Consistent category links

CONTRIBUTING.md

@@ -122,7 +122,7 @@ When contributing new data (categories, functors, properties, implications), ple
 
 - **Special Objects and Morphisms**: For each new category, try to specify its special objects (terminal object, initial object, etc.) in the corresponding table. Also try to specify its special morphisms (isomorphisms, monomorphisms, epimorphisms).
 
-- **Proofs for New Properties**: For every new property, for each existing category or functor, try to find a proof for whether it has this property or not, in case this has not already been deduced automatically via some implication. Use the property detail page to check unknown categories. As mentioned in the section on tests, for a list of selected categories it is actually mandatory to decide their properties.
+- **Proofs for New Properties**: For every new property, for each existing category or functor, try to find a proof for whether it has this property or not, in case this has not already been deduced automatically via some implication. Use the property detail page to check unknown categories. These proofs may also refer to other categories, in which case you may add links to their corresponding pages. As mentioned in the section on tests, for a list of selected categories it is actually mandatory to decide their properties.
 
 - **Counterexamples**: Ensure that at least one category does not satisfy any new property of categories that is added. If no existing category fits, add a new category that does not have the new property. The same remarks apply to properties of functors.
 

databases/catdat/data/categories/2.yaml

@@ -3,7 +3,7 @@ name: discrete category on two objects
 notation: $\2$
 objects: two objects $0$ and $1$
 morphisms: only the two identity morphisms
-description: A concrete representation is the full subcategory of $\CRing$ consisting of the two fields $\IF_2$ and $\IF_3$.
+description: A concrete representation is the full subcategory of <a href="/category/CRing">$\CRing$</a> consisting of the two fields $\IF_2$ and $\IF_3$.
 nlab_link: null
 
 tags:

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

@@ -25,10 +25,10 @@ satisfied_properties:
     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>.
+    reason: This follows in the same way as for <a href="/category/Grp">$\Grp$</a>, 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>.
 
   - property_id: disjoint finite products
-    reason: One can take the same proof as for $\Ring$.
+    reason: One can take the same proof as for <a href="/category/Ring">$\Ring$</a>.
 
 unsatisfied_properties:
   - property_id: balanced
@@ -44,16 +44,16 @@ unsatisfied_properties:
     reason: 'If $\sqcup$ denotes the coproduct of $R$-algebras (see <a href="https://math.stackexchange.com/questions/625874" target="_blank">MSE/625874</a> for their description) and $A$ is an $R$-algebra, the canonical morphism $A \sqcup R^2 \to (A \sqcup R)^2 = A^2$ is usually no isomorphism. For example, for $A = R[X]$ the coproduct on the LHS is not commutative, it has the algebra presentation $\langle X,E : E^2=E \rangle$.'
 
   - property_id: semi-strongly connected
-    reason: This is because already the full subcategory $\CAlg(R)$ of commutative algebras is not semi-strongly connected, see <a href="/category/CAlg(R)">its category page</a> for details.
+    reason: This is because already the full subcategory <a href="/category/CAlg(R)">$\CAlg(R)$</a> of commutative algebras is not semi-strongly connected.
 
   - property_id: co-Malcev
     reason: 'See <a href="https://mathoverflow.net/questions/509552">MO/509552</a>: Consider the forgetful functor $U : \Alg(R) \to \Set$ 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 $R \langle X,Y \rangle / \langle XY-X^2 \rangle$. It is clear that $S$ is reflexive, but not symmetric.'
 
   - property_id: coregular
-    reason: 'We just need to tweak the proof for $\Ring$. 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 $A \to B$, $(x,y) \mapsto \diag(x,y)$ 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.'
+    reason: 'We just need to tweak the proof for <a href="/category/Ring">$\Ring$</a>. 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 $A \to B$, $(x,y) \mapsto \diag(x,y)$ 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.'
 
   - property_id: regular quotient object classifier
-    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).
+    reason: We may copy the proof for <a href="/category/CAlg(R)">$\CRing$</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: >-
@@ -62,7 +62,7 @@ unsatisfied_properties:
       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>.
+    reason: We already know that <a href="/category/CRing">$\CAlg(R)$</a> 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>.
 
   - property_id: effective cocongruences
     reason: 'The counterexample is similar to the one for <a href="/category/Ring">$\Ring$</a>: Let $X := R[p] / (p^2-p)$ with cocongruence $E := R \langle p, q \rangle / (p^2-p, q^2-q, pq-q, qp-p)$.'

databases/catdat/data/categories/Ban.yaml

@@ -24,7 +24,7 @@ satisfied_properties:
     reason: Example 1.48 in <a href="https://ncatlab.org/nlab/show/Locally+Presentable+and+Accessible+Categories" target="_blank">Adamek-Rosicky</a>.
 
   - property_id: cartesian filtered colimits
-    reason: If $X$ is a Banach space and $(Y_i)$ is a filtered diagram of Banach spaces, the canonical map $\colim_i (X \times Y_i) \to X \times \colim_i Y_i$ is the completion of the canonical map in the category of normed vector spaces with non-expansive linear maps. Now the claim follows directly from the <a href="/category/Met">category of metric spaces</a> with non-expansive maps.
+    reason: If $X$ is a Banach space and $(Y_i)$ is a filtered diagram of Banach spaces, the canonical map $\colim_i (X \times Y_i) \to X \times \colim_i Y_i$ is the completion of the canonical map in the category of normed vector spaces with non-expansive linear maps. Now the claim follows directly from <a href="/category/Met">$\Met$</a>.
 
   - property_id: cocartesian cofiltered limits
     reason: 'If $X$ is a Banach space and $(Y_i)$ is a cofiltered diagram of Banach spaces, the canonical map $X \oplus \lim_i Y_i \to \lim_i (X \oplus Y_i)$ is an isomorphism: Since the forgetful functor $\Ban \to \Vect$ preserves finite coproducts and all limits, and $\Vect$ has the claimed property (see <a href="/category-implication/biproducts_cartesian_filtered_colimits">here</a>), the canonical map is bijective. It remains to show that it is isometric. For $(x,y) \in X \oplus \lim_i Y_i$ the norm in the domain is $|x| + \sup_i |y_i|$, and the norm in the codomain is $\sup_i (|x| + |y_i|)$, and these clearly agree.'

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

@@ -26,10 +26,10 @@ satisfied_properties:
     check_redundancy: false
 
   - 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>.
+    reason: This follows in the same way as for <a href="/category/Grp">$\Grp$</a>, 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>.
 
   - property_id: coextensive
-    reason: One can use the same proof as for $\CRing$.
+    reason: One can use the same proof as for <a href="/category/CRing">$\CRing$</a>.
 
 unsatisfied_properties:
   - property_id: balanced
@@ -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 <a href="/category/CRing">$\CRing$</a>: 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>.
@@ -78,7 +78,7 @@ special_morphisms:
     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>.
+    reason: The forgetful functor $\CAlg(R) \to \CRing$ is faithful and hence reflects epimorphisms, but it also preserves epimorphisms since it preserves pushouts (since $\CAlg(R) \cong R / \CRing$). For epimorphisms of commutative rings see the page for <a href="/category/CRing">$\CRing$</a>.
   regular epimorphisms:
     description: surjective homomorphisms
     reason: This holds in every finitary algebraic category.

databases/catdat/data/categories/CMon.yaml

@@ -45,7 +45,7 @@ unsatisfied_properties:
     reason: 'See <a href="https://mathoverflow.net/questions/509552">MO/509552</a>: Consider the forgetful functor $U : \CMon \to \Set$ 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 the free monoid on a single generator and $R$ by the free commutative monoid on two generators $x,y$ subject to the relation $xy=x^2$. It is clear that $R$ is reflexive, but not symmetric.'
 
   - property_id: regular subobject classifier
-    reason: We can use exactly the same proof as for the <a href="/category/Mon">category of monoids</a>.
+    reason: We can use exactly the same proof as for <a href="/category/Mon">$\Mon$</a>.
 
   - property_id: coregular
     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 $u\circ j = v\circ j$.'

databases/catdat/data/categories/CRing.yaml

@@ -29,7 +29,7 @@ satisfied_properties:
     check_redundancy: false
 
   - 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>.
+    reason: This follows in the same way as for <a href="/category/Grp">$\Grp$</a>, 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>.
 
   - property_id: coextensive
     reason: '[Sketch] A ring homomorphism $f : A \times B \to R$ yields the idempotent element $e := f(1,0) \in R$, so that $R \cong eR \times (1-e)R$. Then $f$ decomposes into the ring homomorphisms $f_A : A \to eR$, $f_A(a) := f(a,0)$ and $f_B : B \to (1-e)R$, $f_B(b) := f(0,b)$.'