Quotient object classifiers

.vscode/settings.json

@@ -62,6 +62,7 @@
 		"cogenerator",
 		"cogenerators",
 		"cokernel",
+		"cokernels",
 		"colimit",
 		"colimits",
 		"comonad",
@@ -72,6 +73,8 @@
 		"coprime",
 		"coproduct",
 		"coproducts",
+		"coreflection",
+		"coreflective",
 		"coreflexive",
 		"coregular",
 		"corestricts",
@@ -168,6 +171,7 @@
 		"setoid",
 		"Sheafifiable",
 		"simplicial",
+		"subalgebra",
 		"subconjugated",
 		"submonoid",
 		"subobject",

database/data/003_properties/009_topos-theory.sql

@@ -42,19 +42,45 @@ VALUES
 (
 	'subobject classifier',
 	'has a',
-	'A category $\mathcal{C}$ has a <i>subobject classifier</i> if it has finite limits and a monomorphism* $\top : 1 \to \Omega$ such that for every monomorphism $m : A \to B$ there is a unique morphism $\chi_m : B \to \Omega$ such that $B \leftarrow A \rightarrow 1$ is the pullback of $B \rightarrow \Omega \leftarrow 1$. Equivalently, the functor $\mathrm{Sub} : \mathcal{C}^{\mathrm{op}} \to \mathbf{Set}^+$ is representable.<br>
+	'A category $\mathcal{C}$ has a <i>subobject classifier</i> if it has finite limits and a monomorphism* $\top : 1 \to \Omega$ such that for every monomorphism $m : A \to B$ there is a unique morphism $\chi_m : B \to \Omega$ such that
+	<p>$\begin{array}{ccc} A & \rightarrow & B \\ \downarrow && \downarrow \\ 1 & \rightarrow & \Omega \end{array}$</p>
+	is a pullback diagram. Equivalently, the functor $\mathrm{Sub} : \mathcal{C}^{\mathrm{op}} \to \mathbf{Set}^+$ is representable.<br>
 	*Every morphism $1 \to \Omega$ is a split monomorphism anyway.',
 	'https://ncatlab.org/nlab/show/subobject+classifier',
+	'quotient object classifier',
+	TRUE
+),
+(
+	'quotient object classifier',
+	'has a',
+	'A category $\mathcal{C}$ has a <i>quotient object classifier</i> if its dual has a subobject classifier. This means that it has finite colimits and an epimorphism* $\top : \Psi \to 0$ such that for every epimorphism $e : A \to B$ there is a unique morphism $\psi_e : \Psi \to A$ such that
+	<p>$\begin{array}{ccc} \Psi & \rightarrow & 0 \\ \downarrow && \downarrow \\ A & \rightarrow & B \end{array}$</p>
+	is a pushout diagram. Equivalently, the functor $\mathrm{Quot} : \mathcal{C} \to \mathbf{Set}^+$ is representable.<br>
+	*Every morphism $\Psi \to 0$ is a split epimorphism anyway.',
 	NULL,
+	'subobject classifier',
 	TRUE
 ),
 (
 	'regular subobject classifier',
 	'has a',
-	'A category $\mathcal{C}$ has a <i>regular subobject classifier</i> if it has finite limits and a regular monomorphism* $\top : 1 \to \Omega$ such that for every regular monomorphism $m : A \to B$ there is a unique morphism $\chi_m : B \to \Omega$ such that $B \leftarrow A \rightarrow 1$ is the pullback of $B \rightarrow \Omega \leftarrow 1$. Equivalently, the functor $\mathrm{Sub}_{\mathrm{reg}} : \mathcal{C}^{\mathrm{op}} \to \mathbf{Set}^+$ is representable.<br>
+	'A category $\mathcal{C}$ has a <i>regular subobject classifier</i> if it has finite limits and a regular monomorphism* $\top : 1 \to \Omega$ such that for every regular monomorphism $m : A \to B$ there is a unique morphism $\chi_m : B \to \Omega$ such that
+	<p>$\begin{array}{ccc} A & \rightarrow & B \\ \downarrow && \downarrow \\ 1 & \rightarrow & \Omega \end{array}$</p>
+	is a pullback diagram. Equivalently, the functor $\mathrm{Sub}_{\mathrm{reg}} : \mathcal{C}^{\mathrm{op}} \to \mathbf{Set}^+$ is representable.<br>
 	*Every morphism $1 \to \Omega$ is a split monomorphism and hence regular anyway.',
 	'https://ncatlab.org/nlab/show/subobject+classifier',
+	'regular quotient object classifier',
+	TRUE
+),
+(
+	'regular quotient object classifier',
+	'has a',
+	'A category $\mathcal{C}$ has a <i>regular quotient object classifier</i> if its dual has a regular subobject classifier. This means that it has finite colimits and a regular epimorphism* $\top : \Psi \to 0$ such that for every regular epimorphism $e : A \to B$ there is a unique morphism $\psi_e : \Psi \to A$ such that
+	<p>$\begin{array}{ccc} \Psi & \rightarrow & 0 \\ \downarrow && \downarrow \\ A & \rightarrow & B \end{array}$</p>
+	is a pushout diagram. Equivalently, the functor $\mathrm{Quot}_{\mathrm{reg}} : \mathcal{C} \to \mathbf{Set}^+$ is representable.<br>
+	*Every morphism $\Psi \to 0$ is a split epimorphism anyway.',
 	NULL,
+	'regular subobject classifier',
 	TRUE
 ),
 (

database/data/003_properties/100_related_properties.sql

@@ -119,6 +119,12 @@ VALUES
 ('subobject classifier', 'elementary topos'),
 ('subobject classifier', 'finitely complete'),
 ('subobject classifier', 'regular subobject classifier'),
+('regular subobject classifier', 'subobject classifier'),
+('regular subobject classifier', 'finitely complete'),
+('quotient object classifier', 'finitely cocomplete'),
+('quotient object classifier', 'regular quotient object classifier'),
+('regular quotient object classifier', 'quotient object classifier'),
+('regular quotient object classifier', 'finitely cocomplete'),
 ('infinitary distributive', 'distributive'),
 ('infinitary distributive', 'finite products'),
 ('infinitary distributive', 'coproducts'),
@@ -195,8 +201,6 @@ VALUES
 ('counital', 'finitely cocomplete'),
 ('counital', 'pointed'),
 ('counital', 'co-Malcev'),
-('regular subobject classifier', 'subobject classifier'),
-('regular subobject classifier', 'finitely complete'),
 ('cosifted limits', 'complete'),
 ('cosifted limits', 'cofiltered limits'),
 ('cosifted limits', 'coreflexive equalizers'),

database/data/004_property-assignments/Alg(R).sql

@@ -76,4 +76,10 @@ VALUES
 	'coregular',
 	FALSE,
 	'We just need to tweak the proof for $\mathbf{Ring}$. Since $R \neq 0$, there is an infinite field $K$ with a homomorphism $R \to K$. Since $K$ is infinite, we may choose some $\lambda \in K \setminus \{0,1\}$. Let $B := M_2(K)$ and $A := K \times K$. Then $A \to B$, $(x,y) \mapsto \mathrm{diag}(x,y)$ is a regular monomorphism: A direct calculation shows that a matrix is diagonal iff it commutes with $M := \bigl(\begin{smallmatrix} 1 & 0 \\ 0 & \lambda \end{smallmatrix}\bigr)$, so that $A \to B$ is the equalizer of the identity $B \to B$ and the conjugation $B \to B$, $X \mapsto M X M^{-1}$. Consider the homomorphism $A \to K$, $(a,b) \mapsto a$. We claim that $K \to K \sqcup_A B$ is not a monomorphism, because in fact, the pushout $K \sqcup_A B$ is zero: Since $A \to K$ is surjective with kernel $0 \times K$, the pushout is $B/\langle 0 \times K \rangle$, which is $0$ because $B$ is simple (<a href="https://math.stackexchange.com/questions/22629" target="_blank">proof</a>) or via a direct calculation with elementary matrices.'
+),
+(
+	'Alg(R)',
+	'regular quotient object classifier',
+	FALSE,
+	'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 $\mathbf{Alg}(R)$ would produce one in $\mathbf{CAlg}(R)$ by <a href="/lemma/subobject_classifiers_coreflection">this lemma</a> (dualized).'
 );

database/data/004_property-assignments/CAlg(R).sql

@@ -76,4 +76,10 @@ VALUES
 	'co-Malcev',
 	FALSE,
 	'See <a href="https://mathoverflow.net/questions/509552">MO/509552</a>: Consider the forgetful functor $U : \mathbf{CAlg}(R) \to \mathbf{Set}$ 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.'
+),
+(
+	'CAlg(R)',
+	'regular quotient object classifier',
+	FALSE,
+	'The strategy is similar to the one for $\mathbf{CRing}$: 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.'
 );

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

@@ -69,11 +69,17 @@ VALUES
 	'CMon',
 	'regular subobject classifier',
 	FALSE,
-	'Assume that $\Omega$ is a regular subobject classifier. Since the trivial monoid is a zero object, every regular submonoid $U \subseteq M$ of any commutative monoid $M$ would have the form $\{m \in M : h(m) = 1 \}$ for some homomorphism $M \to \Omega$. Now take any commutative monoid $M$ with zero that has two different homomorphisms with zero $f,g : M \rightrightarrows N$ (for example, let $M = N = \{0\} \cup \{x^n : n \geq 0\}$ be the free monoid with zero on one generator, $f(x) = 0$,and $g(x) = x$). Take their equalizer $U \subseteq M$, and choose a homomorphism $h : M \to \Omega$ with $U = \{m \in M : h(m) = 1\}$. Since $0 \in U$, we have $h(0)=1$. But then for all $m \in M$ we have $h(m) = h(m) h(0) = h(m 0) = h(0) = 1$, i.e. $U = M$, which yields the contradiction $f = g$.'
+	'We can use exactly the same proof as for the <a href="/category/Mon">category of monoids</a>.'
 ),
 (
 	'CMon',
 	'coregular',
 	FALSE,
 	'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 $\mathbb{N}^2$ and its submonoid $U\coloneqq\{(m,n)\mid m\ge n\}$ with the inclusion $i\colon U\hookrightarrow\mathbb{N}^2$. 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 $\mathbf{CMon}$ 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 $u\circ j = v\circ j$.'
+),
+(
+	'CMon',
+	'regular quotient object classifier',
+	FALSE,
+	'If $P \in \mathbf{CMon}$ 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 $f : (\mathbb{N},+) \to (\{0,1\},\vee)$ 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.'
 );

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

@@ -76,4 +76,10 @@ VALUES
 	'co-Malcev',
 	FALSE,
 	'See <a href="https://mathoverflow.net/questions/509552">MO/509552</a>: Consider the forgetful functor $U : \mathbf{CRing} \to \mathbf{Set}$ and the relation $R \subseteq U^2$ defined by $R(A) := \{(a,b) \in U(A)^2 : ab = a^2\}$. Both are representable: $U$ by $\mathbb{Z}[X]$ and $R$ by $\mathbb{Z}[X,Y] / \langle XY-X^2 \rangle$. It is clear that $R$ is reflexive, but not symmetric.'
+),
+(
+	'CRing',
+	'regular quotient object classifier',
+	FALSE,
+	'Assume that $P \to \mathbb{Z}$ is a regular quotient object classifier. If $J$ denotes its kernel, this means that every ideal $I \subseteq A$ of any commutative ring 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 ring of $\sigma$. But then $I$ is generated by elements in the fixed ring. This fails for $A = \mathbb{Z}[X]$, $I = \langle X \rangle$, $\sigma(X)=-X$. The fixed ring is $\mathbb{Z}[X^2]$, and if $I$ was generated by elements $f \in \mathbb{Z}[X^2] \cap I$, they would be multiples of $X^2$, but $X$ is not a multiple of $X^2$.'
 );

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

@@ -70,4 +70,10 @@ VALUES
 	'counital',
 	FALSE,
 	'The canonical morphism $F_2 = \mathbb{Z} \sqcup \mathbb{Z} \to \mathbb{Z} \times \mathbb{Z}$ is not a monomorphism since $F_2$ is not abelian.'
-);
+),
+(
+	'Grp',
+	'regular quotient object classifier',
+	FALSE,
+	'Assume that $\mathbf{Grp}$ 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 $G$.'
+);

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

@@ -70,5 +70,11 @@ VALUES
 	'regular subobject classifier',
 	FALSE,
 	'Assume that $\Omega$ is a regular subobject classifier. Since the trivial monoid is a zero object, every regular submonoid $U \subseteq M$ of any monoid $M$ would have the form $\{m \in M : h(m) = 1 \}$ for some homomorphism $M \to \Omega$. Now take any monoid $M$ with zero that has two different homomorphisms with zero $f,g : M \rightrightarrows N$ (for example, let $M = N = \{0\} \cup \{x^n : n \geq 0\}$ be the free monoid with zero on one generator, $f(x) = 0$,and $g(x) = x$). Take their equalizer $U \subseteq M$, and choose a homomorphism $h : M \to \Omega$ with $U = \{m \in M : h(m) = 1\}$. Since $0 \in U$, we have $h(0)=1$. But then for all $m \in M$ we have $h(m) = h(m) h(0) = h(m 0) = h(0) = 1$, i.e. $U = M$, which yields the contradiction $f = g$.'
+),
+(
+	'Mon',
+	'regular quotient object classifier',
+	FALSE,
+	'We can just copy the proof for the <a href="/category/CMon">category of commutative monoids</a>. Alternatively, we may use <a href="/lemma/subobject_classifiers_coreflection">this lemma</a> (dualized).'
 );
 

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

@@ -76,4 +76,10 @@ VALUES
 	'coregular',
 	FALSE,
 	'Let $B := M_2(\mathbb{Q})$ and $A := \mathbb{Q}^2$. Then $A \to B$, $(x,y) \mapsto \mathrm{diag}(x,y)$ is a regular monomorphism: A direct calculation shows that a matrix is diagonal iff it commutes with $M := \bigl(\begin{smallmatrix} 1 & 0 \\ 0 & 2 \end{smallmatrix}\bigr)$, so that $A \to B$ is the equalizer of the identity $B \to B$ and the conjugation $B \to B$, $X \mapsto M X M^{-1}$. Consider the homomorphism $A \to K$, $(a,b) \mapsto a$. We claim that $K \to K \sqcup_A B$ is not a monomorphism, because in fact, the pushout $K \sqcup_A B$ is zero: Since $A \to K$ is surjective with kernel $0 \times K$, the pushout is $B/\langle 0 \times K \rangle$, which is $0$ because $B$ is simple (<a href="https://math.stackexchange.com/questions/22629" target="_blank">proof</a>) or via a direct calculation with elementary matrices.'
+),
+(
+	'Ring',
+	'regular quotient object classifier',
+	FALSE,
+	'We may copy the proof for the <a href="/category/CRing">category of commutative rings</a> (since the proof there did not use that $P$ is commutative). Alternatively, any regular quotient object classifier in $\mathbf{Ring}$ would produce one in $\mathbf{CRing}$ by <a href="/lemma/subobject_classifiers_coreflection">this lemma</a> (dualized).'
 );

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

@@ -64,4 +64,10 @@ VALUES
 	'coregular',
 	FALSE,
 	'We can copy the proof for the <a href="/category/Ring">category of rings</a>. In short, the inclusion of diagonal matrices $\mathbb{Q}^2 \hookrightarrow M_2(\mathbb{Q})$ is a regular monomorphism, but becomes zero after taking the pushout with $p_1 : \mathbb{Q}^2 \twoheadrightarrow \mathbb{Q}$ because $M_2(\mathbb{Q})$ is simple.'
+),
+(
+	'Rng',
+	'regular quotient object classifier',
+	FALSE,
+	'Assume that $\mathbf{Rng}$ has a regular quotient object classifier $P$. Consider the functor $N : \mathbf{Ab} \to \mathbf{Rng}$ that equips an abelian group with zero multiplication. It is fully faithful and has a left adjoint mapping a rng $R$ to the abelian group $R/R^2$. If $R$ is a rng with zero multiplication and $R \to S$ is a surjective homomorphism, then $S$ has zero multiplication. Therefore, the assumptions of <a href="/lemma/subobject_classifiers_coreflection">this lemma</a> (dualized) apply and we conclude that $P/P^2$ is a regular quotient object classifier of $\mathbf{Ab}$. But we already know that <a href="/category/Ab">this category</a> has no such object (in fact, the only additive categories with such an object are trivial by <a href="https://math.stackexchange.com/questions/4086192" target="_blank">MSE/4086192</a>).'
 );

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

@@ -64,4 +64,10 @@ VALUES
 	'unital',
 	FALSE,
 	'The joint image of $X \to X \times Y \leftarrow Y$ is just $\{(x,0) : x \in X\} \cup \{(0,y) : y \in Y\}$ (where $0$ denotes the base point), which is clearly a proper subset of $X \times Y$ when both $X,Y$ are non-trivial.'
+),
+(
+	'Set*',
+	'quotient object classifier',
+	FALSE,
+	'If there was a quotient object classifier, every surjective pointed map would be a cokernel. However, every cokernel is "injective away from the base point". Formally, if $p : A \to B$ is a cokernel in $\mathbf{Set}_*$, it has the property that $p(x)=p(y) \neq 0$ implies $x=y$ (where $0$ denotes the base point). Clearly this is not satisfied for every surjective pointed map, consider $(\mathbb{N},0) \to (\{0,1\},0)$ defined by $0 \mapsto 0$ and $x \mapsto 1$ for $x > 0$.'
 );

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

@@ -124,6 +124,12 @@ VALUES
 	'unital',
 	FALSE,
 	'The joint image of $X \to X \times Y \leftarrow Y$ is just $\{(x,0) : x \in X\} \cup \{(0,y) : y \in Y\}$ (where $0$ denotes the base point), which is clearly a proper subset of $X \times Y$ when both $X,Y$ are non-trivial.'
+),
+(
+	'Top*',
+	'regular quotient object classifier',
+	FALSE,
+	'We can recycle the proof for the <a href="/category/Set*">category of pointed sets</a> using discrete topological spaces.'
 );
 
 

database/data/005_implications/008_topos-theory-implications.sql

@@ -56,17 +56,17 @@ VALUES
 	FALSE
 ),
 (
-	'subobject_thin',
-	'["thin", "subobject classifier"]',
+	'subobject_classifier_collapse',
+	'["subobject classifier", "strict terminal object"]',
 	'["trivial"]',
-	'Let $\mathcal{C}$ be a thin category with a subobject classifier $\Omega$. Every object has a subobject, hence a morphism to $\Omega$. It must be unique since $\mathcal{C}$ is thin. Hence, $\Omega$ is terminal. If $X$ is any object, then $1 \cong \mathrm{Hom}(X,\Omega) \cong \mathrm{Sub}(X)$ shows that every morphism $Y \to X$ is isomorphic to $\mathrm{id}_X : X \to X$. Apply this to $X = \Omega$ to finish the proof.',
+	'Since $1 \to \Omega$ is an isomorphism, every monomorphism must be an isomorphism. Applying this to the equalizer of a pair of morphisms, we see that the category is thin. But in a thin category, every morphism is a monomorphism. So every object $X$ has a unique isomorphism $X \to 1$.',
 	FALSE
 ),
 (
 	'additive_trivial_condition',
 	'["regular subobject classifier", "additive"]',
 	'["trivial"]',
-	'See <a href="https://math.stackexchange.com/questions/4086192" target="_blank">MSE/4086192</a> (the proof works with these assumptions).',
+	'See <a href="https://math.stackexchange.com/a/5132767/1650" target="_blank">MSE/4086192</a>.',
 	FALSE
 ),
 (