Various UI improvements

.cspell.json

@@ -193,6 +193,7 @@
 		"morphisms",
 		"Multialgébriques",
 		"naturality",
+		"Nekoma",
 		"Neves",
 		"Niefield",
 		"nilradical",

content/Meas_not_regular.md

@@ -0,0 +1,28 @@
+---
+title: The category of measurable spaces is not regular
+description: An example of a quotient measurable map is given whose product with itself is not a quotient map anymore.
+author: Nekoma
+---
+
+## $\Meas$ is not regular
+
+::: Claim
+The category $\Meas$ of measurable spaces and measurable maps is not regular.
+:::
+
+_Proof._
+In a regular category, regular epimorphisms are stable under pullbacks and compositions (see Prop. 3.7 at the [nLab](https://ncatlab.org/nlab/show/regular+epimorphism)), which implies that for every regular epimorphism $f : X \to Y$ also $f \times f : X \times X \to Y \times Y$ is a regular epimorphism. We will show that this fails in $\Meas$.
+
+Let $X \coloneqq [0, 1)$ equipped with the standard Borel $\sigma$-algebra $\B$. Consider the equivalence relation $x \sim y \iff x-y \in \IQ$, let $Y \coloneqq X /{\sim}$ be the set of equivalence classes, and $f: X \to Y$ be the natural projection map. Equip $Y$ with the quotient $\sigma$-algebra $\Sigma_Y$, so that $f$ is a regular epimorphism.
+
+Now consider the diagonal in the quotient space $\Delta_Y \coloneqq \{(y, y) \mid y \in Y\}$. Then
+$$\textstyle (f \times f)^{-1}(\Delta_Y) = \{(x_1, x_2) \in [0, 1)^2 \mid x_1 - x_2  \in \IQ\} \eqqcolon \bigcup_{q \in \IQ} L_q$$
+where each $L_q$ is the intersection of the diagonal level sets of $x_1 - x_2$ with $[0, 1)^2$. Because each line is closed in $\IR^2$, its intersection with $[0, 1)^2$ is a Borel set in $X \times X$. Since a countable union of Borel sets is Borel, $(f \times f)^{-1}(\Delta_Y) \in \B \otimes \B$.
+
+Now take any set $B \in \Sigma_Y$. Its preimage $f^{-1}(B)$ is a Borel set in $[0, 1)$ that is invariant under rational translations modulo 1. Because the action of $\IQ / \IZ$ on $[0, 1)$ is ergodic, the Lebesgue measure $\lambda(f^{-1}(B))$ must be exactly $0$ or $1$. Assume for contradiction that $\Sigma_Y$ is countably separated, i.e. there exists a countable sequence of measurable sets $(B_n)_{n \geq 1}$ in $\Sigma_Y$ that separates the points of $Y$. Let $A_n \coloneqq f^{-1}(B_n)$. Every $A_n$ has $\lambda(A_n) = 0$ or $\lambda(A_n) = 1$.
+
+Define a "bad set" $N \subseteq [0, 1)$ as
+$$\textstyle N \coloneqq \left( \bigcup_{\lambda(A_n)=0} A_n \right) \cup \left( \bigcup_{\lambda(A_n)=1} A_n^c \right)$$
+Because $N$ is a countable union of sets with measure $0$, we have $\lambda(N) = 0$, and thus $\lambda([0, 1) \setminus N)=1$. For any two points $x, y \in [0, 1) \setminus N$, clearly $x \in A_n \iff y \in A_n$ for every $n$. Consequently, the sequence $(B_n)$ fails to separate $f(x)$ and $f(y)$. Hence, $x \sim y$. Since $[0, 1) \setminus N$ has measure $1$, it is uncountable. Because each equivalence class is only countable, these uncountably many points must belong to uncountably many different equivalence classes. Thus, we can easily pick $x, y \in [0, 1) \setminus N$ where $x \not\sim y$. Thus $\Sigma_Y$ is not countably separated.
+
+Hence by Theorem 6.5.7 in Bogachev's [Measure theory](https://link.springer.com/book/10.1007/978-3-540-34514-5) $\Delta_Y \notin \Sigma_Y \otimes \Sigma_Y$. We have identified a non-measurable subset of $Y \times Y$ whose preimage under $f \times f$ is measurable. Therefore, $f \times f$ is not a regular epimorphism. <span class="qed">$\square$</span>

content/aleph1-filtered-colimits-in-deloopings.md

@@ -0,0 +1,119 @@
+---
+title: ℵ₁-filtered colimits in deloopings
+description: We give a detailed proof that the delooping of the monoid of natural numbers, and likewise the delooping of the large monoid of ordinal numbers, has colimits indexed by ℵ₁-filtered categories.
+author: Martin Brandenburg
+---
+
+# $\aleph_1$-filtered colimits in deloopings
+
+Every (possibly large) monoid $M$ induces a category $BM$ with just one object. We will show that this category has $\aleph_1$-filtered colimits in the cases $M = \IN$ and $M = \On$ (both respect to addition).
+
+::: Proposition 1
+The category $B\IN$ has $\aleph_1$-filtered colimits.
+:::
+
+_Proof._
+Let $D : \I \to B\IN$ be an $\aleph_1$-filtered diagram. Every two parallel morphisms $i \rightrightarrows j$ are mapped to the same morphism in $B \IN$, because they are coequalized by some morphism and $(\IN,+)$ is cancellative. Hence, $D$ factors through the preorder reflection of $\I$, and we may therefore assume that $\I$ itself is a preorder. Thus, the diagram consists of numbers $D(i,j) \in \IN$ for all $i \leq j$ satisfying
+
+$$D(j,k) + D(i,j) = D(i,k)$$
+
+for all $i \leq j \leq k$. In particular, $D(i,j) \leq D(i,k)$.
+
+Let $i \in \I$. The set of natural numbers $\{D(i,k) : k \geq i\}$ is bounded above. Otherwise, for every $n \in \IN$ we could find $k_n \in \I$ with $k_n \geq i$ and $D(i,k_n) \geq n$. Since $\I$ is $\aleph_1$-filtered, the family $(k_n)_{n \in \IN}$ has an upper bound $k_\infty \in \I$. But then
+
+$$D(i, k_\infty) = D(k_n,k_\infty) + D(i, k_n) \geq D(i, k_n) \geq n$$
+
+for all $n \in \IN$, contradicting the fact that $D(i,k_\infty) \in \IN$.
+
+Therefore, the maximum
+
+$$u_i \coloneqq \max \{D(i,k) : k \geq i\} \in \IN$$
+
+is well-defined, which we regard as a morphism in $B\IN$. For $i \leq j$ we compute
+
+$$
+\begin{align*}
+u_i & = \max \{D(i,k) : k \geq i\} \\
+& = \max \{D(i,k) : k \geq j\} \\
+& = \max \{ D(j,k) + D(i,j) : k \geq j\} \\
+& = \max \{D(j,k) : k \geq j\}  + D(i,j)\\
+& = u_j + D(i,j),
+\end{align*}
+$$
+
+showing that $(u_i)$ defines a cocone. It is universal: let $(v_i)$ be another cocone, i.e. $v_i \in \IN$ and $v_i = v_j + D(i,j)$ for all $i \leq j$. Then $v_i \geq D(i,j)$ for all $i \leq j$, hence $v_i \geq u_i$. Write $v_i = w_i + u_i$ for some uniquely determined $w_i \in \IN$. For $i \leq j$ we compute
+
+$$w_j + u_j + D(i,j) = v_j + D(i,j) = v_i = w_i + u_i = w_i + u_j + D(i,j),$$
+
+hence $w_j = w_i$. Therefore, the $w_i$ are constant, and the required factorization follows. <span class="qed">$\square$</span>
+
+::: Proposition 2
+The category $B\IN$ is $\aleph_1$-accessible.
+:::
+
+_Proof._
+Based on Proposition 1, it remains to show that the unique object $*$ is $\aleph_1$-presentable, i.e. that for every diagram $D : \I \to B\IN$ as above, the canonical map
+
+$$\alpha : \colim_{i \in \I} \Hom(*,D(i)) \to \Hom(*,\colim_{i \in \I} D(i))$$
+
+is bijective. On objects, we necessarily have $D(i)=*$ and $\colim_{i \in \I} D(i)=*$. Hence, the codomain of $\alpha$ is simply $\IN$, while the domain consists of equivalence classes $[i,n]$ of pairs $(i,n) \in \I \times \IN$, where $(i,n) \sim (j,m)$ iff there exists some $k \geq i,j$ such that
+
+$$D(i,k) + n = D(j,k) + m.$$
+
+By the construction of the colimit cocone, we have
+
+$$\alpha([i,n]) = u_i + n = \max \{D(i,j) : j \geq i\} + n.$$
+
+(1) **The map $\alpha$ is surjective:** Pick some $i \in \I$. Choose $j \geq i$ such that $u_i = D(i,j)$. For all $k \geq j$ we then have
+
+$$u_i \geq D(i,k) = D(j,k) + D(i,j) = D(j,k) + u_i,$$
+
+hence $D(j,k)=0$. Therefore, $u_j=0$, and thus $\alpha([j,n]) = n$ for all $n \in \IN$.
+
+(2) **The map $\alpha$ is injective:** Assume that $[i,n]$ and $[j,m]$ have the same image. Since $\I$ is filtered, we may assume $i=j$. The condition then becomes $u_i + n = u_i + m$, and therefore $n=m$. This completes the proof. <span class="qed">$\square$</span>
+
+::: Proposition 3
+The category $B\On$ has $\aleph_1$-filtered colimits.
+:::
+
+_Proof._
+The proof is similar to $B\IN$. Let $\I$ be an $\aleph_1$-filtered small category and $D : \I \to B\On$ a diagram. A cocone $\lambda = (\lambda_i)_{i \in \I}$ for $D$ is a family of ordinals satisfying $\lambda_i = \lambda_j + D(f)$ for every morphism $f: i \to j$ in $\I$.
+
+We first observe that $D$ factors uniquely through the preorder reflection of $\I$. Indeed, any two parallel morphisms in $\I$ are coequalized by some morphism, and $B\On$ is left cancellative. Thus, we may assume that $\I$ is a preorder. Each inequality $i \leq j$ in $\I$ is mapped to an ordinal number $\alpha_{i,j} \coloneqq D(i \to j)$, and these numbers satisfy
+$$\alpha_{i,k} = \alpha_{j,k} + \alpha_{i,j}$$
+for all $i \leq j \leq k$. In particular, $\alpha_{i,j} \leq \alpha_{i,k}$.
+
+For fixed $i \in \I$, the collection $\{\alpha_{i,j} : j \geq i\}$ is a set of ordinals because $\I$ is small, hence bounded above in $\On$. We claim that it has a maximum element. Otherwise, we can find a countable chain $i = j_0 \leq j_1 \leq j_2 \leq \dotsc$ in $\I$ such that $\alpha_{i,j_n} < \alpha_{i,j_{n+1}}$ for all $n \in \IN$. Since $\I$ is $\aleph_1$-filtered, there is an upper bound $j_\infty \in \I$ of $(j_n)_{n \in \IN}$. For each $n \in \IN$, the equation
+$$\alpha_{i,j_{n+1}} = \alpha_{j_n,j_{n+1}} + \alpha_{i,j_n}$$
+implies that $\alpha_{j_n,j_{n+1}} > 0$. Hence,
+$$\alpha_{j_n,j_\infty} = \alpha_{j_{n+1},j_\infty} + \alpha_{j_n,j_{n+1}} > \alpha_{j_{n+1},j_\infty},$$
+so $(\alpha_{j_n,j_\infty})_{n \in \IN}$ is a strictly decreasing infinite sequence of ordinals, contradicting the well-foundedness of $\On$. Thus, the maximum
+$$u_i \coloneqq \max \{ \alpha_{i,j} : j \geq i \}$$
+is a well-defined ordinal number, which we regard as a morphism in $B\On$. The family $(u_i)_{i \in \I}$ forms a cocone for $D$, since for all $i \leq j$ we have
+
+$$
+\begin{align*}
+u_i & = \max \{ \alpha_{i,k} : k \geq i \} \\
+& = \max \{ \alpha_{i,k} : k \geq j \} \\
+& = \max \{ \alpha_{j,k} + \alpha_{i,j} : k \geq j \} \\
+& = \max \{ \alpha_{j,k} : k \geq j \} + \alpha_{i,j} \\
+& = u_j + \alpha_{i,j}.
+\end{align*}
+$$
+
+To establish the universal property, let $(\lambda_i)_{i \in \I}$ be any cocone for $D$, so that $\lambda_i = \lambda_j + \alpha_{i,j}$ for all $i \leq j$. The cocone relation $u_i = u_j + \alpha_{i,j}$ implies that $u_i \geq u_j$ whenever $i \leq j$. By the well-foundedness of $\On$, there exists $i_0 \in \I$ such that $u_j = u_{i_0}$ for all $j \geq i_0$. For such $j$, the relation
+$$u_{i_0} = u_j + \alpha_{i_0,j} = u_{i_0} + \alpha_{i_0,j}$$
+forces $\alpha_{i_0,j} = 0$. Consequently,
+$$u_{i_0} = \max \{ \alpha_{i_0,j} : j \geq i_0 \} = 0.$$
+Define the mediating morphism to be the ordinal $\kappa \coloneqq \lambda_{i_0}$. We must show that $\lambda_i = \kappa + u_i$ for all $i \in \I$. Choose $j \in \I$ with $j \geq i$ and $j \geq i_0$. Since $j \geq i_0$, we have $u_j = 0$ and $\alpha_{i_0,j} = 0$. The cocone condition for $\lambda$ gives
+$$\kappa = \lambda_{i_0} = \lambda_j + \alpha_{i_0,j} = \lambda_j.$$
+Applying the cocone conditions for $u$ and $\lambda$ to $i \leq j$, we obtain
+$$u_i = u_j + \alpha_{i,j} = 0 + \alpha_{i,j} = \alpha_{i,j}$$
+and
+$$\lambda_i = \lambda_j + \alpha_{i,j} = \kappa + \alpha_{i,j} = \kappa + u_i.$$
+This proves the existence of the mediating morphism.
+
+For uniqueness, suppose $\kappa'$ is any ordinal satisfying $\lambda_i = \kappa' + u_i$ for all $i \in \I$. Evaluating at $i_0$ yields
+$$\lambda_{i_0} = \kappa' + u_{i_0} = \kappa' + 0 = \kappa',$$
+hence $\kappa' = \kappa$. Therefore, the cocone $(u_i)_{i \in \I}$ is the colimit of $D$ in $B\On$.
+<span class="qed">$\square$</span>

content/thin_algebraic_categories.md

@@ -7,8 +7,8 @@ author: Martin Brandenburg
 ## Algebraic categories are "never" thin
 
 ::: Lemma
-Let $\C$ be a [thin](/category-property/thin) and [finitary algebraic](/category-property/finitary_algebraic) category. Then $\C \simeq 1$ or $\C \simeq \{0 < 1\}$.
+Let $\C$ be a [thin](/category-property/thin) and [finitary algebraic](/category-property/finitary_algebraic) category. Then $\C \simeq 1$ or $\C \simeq I$, where $I$ is the walking morphism.
 :::
 
 _Proof._
-Let $F : \Set \to \C$ denote the free algebra functor. Every object $A \in \C$ admits a regular epimorphism $F(X) \twoheadrightarrow A$ for some set $X$. But since $\C$ is thin, every regular epimorphism must be an isomorphism. Thus, $A \cong F(X)$. Also, $F(X)$ is a coproduct of copies of $F(1)$, which means it is either the initial object $0$ or $F(1)$ itself (since $\C$ is thin). If $F(1) \cong 0$, then every object is isomorphic to the initial object $0$, and hence $\C$ is trivial. If not, then $\C$ has exactly two objects up to isomorphism, $0$ and $F(1)$, there is a morphism $0 \to F(1)$, but no morphism $F(1) \to 0$. Since $\C$ is thin, we conclude $\C \simeq \{0 \to 1\}$. <span class="qed">$\square$</span>
+Let $F : \Set \to \C$ denote the free algebra functor. Every object $A \in \C$ admits a regular epimorphism $F(X) \twoheadrightarrow A$ for some set $X$. But since $\C$ is thin, every regular epimorphism must be an isomorphism. Thus, $A \cong F(X)$. Also, $F(X)$ is a coproduct of copies of $F(1)$, which means it is either the initial object $0$ or $F(1)$ itself (since $\C$ is thin). If $F(1) \cong 0$, then every object is isomorphic to the initial object $0$, and hence $\C$ is trivial. If not, then $\C$ has exactly two objects up to isomorphism, $0$ and $F(1)$, there is a morphism $0 \to F(1)$, but no morphism $F(1) \to 0$. Since $\C$ is thin, we conclude $\C \simeq I$. <span class="qed">$\square$</span>

databases/catdat/data/categories/BN.yaml

@@ -40,44 +40,8 @@ satisfied_properties:
   - property: locally cartesian closed
     proof: The slice category $B\IN / *$ is isomorphic to the poset $(\IN,\geq)$ (not to $(\IN,\leq)$). This category is thin and and semi-strongly connected, <a href="/category-implication/sequential_implies_lcc">hence</a> cartesian closed.
 
-  - property: ℵ₁-filtered colimits
-    proof: >-
-      Let $D : \I \to B\IN$ be an $\aleph_1$-filtered diagram. Every two parallel morphisms $i \rightrightarrows j$ are mapped to the same morphism in $B \IN$, because they are coequalized by some morphism and $(\IN,+)$ is cancellative. Hence, $D$ factors through the preorder reflection of $\I$, and we may therefore assume that $\I$ itself is a preorder. Thus, the diagram consists of numbers $D(i,j) \in \IN$ for all $i \leq j$ satisfying
-      $$D(j,k) + D(i,j) = D(i,k)$$
-      for all $i \leq j \leq k$. In particular, $D(i,j) \leq D(i,k)$.
-
-      Let $i \in \I$. The set of natural numbers $\{D(i,k) : k \geq i\}$ is bounded above. Otherwise, for every $n \in \IN$ we could find $k_n \in \I$ with $k_n \geq i$ and $D(i,k_n) \geq n$. Since $\I$ is $\aleph_1$-filtered, the family $(k_n)_{n \in \IN}$ has an upper bound $k_\infty \in \I$. But then
-      $$D(i, k_\infty) = D(k_n,k_\infty) + D(i, k_n) \geq D(i, k_n) \geq n$$
-      for all $n \in \IN$, contradicting the fact that $D(i,k_\infty) \in \IN$.
-
-      Therefore, the maximum
-      $$u_i \coloneqq \max \{D(i,k) : k \geq i\} \in \IN$$
-      is well-defined, which we regard as a morphism in $B\IN$. For $i \leq j$ we compute
-      $$\begin{align*}
-      u_i & = \max \{D(i,k) : k \geq i\} \\
-      & = \max \{D(i,k) : k \geq j\} \\
-      & = \max \{ D(j,k) + D(i,j) : k \geq j\} \\
-      & = \max \{D(j,k) : k \geq j\}  + D(i,j)\\
-      & = u_j + D(i,j),
-      \end{align*}$$
-      showing that $(u_i)$ defines a cocone. It is universal: let $(v_i)$ be another cocone, i.e. $v_i \in \IN$ and $v_i = v_j + D(i,j)$ for all $i \leq j$. Then $v_i \geq D(i,j)$ for all $i \leq j$, hence $v_i \geq u_i$. Write $v_i = w_i + u_i$ for some uniquely determined $w_i \in \IN$. For $i \leq j$ we compute
-      $$w_j + u_j + D(i,j) = v_j + D(i,j) = v_i = w_i + u_i = w_i + u_j + D(i,j),$$
-      hence $w_j = w_i$. Therefore, the $w_i$ are constant, and the required factorization follows.
-    check_redundancy: false
-
   - property: ℵ₁-accessible
-    proof: >-
-      We continue the proof that $\aleph_1$-filtered colimits exist. It remains to show that the unique object $*$ is $\aleph_1$-presentable, i.e. that for every diagram $D : \I \to B\IN$ as above, the canonical map
-      $$\alpha : \colim_{i \in \I} \Hom(*,D(i)) \to \Hom(*,\colim_{i \in \I} D(i))$$
-      is bijective. On objects, we necessarily have $D(i)=*$ and $\colim_{i \in \I} D(i)=*$. Hence, the codomain of $\alpha$ is simply $\IN$, while the domain consists of equivalence classes $[i,n]$ of pairs $(i,n) \in \I \times \IN$, where $(i,n) \sim (j,m)$ iff there exists some $k \geq i,j$ such that
-      $$D(i,k) + n = D(j,k) + m.$$
-      By the construction of the colimit cocone, we have
-      $$\alpha([i,n]) = u_i + n = \max \{D(i,j) : j \geq i\} + n.$$
-      (1) The map $\alpha$ is surjective: Pick some $i \in \I$. Choose $j \geq i$ such that $u_i = D(i,j)$. For all $k \geq j$ we then have
-      $$u_i \geq D(i,k) = D(j,k) + D(i,j) = D(j,k) + u_i,$$
-      hence $D(j,k)=0$. Therefore, $u_j=0$, and thus $\alpha([j,n]) = n$ for all $n \in \IN$.
-
-      (2) The map $\alpha$ is injective: Assume that $[i,n]$ and $[j,m]$ have the same image. Since $\I$ is filtered, we may assume $i=j$. The condition then becomes $u_i + n = u_i + m$, and therefore $n=m$. This completes the proof.
+    proof: A proof can be found <a href="/content/aleph1-filtered-colimits-in-deloopings">here</a> as Proposition 2.
 
 unsatisfied_properties:
   - property: one-way