_n$, the longest term is $XY XY^2 \cdots X Y^n$ of "free product" length $2n$, which is unbounded.
+    reason: 'Consider the ring $A = R[X]$ and the sequence of rings $B_n = R[Y]/(Y^{n+1})$ with projections $B_{n+1} \to B_n$, whose limit is $R[[Y]]$. Every element in the coproduct of rings $R[X] \sqcup R[[Y]]$ has a finite "free product" length. Now consider the elements $$w_n = (1 + XY) (1+XY^2) \cdots (1+X Y^n) \in A \sqcup B_n.$$ Because of $w_n \equiv w_{n-1} \bmod Y^n$ these form an element $w \in \lim_n (A \sqcup B_n)$. Expanding $w_n$, the longest term is $XY XY^2 \cdots X Y^n$ of "free product" length $2n$, which is unbounded.'
 
   - property_id: cofiltered-limit-stable epimorphisms
     reason: We already know that $\CAlg(R)$ does not have this property. Now apply the contrapositive of the dual of <a href="/lemma/filtered-monos">this lemma</a> to the forgetful functor $\CAlg(R) \to \Alg(R)$. It preserves epimorphisms by <a href="https://math.stackexchange.com/questions/5133488" target="_blank">MSE/5133488</a>.

databases/catdat/data/categories/Cat.yaml

@@ -59,10 +59,7 @@ unsatisfied_properties:
     reason: We already know that $\Set$ does not have this property. Now apply the contrapositive of the dual of <a href="/lemma/filtered-monos">this lemma</a> to the functor $\Set \to \Cat$ that maps a set to its discrete category.
 
   - property_id: effective cocongruences
-    reason: |-
-      The counterexample is similar to the one for <a href="/category/Mon">$\Mon$</a>: Let $X$ be the <a href="/category/walking_idempotent">walking idempotent</a>, and let $E$ be the delooping of the monoid with presentation
-      	$$\langle p, q \mid p^2=p,\, q^2=q,\, pq=q,\, qp=p \rangle.$$
-      	The induced relation on functors in $[X, \C]$ is that $F \sim G$ if and only if $F$ and $G$ send the object of $X$ to the same object of $\C$, and they send the idempotent of $X$ to idempotent morphisms $a, b$ in $\C$ satisfying $ab=b$, $ba=a$. From here, the proof that this gives a cocongruence on $\Cat$ which is not effective is similar to the one in $\Mon$.
+    reason: 'The counterexample is similar to the one for <a href="/category/Mon">$\Mon$</a>: Let $X$ be the <a href="/category/walking_idempotent">walking idempotent</a>, and let $E$ be the delooping of the monoid with presentation $$\langle p, q \mid p^2=p,\, q^2=q,\, pq=q,\, qp=p \rangle.$$ The induced relation on functors in $[X, \C]$ is that $F \sim G$ if and only if $F$ and $G$ send the object of $X$ to the same object of $\C$, and they send the idempotent of $X$ to idempotent morphisms $a, b$ in $\C$ satisfying $ab=b$, $ba=a$. From here, the proof that this gives a cocongruence on $\Cat$ which is not effective is similar to the one in $\Mon$.'
 
 special_objects:
   initial object:

databases/catdat/data/categories/Delta.yaml

@@ -52,12 +52,7 @@ satisfied_properties:
     reason: The <a href="/category/FinOrd">proof for $\FinOrd$</a> also works for $\FinSet \setminus \{\varnothing\}$.
 
   - property_id: cosifted
-    reason: |-
-      Let $X,Y \in \Delta$. We may pick $x \in X$, $y \in Y$. Then there is a "point span" $X \xleftarrow{x} [0] \xrightarrow{y} Y$. Every span $X \xleftarrow{f} Z \xrightarrow{g} Y$ is connected to such a point span: Pick $z \in Z$. This defines a morphism of spans:
-          $$\begin{CD} X @<{f(z)}<< [0] @>{g(z)}>> Y \\ @| @VV{z}V @| \\ X @<<{f}< Z @>>{g}> Y \end{CD}$$
-          It remains to show that all point spans are connected to each other. Assume $x_0,x_1 \in X$ and $y \in Y$, w.l.o.g. $x_0 \leq x_1$. Define the map $f : [1] \to X$ by $f(0) = x_0$, $f(1) = x_1$, and the map $g : [1] \to Y$ by $g(0)=g(1)=y$. They are order-preserving and fit into a zig-zag of spans:
-          $$\begin{CD} X @<{x_0}<< [0] @>{y}>> Y \\ @| @V{0}VV @| \\ X @<{f}<< [1] @>{g}>> Y \\ @| @A{1}AA @| \\ X @<{x_1}<< [0] @>{y}>> Y \end{CD}$$
-          This shows that the choice of $x \in X$ does not matter, and for $y \in Y$ the proof is the same.
+    reason: 'Let $X,Y \in \Delta$. We may pick $x \in X$, $y \in Y$. Then there is a "point span" $X \xleftarrow{x} [0] \xrightarrow{y} Y$. Every span $X \xleftarrow{f} Z \xrightarrow{g} Y$ is connected to such a point span: Pick $z \in Z$. This defines a morphism of spans: $$\begin{CD} X @<{f(z)}<< [0] @>{g(z)}>> Y \\ @| @VV{z}V @| \\ X @<<{f}< Z @>>{g}> Y \end{CD}$$ It remains to show that all point spans are connected to each other. Assume $x_0,x_1 \in X$ and $y \in Y$, w.l.o.g. $x_0 \leq x_1$. Define the map $f : [1] \to X$ by $f(0) = x_0$, $f(1) = x_1$, and the map $g : [1] \to Y$ by $g(0)=g(1)=y$. They are order-preserving and fit into a zig-zag of spans: $$\begin{CD} X @<{x_0}<< [0] @>{y}>> Y \\ @| @V{0}VV @| \\ X @<{f}<< [1] @>{g}>> Y \\ @| @A{1}AA @| \\ X @<{x_1}<< [0] @>{y}>> Y \end{CD}$$ This shows that the choice of $x \in X$ does not matter, and for $y \in Y$ the proof is the same.'
 
 unsatisfied_properties:
   - property_id: strict terminal object

databases/catdat/data/categories/FS.yaml

@@ -65,14 +65,7 @@ unsatisfied_properties:
     reason: Assume that the copower $X := 2+2$ exists. Since we have a surjective map $2 \to X$, the set $X$ has at most $2$ elements. The codiagonal $X \to 2$ shows that $X$ has at least $2$ elements. Thus, $X \cong 2$. For all finite sets $Y$ we get a bijection $\Hom(2,Y) \cong \Hom(2,Y)^2$, in particular the cardinalities are the same. For $Y=2$ this gives the contradiction $2 = 4$.
 
   - property_id: locally cocartesian coclosed
-    reason: |-
-      If $X$ is a finite set, the coslice category $X / \FS$ is thin and in fact equivalent to the lattice of equivalence relations on $X$. If $X$ has $\geq 3$ elements, it is not codistributive* and <a href="/category-implication/dual_distributive_criterion">hence</a> not cocartesian coclosed: For simplicity assume $X = \{a,b,c\}$. The bottom element $\bot$ corresponds to the partition $\{\{a\},\{b\},\{c\}\}$, the top element $\top$ to the partition $\{\{a,b,c\}\}$. Now consider the three equivalence relations $E_1,E_2,E_3$ corresponding to the three partitions
-      	$$\{\{a,b\},\{c\}\}, \, \{\{a,c\},\{b\}\}, \, \{\{b,c\},\{a\}\}.$$
-      	Then
-      	$$E_1 \vee (E_2 \wedge E_3) = E_1 \vee \bot = E_1,$$
-      	but
-      	$$(E_1 \vee E_2) \wedge (E_1 \vee E_3) = \top \wedge \top = \top.$$
-      	*For thin categories, the properties codistributive and distributive <a href="/category-implication/distributive_duality">are equivalent</a>.
+    reason: 'If $X$ is a finite set, the coslice category $X / \FS$ is thin and in fact equivalent to the lattice of equivalence relations on $X$. If $X$ has $\geq 3$ elements, it is not codistributive* and <a href="/category-implication/dual_distributive_criterion">hence</a> not cocartesian coclosed: For simplicity assume $X = \{a,b,c\}$. The bottom element $\bot$ corresponds to the partition $\{\{a\},\{b\},\{c\}\}$, the top element $\top$ to the partition $\{\{a,b,c\}\}$. Now consider the three equivalence relations $E_1,E_2,E_3$ corresponding to the three partitions $$\{\{a,b\},\{c\}\}, \, \{\{a,c\},\{b\}\}, \, \{\{b,c\},\{a\}\}.$$ Then $$E_1 \vee (E_2 \wedge E_3) = E_1 \vee \bot = E_1,$$ but $$(E_1 \vee E_2) \wedge (E_1 \vee E_3) = \top \wedge \top = \top.$$ *For thin categories, the properties codistributive and distributive <a href="/category-implication/distributive_duality">are equivalent</a>.'
 
   - property_id: multi-initial object
     reason: If a multi-initial object exists, then the connected component consisting of non-empty finite sets has an initial object $X$. Then, any non-empty finite set cannot have a cardinality strictly greater than $X$,  which is a contradiction.

databases/catdat/data/categories/FinSet.yaml

@@ -51,10 +51,7 @@ unsatisfied_properties:
     reason: This is trivial.
 
   - property_id: natural numbers object
-    reason: |-
-      If $(N,z,s)$ is a natural numbers object, then
-      	$$1 \xrightarrow{z} N \xleftarrow{s} N$$
-      	is a coproduct cocone by <a href="https://ncatlab.org/nlab/show/Sketches+of+an+Elephant" target="_blank">Johnstone</a>, Part A, Lemma 2.5.5. But there is no finite set $N$ with $N \cong 1 + N$.
+    reason: 'If $(N,z,s)$ is a natural numbers object, then $$1 \xrightarrow{z} N \xleftarrow{s} N$$ is a coproduct cocone by <a href="https://ncatlab.org/nlab/show/Sketches+of+an+Elephant" target="_blank">Johnstone</a>, Part A, Lemma 2.5.5. But there is no finite set $N$ with $N \cong 1 + N$.'
 
 special_objects:
   initial object:

databases/catdat/data/categories/Fld.yaml

@@ -63,10 +63,7 @@ unsatisfied_properties:
     reason: Every field has a non-trivial extension, for instance, the rational function field over itself in one variable. Hence, a multi-terminal object never exists.
 
   - property_id: cofiltered-limit-stable epimorphisms
-    reason: |-
-      Inside of $\IF_p(X)$ consider the descending sequence of subfields
-      	$$\IF_p(X) \supseteq \IF_p(X^p) \supseteq \IF_p(X^{p^2}) \supseteq \cdots,$$
-      	whose intersection is $\IF_p$. Each $\IF_p(X^{p^n}) \hookrightarrow \IF_p(X)$ is purely inseparable, hence an epimorphism, but in the limit we get $\IF_p \hookrightarrow \IF_p(X)$, which is not even algebraic.
+    reason: 'Inside of $\IF_p(X)$ consider the descending sequence of subfields $$\IF_p(X) \supseteq \IF_p(X^p) \supseteq \IF_p(X^{p^2}) \supseteq \cdots,$$ whose intersection is $\IF_p$. Each $\IF_p(X^{p^n}) \hookrightarrow \IF_p(X)$ is purely inseparable, hence an epimorphism, but in the limit we get $\IF_p \hookrightarrow \IF_p(X)$, which is not even algebraic.'
 
 special_objects: {}
 

databases/catdat/data/categories/FreeAb.yaml

@@ -34,10 +34,7 @@ satisfied_properties:
     reason: It is easy to check that $\IZ$ is a cogenerator for free abelian groups.
 
   - property_id: regular
-    reason: |-
-      This follows formally from the fact that $\Ab$ is regular and $\FreeAb$ is closed under subobjects and finite products: By Prop. 2.5 in the <a href="https://ncatlab.org/nlab/show/regular+category">nlab</a> it suffices to prove that there are pullback-stable (reg epi, mono)-factorizations. Every homomorphism $f  : A \to B$ in $\FreeAb$ factors as $f = i \circ p : A \twoheadrightarrow C \hookrightarrow B$, where $C$ is a subgroup, hence free, and $A \to C$ is surjective. Clearly, surjective homomorphisms are pullback-stable. It remains to show that they coincide with the regular epimorphisms.<br>
-      	(1) If $f : A \to B$ is surjective, it is the coequalizer of $A \times_B A \rightrightarrows A$ in $\Ab$. Since $A \times_B A$ is free abelian, $f$ is also an coequalizer in $\FreeAb$.<br>
-      	(2) If $f : A \to B$ is a regular epimorphism in $\FreeAb$, consider the factorization $f = i \circ p$ as above. Since $f$ is an extremal epimorphism, $i$ must be an isomorphism, so that $f$ is surjective.
+    reason: 'This follows formally from the fact that $\Ab$ is regular and $\FreeAb$ is closed under subobjects and finite products: By Prop. 2.5 in the <a href="https://ncatlab.org/nlab/show/regular+category">nlab</a> it suffices to prove that there are pullback-stable (reg epi, mono)-factorizations. Every homomorphism $f : A \to B$ in $\FreeAb$ factors as $f = i \circ p : A \twoheadrightarrow C \hookrightarrow B$, where $C$ is a subgroup, hence free, and $A \to C$ is surjective. Clearly, surjective homomorphisms are pullback-stable. It remains to show that they coincide with the regular epimorphisms.<br> (1) If $f : A \to B$ is surjective, it is the coequalizer of $A \times_B A \rightrightarrows A$ in $\Ab$. Since $A \times_B A$ is free abelian, $f$ is also an coequalizer in $\FreeAb$.<br> (2) If $f : A \to B$ is a regular epimorphism in $\FreeAb$, consider the factorization $f = i \circ p$ as above. Since $f$ is an extremal epimorphism, $i$ must be an isomorphism, so that $f$ is surjective.'
 
 unsatisfied_properties:
   - property_id: countable powers
@@ -53,9 +50,7 @@ unsatisfied_properties:
     reason: See <a href="https://mathoverflow.net/questions/509715" target="_blank">MO/509715</a>.
 
   - property_id: effective cocongruences
-    reason: |-
-      We will let $E$ be the abelian group with presentation $\langle a, b, c \mid a - b = 2c \rangle$, with two morphisms $\IZ \rightrightarrows E$ given by $1\mapsto a$, $1\mapsto b$. Note that $E$ is free with basis $\{ b, c \}$. Then $\Hom(E, G) \cong \{ (x, y, z) \in G^3 \mid x - y = 2z \}$. Observe that since $G$ is torsion-free, the projection onto the first two coordinates is injective; and $(x, y)$ is in the image precisely when $x \equiv y \pmod{2G}$, which gives an equivalence relation. Therefore, $E$ gives a cocongruence on $\IZ$.<br>
-      	On the other hand, if $E$ were the cokernel pair of $h : H \to \IZ$, that would mean that for $x, y : \IZ \to G$, $x \equiv y \pmod{2G}$ if and only if $x \circ h = y \circ h$. In particular, from the case $G := \IZ$, $x := 2 \id$, $y := 0$, we would have $2h = 0$. That implies $h = 0$, but then that would give $\id_{\IZ} \equiv 0 \pmod{2}$, resulting in a contradiction.
+    reason: 'We will let $E$ be the abelian group with presentation $\langle a, b, c \mid a - b = 2c \rangle$, with two morphisms $\IZ \rightrightarrows E$ given by $1\mapsto a$, $1\mapsto b$. Note that $E$ is free with basis $\{ b, c \}$. Then $\Hom(E, G) \cong \{ (x, y, z) \in G^3 \mid x - y = 2z \}$. Observe that since $G$ is torsion-free, the projection onto the first two coordinates is injective; and $(x, y)$ is in the image precisely when $x \equiv y \pmod{2G}$, which gives an equivalence relation. Therefore, $E$ gives a cocongruence on $\IZ$.<br> On the other hand, if $E$ were the cokernel pair of $h : H \to \IZ$, that would mean that for $x, y : \IZ \to G$, $x \equiv y \pmod{2G}$ if and only if $x \circ h = y \circ h$. In particular, from the case $G := \IZ$, $x := 2 \id$, $y := 0$, we would have $2h = 0$. That implies $h = 0$, but then that would give $\id_{\IZ} \equiv 0 \pmod{2}$, resulting in a contradiction.'
 
 category_property_comments:
   - property_id: accessible

databases/catdat/data/categories/Grp.yaml

@@ -62,12 +62,7 @@ unsatisfied_properties:
     reason: 'Assume that $\Grp$ has a (regular) quotient object classifier, i.e. a group $P$ such that every surjective homomorphism $G \to H$ is the cokernel of a unique homomorphism $\varphi : P \to G$. Equivalently, every normal subgroup $N \subseteq G$ is $\langle \langle \varphi(P) \rangle \rangle$ for a unique homomorphism $\varphi : P \to G$, where $\langle \langle - \rangle \rangle$ denotes the normal closure. If $c_g : G \to G$ denotes the conjugation with $g \in G$, then the images of $\varphi$ and $c_g \circ \varphi$ have the same normal closures, so the homomorphisms must be equal. In other words, $\varphi$ factors through the center $Z(G)$. But then every normal subgroup of $G$, in particular $G$ itself, would be contained in $Z(G)$, which is wrong for every non-abelian group $G$.'
 
   - property_id: cocartesian cofiltered limits
-    reason: |-
-      For cofiltered diagrams of groups $(H_i)$ and a group $G$ the canonical homomorphism
-      	$$\textstyle \alpha : G \sqcup \lim_i H_i \to \lim_i (G \sqcup H_i)$$
-      	is injective, but often fails to be surjective because the components of an element in the image have bounded <i>free product length</i> (the number of factors appearing in the reduced form). Specifically, consider the free groups $G = \langle y \rangle$ and $H_n = \langle x_1,\dotsc,x_n \rangle$ for $n \in \IN$ with the truncation maps $H_{n+1} \to H_n$, $x_{n+1} \mapsto 1$. Define
-      	$$p_n := x_1 \, y \, x_2 \, y \, \cdots \, x_{n-1} \, y \, x_n \, y^{-(n-1)} \in G \sqcup H_n.$$
-      	If we substitute $x_{n+1}=1$ in $p_{n+1}$, we get $p_n$. Thus, we have $p = (p_n) \in \lim_n (G \sqcup H_n)$. This element does not lie in the image of $\alpha$ since the free product length of $p_n$ (which is well-defined) is $2n$, which is unbounded.
+    reason: 'For cofiltered diagrams of groups $(H_i)$ and a group $G$ the canonical homomorphism $$\textstyle \alpha : G \sqcup \lim_i H_i \to \lim_i (G \sqcup H_i)$$ is injective, but often fails to be surjective because the components of an element in the image have bounded <i>free product length</i> (the number of factors appearing in the reduced form). Specifically, consider the free groups $G = \langle y \rangle$ and $H_n = \langle x_1,\dotsc,x_n \rangle$ for $n \in \IN$ with the truncation maps $H_{n+1} \to H_n$, $x_{n+1} \mapsto 1$. Define $$p_n := x_1 \, y \, x_2 \, y \, \cdots \, x_{n-1} \, y \, x_n \, y^{-(n-1)} \in G \sqcup H_n.$$ If we substitute $x_{n+1}=1$ in $p_{n+1}$, we get $p_n$. Thus, we have $p = (p_n) \in \lim_n (G \sqcup H_n)$. This element does not lie in the image of $\alpha$ since the free product length of $p_n$ (which is well-defined) is $2n$, which is unbounded.'
 
   - property_id: CSP
     reason: The canonical homomorphism $\coprod_{n \geq 0} \IZ \to \prod_{n \geq 0} \IZ$ is not surjective because its domain is countable and its codomain is uncountable. Hence it is no epimorphism.

databases/catdat/data/categories/J2.yaml

@@ -28,13 +28,7 @@ unsatisfied_properties:
     reason: This is trivial.
 
   - property_id: semi-strongly connected
-    reason: |-
-      There is a bijection $\alpha = (\lambda,\rho) : \IN \to \IN \times \IN$ such that $\lambda$ has a fixed point, but $\rho$ does not (see below). Then the isomorphism $\beta := (\rho,\lambda)$ has the opposite property. There cannot be any morphism $(\IN,\alpha) \to (\IN,\beta)$, as it would map the fixed point of $\lambda$ to a fixed point of $\rho$, and likewise there is no morphism $(\IN,\beta) \to (\IN,\alpha)$.<br>
-      	To construct $\alpha$ or rather $\alpha^{-1} : \IN \times \IN \to \IN$, we can alter the standard bijection $(n,m) \mapsto 2^n (2m+1) - 1$ as follows:
-      	$$\alpha^{-1}(n,m) = \begin{cases} 2 & (n,m) = (0,0) \\ 0 & (n,m) = (0,1) \\ 2^n (2m+1) - 1 & \text{otherwise} \end{cases}$$
-      	Then $\alpha(0)=(0,1)$, i.e. $\lambda(0)=0$. The function $\rho$ has no fixed point, i.e. $\alpha^{-1}(n,m) \neq m$ for all $n,m$. Namely, if $(n,m)=(0,0)$, then $\alpha^{-1}(n,m)=2 \neq m$. If $(n,m)=(0,1)$, then $\alpha^{-1}(n,m)=0 \neq m$. Otherwise,
-      	$$\alpha^{-1}(n,m) = 2^n (2m+1) - 1 \geq (2m+1)-1 = 2m \geq m,$$
-      	and equality can only hold if $m=0$ and $n=0$, which we already excluded.
+    reason: 'There is a bijection $\alpha = (\lambda,\rho) : \IN \to \IN \times \IN$ such that $\lambda$ has a fixed point, but $\rho$ does not (see below). Then the isomorphism $\beta := (\rho,\lambda)$ has the opposite property. There cannot be any morphism $(\IN,\alpha) \to (\IN,\beta)$, as it would map the fixed point of $\lambda$ to a fixed point of $\rho$, and likewise there is no morphism $(\IN,\beta) \to (\IN,\alpha)$.<br> To construct $\alpha$ or rather $\alpha^{-1} : \IN \times \IN \to \IN$, we can alter the standard bijection $(n,m) \mapsto 2^n (2m+1) - 1$ as follows: $$\alpha^{-1}(n,m) = \begin{cases} 2 & (n,m) = (0,0) \\ 0 & (n,m) = (0,1) \\ 2^n (2m+1) - 1 & \text{otherwise} \end{cases}$$ Then $\alpha(0)=(0,1)$, i.e. $\lambda(0)=0$. The function $\rho$ has no fixed point, i.e. $\alpha^{-1}(n,m) \neq m$ for all $n,m$. Namely, if $(n,m)=(0,0)$, then $\alpha^{-1}(n,m)=2 \neq m$. If $(n,m)=(0,1)$, then $\alpha^{-1}(n,m)=0 \neq m$. Otherwise, $$\alpha^{-1}(n,m) = 2^n (2m+1) - 1 \geq (2m+1)-1 = 2m \geq m,$$ and equality can only hold if $m=0$ and $n=0$, which we already excluded.'
 
 special_objects:
   initial object: