format yaml files: replace " with ', unescape backslashes

Author: ScriptRaccoon

databases/catdat/data/categories/Ab.yaml

@@ -30,7 +30,7 @@ satisfied_properties:
 
 unsatisfied_properties:
   - property_id: split abelian
-    reason: "The short exact sequence $0 \\xrightarrow{} \\IZ \\xrightarrow{p} \\IZ \\xrightarrow{} \\IZ/p \\xrightarrow{} 0$ does not split. "
+    reason: 'The short exact sequence $0 \xrightarrow{} \IZ \xrightarrow{p} \IZ \xrightarrow{} \IZ/p \xrightarrow{} 0$ does not split. '
 
   - property_id: skeletal
     reason: This is trivial.
@@ -54,10 +54,10 @@ special_morphisms:
     reason: This characterization holds in every algebraic category.
   monomorphisms:
     description: injective homomorphisms
-    reason: "This holds in every finitary algebraic category: the forgetful functor to $\\Set$ is faithful, hence reflects monomorphisms, and it is continuous (even representable), hence preserves monomorphisms."
+    reason: 'This holds in every finitary algebraic category: the forgetful functor to $\Set$ is faithful, hence reflects monomorphisms, and it is continuous (even representable), hence preserves monomorphisms.'
   epimorphisms:
     description: surjective homomorphisms
-    reason: "For the non-trivial direction, if $f : A \\to B$ is an epimorphism, then $p \\circ f = 0$ for the projection  $p : B \\to B/f(A)$ implies that $p = 0$, so that $B = f(A)$."
+    reason: 'For the non-trivial direction, if $f : A \to B$ is an epimorphism, then $p \circ f = 0$ for the projection  $p : B \to B/f(A)$ implies that $p = 0$, so that $B = f(A)$.'
   regular monomorphisms:
     description: same as monomorphisms
     reason: This is because the category is mono-regular.

databases/catdat/data/categories/Ab_fg.yaml

@@ -60,7 +60,7 @@ special_morphisms:
     reason: This follows exactly as for abelian groups.
   monomorphisms:
     description: injective homomorphisms
-    reason: "Let $f : A \\to B$ be a monomorphism of finitely generated abelian groups. Let $a \\in A$ be in the kernel of $a$. Then we may view $a$ as a morphism $a : \\IZ \\to A$ with $f \\circ a = 0$, and $\\IZ$ is finitely generated. Hence, $a = 0$."
+    reason: 'Let $f : A \to B$ be a monomorphism of finitely generated abelian groups. Let $a \in A$ be in the kernel of $a$. Then we may view $a$ as a morphism $a : \IZ \to A$ with $f \circ a = 0$, and $\IZ$ is finitely generated. Hence, $a = 0$.'
   epimorphisms:
     description: surjective homomorphisms
     reason: Use the same proof as for abelian groups.

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

@@ -22,7 +22,7 @@ satisfied_properties:
     reason: Take the algebraic theory of an $R$-algebra.
 
   - property_id: strict terminal object
-    reason: "If $f : 0 \\to A$ is an algebra homomorphism, then $A$ satisfies $1=f(1)=f(0)=0$, so that $A=0$."
+    reason: 'If $f : 0 \to A$ is an algebra homomorphism, then $A$ satisfies $1=f(1)=f(0)=0$, so that $A=0$.'
 
   - property_id: Malcev
     reason: This follows in the same way as for groups, see also Example 2.2.5 in <a href="https://ncatlab.org/nlab/show/Malcev,+protomodular,+homological+and+semi-abelian+categories" target="_blank">Malcev, protomodular, homological and semi-abelian categories</a>.
@@ -83,7 +83,7 @@ special_morphisms:
     reason: This characterization holds in every algebraic category.
   monomorphisms:
     description: injective ring homomorphisms
-    reason: "This holds in every finitary algebraic category: the forgetful functor to $\\Set$ is faithful, hence reflects monomorphisms, and it is continuous (even representable), hence preserves monomorphisms."
+    reason: 'This holds in every finitary algebraic category: the forgetful functor to $\Set$ is faithful, hence reflects monomorphisms, and it is continuous (even representable), hence preserves monomorphisms.'
   regular epimorphisms:
     description: surjective homomorphisms
     reason: This holds in every finitary algebraic category.

databases/catdat/data/categories/Ban.yaml

@@ -58,15 +58,15 @@ unsatisfied_properties:
     reason: By using the concrete description of products and coproducts, for the constant family $X_n = \IC$ the canonical morphism $\coprod_n X_n \to \prod_n X_n$ becomes the canonical inclusion map $\ell^1 \hookrightarrow \ell^\infty$. This is not an epimorphism (i.e., has no dense image) since the closure of the image is precisely $c_0$. So for example, $(1,1,\dotsc)$ is not contained in the closure of the image.
 
   - property_id: cofiltered-limit-stable epimorphisms
-    reason: "We show that epimorphisms are not stable under sequential limits. Let $X_n = Y_n = \\IC$ for all $n \\geq 0$. The transition morphism $Y_{n+1} \\to Y_n$ is the identity, and the transition morphism $X_{n+1} \\to X_n$ is $x \\mapsto x/2$. The morphisms $X_n \\to Y_n$, $x \\mapsto x/2^n$ are compatible with the transitions, and they are surjective, hence epimorphisms. Now we check $\\lim_n X_n = 0$: An element $(x_n) \\in \\lim_n X_n$ is a family of complex numbers satisfying $x_n = x_{n+1}/2$ <i>and</i> $\\sup_n |x_n| < \\infty$. But then $x_n = 2^n x_0$ and this can only be bounded when $x_0=0$. Hence, $0 = \\lim_n X_n \\to \\lim_n Y_n = \\IC$ is no epimorphism."
+    reason: 'We show that epimorphisms are not stable under sequential limits. Let $X_n = Y_n = \IC$ for all $n \geq 0$. The transition morphism $Y_{n+1} \to Y_n$ is the identity, and the transition morphism $X_{n+1} \to X_n$ is $x \mapsto x/2$. The morphisms $X_n \to Y_n$, $x \mapsto x/2^n$ are compatible with the transitions, and they are surjective, hence epimorphisms. Now we check $\lim_n X_n = 0$: An element $(x_n) \in \lim_n X_n$ is a family of complex numbers satisfying $x_n = x_{n+1}/2$ <i>and</i> $\sup_n |x_n| < \infty$. But then $x_n = 2^n x_0$ and this can only be bounded when $x_0=0$. Hence, $0 = \lim_n X_n \to \lim_n Y_n = \IC$ is no epimorphism.'
 
 special_objects:
   initial object:
     description: trivial Banach space
   terminal object:
     description: trivial Banach space
   coproducts:
-    description: "The coproduct of a family of Banach spaces $(B_i)$ is the subspace $\\bigl\\{x \\in \\prod_i B_i : \\sum_i |x_i| < \\infty\\bigr\\}$ equipped with the $1$-norm $|x| := \\sum_i |x_i|$."
+    description: 'The coproduct of a family of Banach spaces $(B_i)$ is the subspace $\bigl\{x \in \prod_i B_i : \sum_i |x_i| < \infty\bigr\}$ equipped with the $1$-norm $|x| := \sum_i |x_i|$.'
   products:
     description: direct products with the $\sup$-norm
 
@@ -76,10 +76,10 @@ special_morphisms:
     reason: This is easy.
   monomorphisms:
     description: injective linear contractions
-    reason: "The unit ball functor $U : \\Ban \\to \\Set$ is faithful and representable (by $\\IC$), hence reflects and preserves monomorphisms."
+    reason: 'The unit ball functor $U : \Ban \to \Set$ is faithful and representable (by $\IC$), hence reflects and preserves monomorphisms.'
   epimorphisms:
     description: linear contractions with dense image
-    reason: "Let $f : X \\to Y$ be an epimorphism of Banach spaces. The subspace $U := \\overline{f(X)} \\subseteq Y$ is closed. It is well-known that the quotient $Y/U$ is also a Banach space with a projection $p : Y \\to Y/U$. Since $p \\circ f = 0 = 0 \\circ f$, we infer $p = 0$, so that $U = Y$."
+    reason: 'Let $f : X \to Y$ be an epimorphism of Banach spaces. The subspace $U := \overline{f(X)} \subseteq Y$ is closed. It is well-known that the quotient $Y/U$ is also a Banach space with a projection $p : Y \to Y/U$. Since $p \circ f = 0 = 0 \circ f$, we infer $p = 0$, so that $U = Y$.'
   regular monomorphisms:
     description: closed embeddings
     reason: The non-trivial direction follows from the <a href="https://math.stackexchange.com/questions/319867" target="_blank">well-known fact</a> that for every closed subspace of a Banach space its quotient space is again a Banach space.

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

@@ -22,7 +22,7 @@ satisfied_properties:
     reason: Take the algebraic theory of a commutative ring.
 
   - property_id: strict terminal object
-    reason: "If $f : 0 \\to R$ is a homomorphism, then $R$ satisfies $1=f(1)=f(0)=0$, so that $R=0$."
+    reason: 'If $f : 0 \to R$ is a homomorphism, then $R$ satisfies $1=f(1)=f(0)=0$, so that $R=0$.'
     check_redundancy: false
 
   - property_id: Malcev
@@ -42,7 +42,7 @@ unsatisfied_properties:
     reason: This is trivial.
 
   - property_id: countably codistributive
-    reason: "The canonical homomorphism $A \\otimes_R R^{\\IN} \\to A^{\\IN}$ is given by $a \\otimes (r_n)_n \\mapsto (r_n a)_n$ and does not have to be surjective: Since $R \\neq 0$, there is a commutative $R$-algebra $K$ which is a field. Now take $A := K[X]$ and consider the sequence $(X^n)_{n} \\in A^{\\IN}$."
+    reason: 'The canonical homomorphism $A \otimes_R R^{\IN} \to A^{\IN}$ is given by $a \otimes (r_n)_n \mapsto (r_n a)_n$ and does not have to be surjective: Since $R \neq 0$, there is a commutative $R$-algebra $K$ which is a field. Now take $A := K[X]$ and consider the sequence $(X^n)_{n} \in A^{\IN}$.'
 
   - property_id: semi-strongly connected
     reason: Choose a maximal ideal $\mathfrak{m}$ of $R$, so $K := R/\mathfrak{m}$ is a field. If $\CAlg(R)$ is semi-strongly connected, then also $\CAlg(K)$ is semi-strongly connected. This has been disproven in <a href="https://math.stackexchange.com/questions/5129689" target="_blank">MSE/5129689</a>.
@@ -54,7 +54,7 @@ unsatisfied_properties:
     reason: 'See <a href="https://mathoverflow.net/questions/509552">MO/509552</a>: Consider the forgetful functor $U : \CAlg(R) \to \Set$ and the relation $S \subseteq U^2$ defined by $S(A) := \{(a,b) \in U(A)^2 : ab = a^2\}$. Both are representable: $U$ by $R[X]$ and $S$ by $R[X,Y] / \langle XY-X^2 \rangle$. It is clear that $S$ is reflexive, but not symmetric.'
 
   - property_id: regular quotient object classifier
-    reason: "The strategy is similar to the one for $\\CRing$: Assume that $P \\to R$ is a regular quotient object classifier. If $J$ denotes the kernel of $P \\to R$, every ideal $I \\subseteq A$ of any commutative $R$-algebra has the form $I = \\langle \\varphi(J) \\rangle$ for a unique homomorphism $\\varphi : P \\to A$. If $\\sigma : A \\to A$ is an automorphism with $\\sigma(I)=I$, then uniqueness gives us $\\sigma \\circ \\varphi = \\varphi$, which means that $\\varphi(J)$ lies in $A^{\\sigma}$, the fixed algebra of $\\sigma$. But then $I$ is generated by elements in $A^{\\sigma} \\cap I$. If $K$ is a residue field of $R$, this fails for $A = K[X,Y]$, $I = \\langle X,Y \\rangle$, $\\sigma(X)=Y$, $\\sigma(Y)=X$. The fixed algebra is the subalgebra of symmetric polynomials, which is $K[X+Y,XY]$. So $\\langle X,Y \\rangle$ is generated by symmetric polynomials without constant term, which implies $\\langle X,Y \\rangle \\subseteq \\langle X+Y,XY \\rangle$ in $K[X,Y]$. But reducing an equation like $X = a(X,Y) \\cdot (X+Y) + b(X,Y) \\cdot (XY)$ modulo $\\langle X^2,Y^2,XY \\rangle$ yields a contradiction."
+    reason: 'The strategy is similar to the one for $\CRing$: Assume that $P \to R$ is a regular quotient object classifier. If $J$ denotes the kernel of $P \to R$, every ideal $I \subseteq A$ of any commutative $R$-algebra has the form $I = \langle \varphi(J) \rangle$ for a unique homomorphism $\varphi : P \to A$. If $\sigma : A \to A$ is an automorphism with $\sigma(I)=I$, then uniqueness gives us $\sigma \circ \varphi = \varphi$, which means that $\varphi(J)$ lies in $A^{\sigma}$, the fixed algebra of $\sigma$. But then $I$ is generated by elements in $A^{\sigma} \cap I$. If $K$ is a residue field of $R$, this fails for $A = K[X,Y]$, $I = \langle X,Y \rangle$, $\sigma(X)=Y$, $\sigma(Y)=X$. The fixed algebra is the subalgebra of symmetric polynomials, which is $K[X+Y,XY]$. So $\langle X,Y \rangle$ is generated by symmetric polynomials without constant term, which implies $\langle X,Y \rangle \subseteq \langle X+Y,XY \rangle$ in $K[X,Y]$. But reducing an equation like $X = a(X,Y) \cdot (X+Y) + b(X,Y) \cdot (XY)$ modulo $\langle X^2,Y^2,XY \rangle$ yields a contradiction.'
 
   - property_id: cofiltered-limit-stable epimorphisms
     reason: Let $K$ be a field over $R$. Consider the sequence of projections $\cdots \to K[X]/\langle X^2 \rangle \to K[X]/\langle X \rangle$ and the constant sequence $\cdots \to K[X] \to K[X]$. The surjective homomorphisms $K[X] \to K[X]/\langle X^n \rangle$ induce the inclusion $K[X] \hookrightarrow K[[X]]$ in the limit, where $K[[X]]$ is the algebra of formal power series. It is clearly not surjective, but this is not sufficient, we need to argue that it is not an epimorphism in $\CAlg(R)$, or equivalently, in $\CRing$. For a proof, see <a href="https://math.stackexchange.com/questions/2391187" target="_blank">MSE/2391187</a>.
@@ -75,7 +75,7 @@ special_morphisms:
     reason: This characterization holds in every algebraic category.
   monomorphisms:
     description: injective homomorphisms
-    reason: "This holds in every finitary algebraic category: the forgetful functor to $\\Set$ is faithful, hence reflects monomorphisms, and it is continuous (even representable), hence preserves monomorphisms."
+    reason: 'This holds in every finitary algebraic category: the forgetful functor to $\Set$ is faithful, hence reflects monomorphisms, and it is continuous (even representable), hence preserves monomorphisms.'
   epimorphisms:
     description: a homomorphism of algebras which is an epimorphism of commutative rings
     reason: The forgetful functor $\CAlg(R) \to \Ring$ is faithful and hence reflects epimorphisms, but it also preserves epimorphisms since it preserves pushouts (since $\CAlg(R) \cong R / \Ring$). For epimorphisms of commutative rings see their <a href="/category/CRing">detail page</a>.

databases/catdat/data/categories/CMon.yaml

@@ -35,7 +35,7 @@ unsatisfied_properties:
     reason: This is trivial.
 
   - property_id: Malcev
-    reason: "Consider the submonoid $\\{(a,b) : a \\leq b \\}$ of $\\IN^2$."
+    reason: 'Consider the submonoid $\{(a,b) : a \leq b \}$ of $\IN^2$.'
 
   - property_id: cogenerator
     reason: See <a href="https://mathoverflow.net/questions/509232" target="_blank">MO/509232</a>.
@@ -50,7 +50,7 @@ unsatisfied_properties:
     reason: 'We can show this analogously to the case of commutative rings <a href="https://math.stackexchange.com/a/3746890" target="_blank">MSE/3746890</a>. Consider the commutative monoid $\IN^2$ and its submonoid $U\coloneqq\{(m,n)\mid m\ge n\}$ with the inclusion $i\colon U\hookrightarrow\IN^2$. Then, the pushout of $i$ along itself is $\langle x,y,z : x+y=x+z \rangle$, and the equalizer of the cokernel pair of $i$ is $D\coloneqq\{(m,n)\mid m=0 \implies n=0 \}$. If the category $\CMon$ were coregular, the canonical inclusion $j\colon U\hookrightarrow D$ would have to be an epimorphism. However, it is not: let $I\coloneqq\{0,1\}$ be the two-element commutative monoid with $1+1=1$, and let $u,v\colon D \rightrightarrows I$ be the morphisms defined by $u^{-1}(0)=\{(0,0)\}$ and $v^{-1}(0)=\{(0,0),(1,2)\}$; then we have $u\circ j = v\circ j$.'
 
   - property_id: regular quotient object classifier
-    reason: "If $P \\in \\CMon$ is a regular quotient object classifier, this means that every surjective homomorphism of commutative monoids $A \\to B$ is the cokernel of a unique homomorphism $P \\to A$. But there are many surjective homomorphisms which are no cokernels at all: Consider the Boolean monoid $(\\{0,1\\},\\vee)$ with $1 \\vee 1 = 1$ and the surjective homomorphism $f : (\\IN,+) \\to (\\{0,1\\},\\vee)$ defined by $f(0)=0$ and $f(n)=1$ for $n \\geq 1$. It has trivial kernel, but is no isomorphism, so it cannot be a cokernel."
+    reason: 'If $P \in \CMon$ is a regular quotient object classifier, this means that every surjective homomorphism of commutative monoids $A \to B$ is the cokernel of a unique homomorphism $P \to A$. But there are many surjective homomorphisms which are no cokernels at all: Consider the Boolean monoid $(\{0,1\},\vee)$ with $1 \vee 1 = 1$ and the surjective homomorphism $f : (\IN,+) \to (\{0,1\},\vee)$ defined by $f(0)=0$ and $f(n)=1$ for $n \geq 1$. It has trivial kernel, but is no isomorphism, so it cannot be a cokernel.'
 
   - property_id: CSP
     reason: First of all, epimorphisms in $\CMon$ are preserved and reflected by the forgetful functor to $\Mon$ (see below). Furthermore, if $M \to N$ is an epimorphism in $\Mon$ and $M$ is infinite, then $\card(N) \leq \card(M)$ (see <a href="https://mathoverflow.net/questions/510431/" target="_blank">MO/510431</a>). This implies that in $\CMon$ the canonical homomorphism $\bigoplus_{n \geq 0} \IN \to \prod_{n \geq 0} \IN$ is not an epimorphism because its domain is countable and its codomain is uncountable.
@@ -71,7 +71,7 @@ special_morphisms:
     reason: This characterization holds in every algebraic category.
   monomorphisms:
     description: injective homomorphisms
-    reason: "This holds in every finitary algebraic category: the forgetful functor to $\\Set$ is faithful, hence reflects monomorphisms, and it is continuous (even representable), hence preserves monomorphisms."
+    reason: 'This holds in every finitary algebraic category: the forgetful functor to $\Set$ is faithful, hence reflects monomorphisms, and it is continuous (even representable), hence preserves monomorphisms.'
   epimorphisms:
     description: A morphism in $\CMon$ is an epimorphism iff it is an epimorphism in $\Mon$, which in turn can be characterized by <a href="https://en.wikipedia.org/wiki/Isbell's_zigzag_theorem" target="_blank">Isbell's zigzag theorem</a>.
     reason: