Add properties: cartesian filtered colimits + cocartesian cofiltered limits

.vscode/settings.json

@@ -124,6 +124,7 @@
 		"injection",
 		"injections",
 		"injective",
+		"injectivity",
 		"Isbell",
 		"Johnstone",
 		"Kashiwara",

database/data/001_categories/100_related-categories.sql

@@ -11,6 +11,7 @@ VALUES
 ('Alg(R)', 'CAlg(R)'),
 ('Alg(R)', 'R-Mod'),
 ('Alg(R)', 'Ring'),
+('Ban','Met'),
 ('B', 'FI'),
 ('B', 'FS'),
 ('BG', 'BG_f'),
@@ -58,6 +59,7 @@ VALUES
 ('Meas', 'Top'),
 ('Met', 'Met_c'),
 ('Met', 'Met_oo'),
+('Met', 'Ban'),
 ('Met_c', 'Met'),
 ('Met_c', 'Met_oo'),
 ('Met_c', 'Top'),

database/data/003_properties/003_limits-colimits-behavior.sql

@@ -82,6 +82,24 @@ VALUES
 	NULL,
 	TRUE
 ),
+(
+	'cartesian filtered colimits',
+	'has',
+	'In a category $\mathcal{C}$, which we assume to have filtered colimits and finite products, we say that <i>filtered colimits are cartesian</i> if for every finite set $I$ the product functor $\prod : \mathcal{C}^I \to \mathcal{C}$ preserves filtered colimits. Equivalently, for every $X \in \mathcal{C}$ the functor $X \times - : \mathcal{C} \to \mathcal{C}$ preserves filtered colimits.<br>
+	This is no standard terminology, it has been suggested in <a href="https://mathoverflow.net/questions/510240" target="_blank">MO/510240</a>. We have added it to the database since it clarifies the relationship between many related properties.',
+	NULL,
+	'cocartesian cofiltered limits',
+	TRUE
+),
+(
+	'cocartesian cofiltered limits',
+	'has',
+	'In a category $\mathcal{C}$, which we assume to have cofiltered limits and finite coproducts, we say that <i>cofiltered limits are cocartesian</i> if for every finite set $I$ the coproduct functor $\coprod : \mathcal{C}^I \to \mathcal{C}$ preserves cofiltered limits. Equivalently, for every $X \in \mathcal{C}$ the functor $X \sqcup - : \mathcal{C} \to \mathcal{C}$ preserves cofiltered limits.<br>
+	This is no standard terminology, its dual has been suggested in <a href="https://mathoverflow.net/questions/510240" target="_blank">MO/510240</a>. We have added it to the database since it clarifies the relationship between many related properties.',
+	NULL,
+	'cartesian filtered colimits',
+	TRUE
+),
 (
 	'disjoint finite coproducts',
 	'has',

database/data/003_properties/100_related-properties.sql

@@ -151,6 +151,12 @@ VALUES
 ('codistributive', 'finite coproducts'),
 ('exact filtered colimits', 'filtered colimits'),
 ('exact filtered colimits', 'finitely complete'),
+('exact filtered colimits', 'cartesian filtered colimits'),
+('cartesian filtered colimits', 'filtered colimits'),
+('cartesian filtered colimits', 'finite products'),
+('cartesian filtered colimits', 'exact filtered colimits'),
+('cocartesian cofiltered limits', 'cofiltered limits'),
+('cocartesian cofiltered limits', 'finite coproducts'),
 ('generator', 'generating set'),
 ('generating set', 'generator'),
 ('Grothendieck abelian', 'abelian'),

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

@@ -82,4 +82,12 @@ VALUES
 	'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).'
+),
+(
+	'Alg(R)',
+	'cocartesian cofiltered limits',
+	FALSE,
+	'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
+	<br>$w_n = (1 + XY) (1+XY^2) \cdots (1+X Y^n) \in A \sqcup B_n$.</br>
+	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.'
 );

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

@@ -23,6 +23,18 @@ VALUES
 	TRUE,
 	'Example 1.48 in <a href="https://ncatlab.org/nlab/show/Locally+Presentable+and+Accessible+Categories" target="_blank">Adamek-Rosicky</a>.'
 ),
+(
+	'Ban',
+	'cartesian filtered colimits',
+	TRUE,
+	'If $X$ is a Banach space and $(Y_i)$ is a filtered diagram of Banach spaces, the canonical map $\mathrm{colim}_i (X \times Y_i) \to X \times \mathrm{colim}_i Y_i$ 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.'
+),
+(
+	'Ban',
+	'cocartesian cofiltered limits',
+	TRUE,
+	'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 $\mathbf{Ban} \to \mathbf{Vect}$ preserves finite coproducts and all limits, and $\mathbf{Vect}$ 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.'
+),
 (
 	'Ban',
 	'cogenerator',

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

@@ -82,4 +82,14 @@ VALUES
 	'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 $G$.'
+),
+(
+	'Grp',
+	'cocartesian cofiltered limits',
+	FALSE,
+	'For cofiltered diagrams of groups $(H_i)$ and a group $G$ the canonical homomorphism
+	<br>$\alpha : G \sqcup \lim_i H_i \to \lim_i (G \sqcup H_i)$<br>
+	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 $n \in \mathbb{N}$ with the truncation maps $H_{n+1} \to H_n$, $x_{n+1} \mapsto 1$. Define
+	<br>$p_n := x_1 \, y \, x_2 \, y \, \cdots \, x_{n-1} \, y \, x_n \, y^{-(n-1)} \in G \sqcup H_n$.<br>
+	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.'
 );

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

@@ -67,7 +67,7 @@ VALUES
 ),
 (
 	'Haus',
-	'cartesian closed',
+	'cartesian filtered colimits',
 	FALSE,
 	'It is shown in <a href="https://math.stackexchange.com/questions/1255678">MSE/1255678</a> that $\mathbb{Q} \times - : \mathbf{Top} \to \mathbf{Top}$ does not preserve sequential colimits (so that it cannot be a left adjoint). The same example also works in $\mathbf{Haus}$: Surely $\mathbb{Q}$ is Hausdorff, $X_n$ is Hausdorff, as is their colimit $X$, and the colimit (taken in $\mathbf{Top}$) of the $X_n \times \mathbb{Q}$ admits a bijective continuous map to a Hausdorff space, therefore is also Hausdorff, meaning it is also the colimit taken in $\mathbf{Haus}$.'
 ),

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

@@ -67,13 +67,7 @@ VALUES
 ),
 (
 	'Meas',
-	'cartesian closed',
-	FALSE,
-	'The functors $X \times -$ do not preserve filtered colimits by <a href="https://math.stackexchange.com/questions/5027218" target="_blank">MSE/5027218</a>, hence cannot be left adjoints.'
-),
-(
-	'Meas',
-	'exact filtered colimits',
+	'cartesian filtered colimits',
 	FALSE,
 	'See <a href="https://math.stackexchange.com/questions/5027218" target="_blank">MSE/5027218</a>.'
 ),

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

@@ -47,6 +47,12 @@ VALUES
 	TRUE,
 	'Given a directed diagram $(X_i)$ of metric spaces, take the directed colimit $X$ of the underlying sets with the following metric: If $x,y \in X$, let $d(x,y)$ be infimum of all $d(x_i,y_i)$, where $x_i,y_i \in X_i$ are some preimages of $x,y$ in some $X_i$. This is only a pseudo-metric, so finally take the associated metric space (Kolmogorov quotient). The definition ensures that each $X_i \to X$ is non-expansive, and the universal property is easy to check.' 
 ),
+(
+	'Met',
+	'cartesian filtered colimits',
+	TRUE,
+	'The canonical map $\mathrm{colim}_i  (X \times Y_i) \to X \times \mathrm{colim}_i Y_i$ is an isomorphism for directed diagrams $(Y_i)$: It is surjective by the concrete description of directed colimits. It is isometric because of the elementary observation $\inf_i \max(r, s_i) = \max(r, \inf_i s_i)$ for $r, s_i \in \mathbb{R}$, where $i \leq j \implies s_i \geq s_j$.'
+),
 (
 	'Met',
 	'strict initial object',
@@ -117,7 +123,7 @@ VALUES
 	'Met',
 	'exact filtered colimits',
 	FALSE,
-	'Remark 2.7 in <a href="https://arxiv.org/abs/2006.01399" target="_blank">this paper</a>'
+	'See Remark 2.7 in <a href="https://arxiv.org/abs/2006.01399" target="_blank">Approximate injectivity and smallness in metric-enriched categories</a> by Adamek-Rosicky.'
 ),
 (
 	'Met',

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

@@ -23,6 +23,12 @@ VALUES
 	TRUE,
 	'Example 4.5 in <a href="https://arxiv.org/abs/1504.02660" target="_blank">this preprint</a>'
 ),
+(
+	'Met_oo',
+	'cartesian filtered colimits',
+	TRUE,
+	'We can use the same proof as for the <a href="/category/Met">category of metric spaces</a> since the equation $\inf_i \max(r, s_i) = \max(r, \inf_i s_i)$ also holds for for $r, s_i \in \mathbb{R} \cup \{\infty\}$.'
+),
 (
 	'Met_oo',
 	'infinitary extensive',
@@ -57,7 +63,7 @@ VALUES
 	'Met_oo',
 	'exact filtered colimits',
 	FALSE,
-	'2.7 in <a href="https://arxiv.org/abs/2006.01399" target="_blank">this paper</a>'
+	'See Remark 2.7 in <a href="https://arxiv.org/abs/2006.01399" target="_blank">Approximate injectivity and smallness in metric-enriched categories</a> by Adamek-Rosicky.'
 ),
 (
 	'Met_oo',

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

@@ -76,5 +76,11 @@ VALUES
 	'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).'
+),
+(
+	'Mon',
+	'cocartesian cofiltered limits',
+	FALSE,
+	'We know that the <a href="/category/Grp">category of groups</a> fails to satisfy this property. The same counterexample works here since the inclusion $\mathbf{Grp} \hookrightarrow \mathbf{Mon}$ preserves limits and colimits (it has a left and a right adjoint) and is conservative. A similar counterexample is given by the free monoids $N_n = \langle x_1,\dotsc,x_n \rangle$ and the Boolean monoid $M = \langle e : e^2=e \rangle$ with the maps $N_{n+1} \to N_n$, $x_{n+1} \mapsto 1$. Then the element $(x_1 e \cdots x_n e) \in \lim_n (M \sqcup N_n)$ does not come from $M \sqcup \lim_n N_n$ because its components have unbounded free product length.'
 );
 

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

@@ -82,4 +82,12 @@ VALUES
 	'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).'
+),
+(
+	'Ring',
+	'cocartesian cofiltered limits',
+	FALSE,
+	'Consider the ring $A = \mathbb{Z}[X]$ and the sequence of rings $B_n = \mathbb{Z}[Y]/(Y^{n+1})$ with projections $B_{n+1} \to B_n$, whose limit is $\mathbb{Z}[[Y]]$. Every element in the coproduct of rings $\mathbb{Z}[X] \sqcup \mathbb{Z}[[Y]]$ has a finite "free product" length. Now consider the elements
+	<br>$w_n = (1 + XY) (1+XY^2) \cdots (1+X Y^n) \in A \sqcup B_n$.</br>
+	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.'
 );

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

@@ -70,4 +70,12 @@ VALUES
 	'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>).'
+),
+(
+	'Rng',
+	'cocartesian cofiltered limits',
+	FALSE,
+	'Consider the ring $A = \mathbb{Z}[X]$ and the sequence of rings $B_n = \mathbb{Z}[Y]/(Y^{n+1})$ with projections $B_{n+1} \to B_n$, whose limit is $\mathbb{Z}[[Y]]$ (both in $\mathbf{Ring}$ and $\mathbf{Rng}$). Every element in the coproduct of rngs $\mathbb{Z}[X] \sqcup \mathbb{Z}[[Y]]$ has a finite "free product" length. Now consider the elements
+	<br>$w_n = (1 + XY) (1+XY^2) \cdots (1+X Y^n) - 1 \in A \sqcup B_n$.</br>
+	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.'
 );

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

@@ -53,6 +53,14 @@ VALUES
 	TRUE,
 	'Malcev categories are closed under slice categories by Prop. 2.2.14 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>. It follows that co-Malcev categories are closed under coslice categories, and $\mathbf{Set}_*$ is a coslice category of $\mathbf{Set}$, which is co-Malcev since every elementary topos is co-Malcev.'
 ),
+(
+	'Set*',
+	'cocartesian cofiltered limits',
+	TRUE,
+	'Let $X$ be a pointed set and $(Y_i)$ be a filtered diagram of pointed sets. Base points will be denoted by $0$. The canonical map $X \vee \lim_i Y_i \to \lim_i (X \vee Y_i)$ is injective since the wedge sum naturally embeds into the product and the natural map $X \vee \prod_i Y_i \to \prod_i (X \times Y_i)$ is injective. Now let $z = (z_i) \in \lim_i (X \vee Y_i)$.
+	<br>Case 1: There is some index $i$ with $z_i \in X \setminus \{0\}$. We claim $z_j \in X$ for any index $j$ and $z_j = z_i$ in $X$, so that $z$ has a preimage in $X$. To see this, choose an index $k \geq i,j$. Since $X \vee Y_i \to X \vee Y_k$ maps $z_i \mapsto z_k$ and is the identity on $X$, we see that $z_k \in X$ and $z_k = z_i$ in $X$. Since $X \vee Y_j \to X \vee Y_k$ maps $z_j \mapsto z_k$, we see that $z_j \notin Y_j$, since otherwise $z_k \in Y_k \cap X = \{0\}$. Hence, $z_j \in X \setminus \{0\}$, and then $z_j = z_k = z_i$.
+	<br>Case 2: We have $z_i \in Y_i$ for all $i$. Then clearly $(z_i) \in \lim_i Y_i$ is a preimage.'
+),
 (
 	'Set*',
 	'skeletal',

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

@@ -73,9 +73,9 @@ VALUES
 ),
 (
 	'Top',
-	'cartesian closed',
+	'cartesian filtered colimits',
 	FALSE,
-	'The functor $\mathbb{Q} \times - : \mathbf{Top} \to \mathbf{Top}$ does not preserve colimits, hence has no right adjoint. See <a href="https://math.stackexchange.com/questions/2969372" target="_blank">MSE/2969372</a> for a proof.'
+	'The functor $\mathbb{Q} \times - : \mathbf{Top} \to \mathbf{Top}$ does not preserve colimits, see <a href="https://math.stackexchange.com/questions/2969372" target="_blank">MSE/2969372</a>.'
 ),
 (
 	'Top',
@@ -95,12 +95,6 @@ VALUES
 	FALSE,
 	'If $X$ is a set, consider the discrete space $X_d$ on $X$ and the indiscrete space $X_i$ on $X$. The identity map $X \to X$ lifts to a continuous map $X_d \to X_i$, which is bijective and therefore both a mono- and an epimorphism, but it is not an isomorphism unless $X$ has at most one element.'
 ),
-(
-	'Top',
-	'exact filtered colimits',
-	FALSE,
-	'See <a href="https://math.stackexchange.com/questions/1255678" target="_blank">MSE/1255678</a>.'
-),
 (
 	'Top',
 	'skeletal',

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

@@ -77,6 +77,13 @@ VALUES
 	TRUE,
 	'Since embeddings are regular monomorphisms in this category (see below) and hence strong monomorphisms, it suffices to prove that the canonical morphism $X \vee Y \hookrightarrow X \times Y$ is an embedding. For a proof, see <a href="https://math.stackexchange.com/questions/4055988" target="_blank">MSE/4055988</a>.'
 ),
+(
+	'Top*',
+	'cocartesian cofiltered limits',
+	TRUE,
+	'We continue the proof for <a href="/category/Set*">$\mathbf{Set}_*$</a> by showing that the natural bijective map <br>$\alpha : X \vee \lim_i Y_i \to \lim_i (X \vee Y_i)$<br>
+	is open. It suffices to consider open sets of two types: (1) If $U \subseteq X$ is open, the $\alpha$-image of $U \vee \lim_i Y_i$ is $p_{i_0}^{-1}(U \vee Y_{i_0})$ for any chosen index $i_0$, hence open. (2) If $i$ is an index and $V_i \subseteq Y_i$ is open, then the $\alpha$-image of $X \vee (p_i^{-1}(V_i) \cap \lim_i Y_i)$ is $p_i^{-1}(X \vee V_i)$, hence open.'
+),
 (
 	'Top*',
 	'coregular',
@@ -103,9 +110,9 @@ VALUES
 ),
 (
 	'Top*',
-	'exact filtered colimits',
+	'cartesian filtered colimits',
 	FALSE,
-	'See <a href="https://math.stackexchange.com/questions/1255678" target="_blank">MSE/1255678</a> (the counterexample also works for pointed spaces).'
+	'The functor $\mathbb{Q} \times - : \mathbf{Top}_* \to \mathbf{Top}_*$ does not preserve colimits, see <a href="https://math.stackexchange.com/questions/2969372" target="_blank">MSE/2969372</a>. The counterexample also works for pointed spaces.'
 ),
 (
 	'Top*',