replace <br> with yaml files with proper line breaks + render them as <br>

Author: ScriptRaccoon

databases/catdat/data/categories/FreeAb.yaml

@@ -34,7 +34,10 @@ satisfied_properties:
     reason: It is easy to check that $\IZ$ is a cogenerator for free abelian groups.
 
   - property_id: regular
-    reason: 'This follows formally from the fact that $\Ab$ is regular and $\FreeAb$ is closed under subobjects and finite products: By Prop. 2.5 in the <a href="https://ncatlab.org/nlab/show/regular+category">nlab</a> it suffices to prove that there are pullback-stable (reg epi, mono)-factorizations. Every homomorphism $f : A \to B$ in $\FreeAb$ factors as $f = i \circ p : A \twoheadrightarrow C \hookrightarrow B$, where $C$ is a subgroup, hence free, and $A \to C$ is surjective. Clearly, surjective homomorphisms are pullback-stable. It remains to show that they coincide with the regular epimorphisms.<br> (1) If $f : A \to B$ is surjective, it is the coequalizer of $A \times_B A \rightrightarrows A$ in $\Ab$. Since $A \times_B A$ is free abelian, $f$ is also an coequalizer in $\FreeAb$.<br> (2) If $f : A \to B$ is a regular epimorphism in $\FreeAb$, consider the factorization $f = i \circ p$ as above. Since $f$ is an extremal epimorphism, $i$ must be an isomorphism, so that $f$ is surjective.'
+    reason: |
+      This follows formally from the fact that $\Ab$ is regular and $\FreeAb$ is closed under subobjects and finite products: By Prop. 2.5 in the <a href="https://ncatlab.org/nlab/show/regular+category">nlab</a> it suffices to prove that there are pullback-stable (reg epi, mono)-factorizations. Every homomorphism $f : A \to B$ in $\FreeAb$ factors as $f = i \circ p : A \twoheadrightarrow C \hookrightarrow B$, where $C$ is a subgroup, hence free, and $A \to C$ is surjective. Clearly, surjective homomorphisms are pullback-stable. It remains to show that they coincide with the regular epimorphisms.
+      (1) If $f : A \to B$ is surjective, it is the coequalizer of $A \times_B A \rightrightarrows A$ in $\Ab$. Since $A \times_B A$ is free abelian, $f$ is also an coequalizer in $\FreeAb$.
+      (2) If $f : A \to B$ is a regular epimorphism in $\FreeAb$, consider the factorization $f = i \circ p$ as above. Since $f$ is an extremal epimorphism, $i$ must be an isomorphism, so that $f$ is surjective.
 
 unsatisfied_properties:
   - property_id: countable powers
@@ -50,7 +53,9 @@ unsatisfied_properties:
     reason: See <a href="https://mathoverflow.net/questions/509715" target="_blank">MO/509715</a>.
 
   - property_id: effective cocongruences
-    reason: 'We will let $E$ be the abelian group with presentation $\langle a, b, c \mid a - b = 2c \rangle$, with two morphisms $\IZ \rightrightarrows E$ given by $1\mapsto a$, $1\mapsto b$. Note that $E$ is free with basis $\{ b, c \}$. Then $\Hom(E, G) \cong \{ (x, y, z) \in G^3 \mid x - y = 2z \}$. Observe that since $G$ is torsion-free, the projection onto the first two coordinates is injective; and $(x, y)$ is in the image precisely when $x \equiv y \pmod{2G}$, which gives an equivalence relation. Therefore, $E$ gives a cocongruence on $\IZ$.<br> On the other hand, if $E$ were the cokernel pair of $h : H \to \IZ$, that would mean that for $x, y : \IZ \to G$, $x \equiv y \pmod{2G}$ if and only if $x \circ h = y \circ h$. In particular, from the case $G := \IZ$, $x := 2 \id$, $y := 0$, we would have $2h = 0$. That implies $h = 0$, but then that would give $\id_{\IZ} \equiv 0 \pmod{2}$, resulting in a contradiction.'
+    reason: |
+      We will let $E$ be the abelian group with presentation $\langle a, b, c \mid a - b = 2c \rangle$, with two morphisms $\IZ \rightrightarrows E$ given by $1\mapsto a$, $1\mapsto b$. Note that $E$ is free with basis $\{ b, c \}$. Then $\Hom(E, G) \cong \{ (x, y, z) \in G^3 \mid x - y = 2z \}$. Observe that since $G$ is torsion-free, the projection onto the first two coordinates is injective; and $(x, y)$ is in the image precisely when $x \equiv y \pmod{2G}$, which gives an equivalence relation. Therefore, $E$ gives a cocongruence on $\IZ$.
+      On the other hand, if $E$ were the cokernel pair of $h : H \to \IZ$, that would mean that for $x, y : \IZ \to G$, $x \equiv y \pmod{2G}$ if and only if $x \circ h = y \circ h$. In particular, from the case $G := \IZ$, $x := 2 \id$, $y := 0$, we would have $2h = 0$. That implies $h = 0$, but then that would give $\id_{\IZ} \equiv 0 \pmod{2}$, resulting in a contradiction.
 
 category_property_comments:
   - property_id: accessible

databases/catdat/data/categories/J2.yaml

@@ -28,7 +28,9 @@ unsatisfied_properties:
     reason: This is trivial.
 
   - property_id: semi-strongly connected
-    reason: 'There is a bijection $\alpha = (\lambda,\rho) : \IN \to \IN \times \IN$ such that $\lambda$ has a fixed point, but $\rho$ does not (see below). Then the isomorphism $\beta := (\rho,\lambda)$ has the opposite property. There cannot be any morphism $(\IN,\alpha) \to (\IN,\beta)$, as it would map the fixed point of $\lambda$ to a fixed point of $\rho$, and likewise there is no morphism $(\IN,\beta) \to (\IN,\alpha)$.<br> To construct $\alpha$ or rather $\alpha^{-1} : \IN \times \IN \to \IN$, we can alter the standard bijection $(n,m) \mapsto 2^n (2m+1) - 1$ as follows: $$\alpha^{-1}(n,m) = \begin{cases} 2 & (n,m) = (0,0) \\ 0 & (n,m) = (0,1) \\ 2^n (2m+1) - 1 & \text{otherwise} \end{cases}$$ Then $\alpha(0)=(0,1)$, i.e. $\lambda(0)=0$. The function $\rho$ has no fixed point, i.e. $\alpha^{-1}(n,m) \neq m$ for all $n,m$. Namely, if $(n,m)=(0,0)$, then $\alpha^{-1}(n,m)=2 \neq m$. If $(n,m)=(0,1)$, then $\alpha^{-1}(n,m)=0 \neq m$. Otherwise, $$\alpha^{-1}(n,m) = 2^n (2m+1) - 1 \geq (2m+1)-1 = 2m \geq m,$$ and equality can only hold if $m=0$ and $n=0$, which we already excluded.'
+    reason: |
+      There is a bijection $\alpha = (\lambda,\rho) : \IN \to \IN \times \IN$ such that $\lambda$ has a fixed point, but $\rho$ does not (see below). Then the isomorphism $\beta := (\rho,\lambda)$ has the opposite property. There cannot be any morphism $(\IN,\alpha) \to (\IN,\beta)$, as it would map the fixed point of $\lambda$ to a fixed point of $\rho$, and likewise there is no morphism $(\IN,\beta) \to (\IN,\alpha)$.
+      To construct $\alpha$ or rather $\alpha^{-1} : \IN \times \IN \to \IN$, we can alter the standard bijection $(n,m) \mapsto 2^n (2m+1) - 1$ as follows: $$\alpha^{-1}(n,m) = \begin{cases} 2 & (n,m) = (0,0) \\ 0 & (n,m) = (0,1) \\ 2^n (2m+1) - 1 & \text{otherwise} \end{cases}$$ Then $\alpha(0)=(0,1)$, i.e. $\lambda(0)=0$. The function $\rho$ has no fixed point, i.e. $\alpha^{-1}(n,m) \neq m$ for all $n,m$. Namely, if $(n,m)=(0,0)$, then $\alpha^{-1}(n,m)=2 \neq m$. If $(n,m)=(0,1)$, then $\alpha^{-1}(n,m)=0 \neq m$. Otherwise, $$\alpha^{-1}(n,m) = 2^n (2m+1) - 1 \geq (2m+1)-1 = 2m \geq m,$$ and equality can only hold if $m=0$ and $n=0$, which we already excluded.
 
 special_objects:
   initial object:

databases/catdat/data/categories/Man.yaml

@@ -45,7 +45,9 @@ satisfied_properties:
     reason: 'This is a consequence of the <a href="https://en.wikipedia.org/wiki/Whitney_embedding_theorem" target="_blank">Whitney embedding theorem</a>. But there is also a more direct proof: Since a manifold is second-countable, it is Lindelöf (<a href="https://topospaces.subwiki.org/wiki/Second-countable_implies_Lindelof" target="_blank">proof</a>). In particular, there is a countable atlas. It is then completely determined by countable many open subsets of Euclidean spaces and the transition maps.'
 
   - property_id: coquotients of cocongruences
-    reason: 'Let $p : X + X \twoheadrightarrow E$ be a cocongruence with coreflexivity morphism $r : E \to X$, so that $r \circ p : X + X \to X$ is the codiagonal. Since $p$ is an epimorphism, it has dense image (see below). We first claim that in fact $p$ also has closed image and therefore is surjective. Because $r \circ (p \circ i_1) : X \to X$ is the identity, the image of $p \circ i_1$ is the equalizer of $\id_E$ and $(p \circ i_1) \circ r$, hence closed. Likewise, the image of $p \circ i_2$ is closed. Thus, the image of $p$, which is the union of these images, is closed.<br> Now, since the pushforward maps of tangent spaces compose to the identity, we see that $p$ must be a local immersion and $r$ must be a submersion. Also, since the fibers of $r$ have one or two points each, we see that the dimension of $E$ must locally be the same as the dimension of $X$. This implies that in fact $p$ and $r$ are local diffeomorphisms. Therefore, the cardinality of the fiber of $r$ is locally constant. Thus, if $U$ is the subset of $X$ where $r$ has fiber of a single point, with the subspace topology, then $U$ is a clopen submanifold of $X$ which serves as the equalizer of $p \circ i_1$ and $p \circ i_2$.'
+    reason: |
+      Let $p : X + X \twoheadrightarrow E$ be a cocongruence with coreflexivity morphism $r : E \to X$, so that $r \circ p : X + X \to X$ is the codiagonal. Since $p$ is an epimorphism, it has dense image (see below). We first claim that in fact $p$ also has closed image and therefore is surjective. Because $r \circ (p \circ i_1) : X \to X$ is the identity, the image of $p \circ i_1$ is the equalizer of $\id_E$ and $(p \circ i_1) \circ r$, hence closed. Likewise, the image of $p \circ i_2$ is closed. Thus, the image of $p$, which is the union of these images, is closed.
+      Now, since the pushforward maps of tangent spaces compose to the identity, we see that $p$ must be a local immersion and $r$ must be a submersion. Also, since the fibers of $r$ have one or two points each, we see that the dimension of $E$ must locally be the same as the dimension of $X$. This implies that in fact $p$ and $r$ are local diffeomorphisms. Therefore, the cardinality of the fiber of $r$ is locally constant. Thus, if $U$ is the subset of $X$ where $r$ has fiber of a single point, with the subspace topology, then $U$ is a clopen submanifold of $X$ which serves as the equalizer of $p \circ i_1$ and $p \circ i_2$.
 
   - property_id: effective cocongruences
     reason: 'From the proof that $\Man$ has coquotients of cocongruences, we know that for any cocongruence $X \rightrightarrows E$, there is a clopen submanifold $U$ of $X$ such that the fibers of $r : E \twoheadrightarrow X$ have one point on $U$, and two points on $X \setminus U$. Therefore, $E$ is the cokernel pair of the inclusion map $U \hookrightarrow X$.'

databases/catdat/data/categories/Met.yaml

@@ -93,7 +93,9 @@ unsatisfied_properties:
     reason: 'Any kernel pair of $h : X \to Z$ in $\Met$ corresponds to a closed subset of $X\times X$. However, there are plenty of non-closed congruences, such as $\Delta \cup (\IQ \times \IQ) \subseteq \IR \times \IR$ with the subspace metric.'
 
   - property_id: effective cocongruences
-    reason: 'We will define a cocongruence on the interval $(0,1) \subseteq \IR$ where $E := (-1, 0) \cup (0, 1) \subseteq \IR$, and the two maps $(0, 1) \rightrightarrows E$ are the inclusion map and $x \mapsto -x$. Then for any metric space $X$, the induced relation on non-expansive maps $(0, 1) \to X$ is that $f \sim g$ if and only if $$d(f(x), g(y)) \le x+y$$ for each $x, y \in (0, 1)$. This is reflexive since $d(f(x), f(y)) \le |x-y| < x+y$, and it is clearly symmetric. For transitivity, suppose $f\sim g$ and $g\sim h$. Then for any $\varepsilon > 0$, we have $$d(f(x), h(y)) \le d(f(x), g(\varepsilon)) + d(g(\varepsilon), h(y)) \le (x + \varepsilon) + (y + \varepsilon).$$ Since this holds for every $\varepsilon > 0$, we conclude $d(f(x), h(y)) \le x+y$.<br> On the other hand, if this cocongruence were effective, then by the dual of <a href="/lemma/effective-congruence-quotients">this result</a>, it would be the cokernel pair of the equalizer of the two inclusion maps. However, that equalizer is empty, so $E$ would have to be a binary copower of $(0,1)$, which does not exist in $\Met$.'
+    reason: |
+      We will define a cocongruence on the interval $(0,1) \subseteq \IR$ where $E := (-1, 0) \cup (0, 1) \subseteq \IR$, and the two maps $(0, 1) \rightrightarrows E$ are the inclusion map and $x \mapsto -x$. Then for any metric space $X$, the induced relation on non-expansive maps $(0, 1) \to X$ is that $f \sim g$ if and only if $$d(f(x), g(y)) \le x+y$$ for each $x, y \in (0, 1)$. This is reflexive since $d(f(x), f(y)) \le |x-y| < x+y$, and it is clearly symmetric. For transitivity, suppose $f\sim g$ and $g\sim h$. Then for any $\varepsilon > 0$, we have $$d(f(x), h(y)) \le d(f(x), g(\varepsilon)) + d(g(\varepsilon), h(y)) \le (x + \varepsilon) + (y + \varepsilon).$$ Since this holds for every $\varepsilon > 0$, we conclude $d(f(x), h(y)) \le x+y$.
+      On the other hand, if this cocongruence were effective, then by the dual of <a href="/lemma/effective-congruence-quotients">this result</a>, it would be the cokernel pair of the equalizer of the two inclusion maps. However, that equalizer is empty, so $E$ would have to be a binary copower of $(0,1)$, which does not exist in $\Met$.
 
 special_objects:
   initial object:

databases/catdat/data/categories/Mon.yaml

@@ -67,7 +67,9 @@ unsatisfied_properties:
     reason: 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 $\Mon$ the canonical homomorphism $\coprod_{n \geq 0} \IN \to \prod_{n \geq 0} \IN$ is not an epimorphism because its domain is countable and its codomain is uncountable.
 
   - property_id: effective cocongruences
-    reason: 'We adapt the counterexample from <a href="https://mathoverflow.net/a/510809" target="_blank">MO/510744</a> for $\Ring$. Namely, consider the monoids $$\begin{align*} X & := \langle p \mid p^2 = p \rangle \cong (\{ 0, 1 \}, \cdot),\\ E & := \langle p, q \mid p^2 = p,\, q^2 = q,\, pq = q,\, qp = p \rangle. \end{align*}$$ Then $X$ represents the functor sending a monoid $M$ to its idempotents, and $E$ represents the relation on idempotents $a, b$ of $M$ that $ab = b$, $ba = a$. The equations are equivalent to $aM = bM$, showing that the relation is indeed an equivalence relation.<br> On the other hand, using the multiplicative map $$E \to M_{2\times 2}(\IZ), \quad p \mapsto \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix},\quad q \mapsto \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix},$$ we can see that $p \ne q$ in $E$, so the equalizer of the two maps $X \rightrightarrows E$ is the trivial submonoid $\{ 1 \}$. Therefore, if $E$ were effective, it would be isomorphic to the coproduct $X \sqcup X$, whose underlying set consists of words in $p,q$ with $p,q$ strictly alternating. In particular, in this coproduct, $pq \ne q$.'
+    reason: |
+      We adapt the counterexample from <a href="https://mathoverflow.net/a/510809" target="_blank">MO/510744</a> for $\Ring$. Namely, consider the monoids $$\begin{align*} X & := \langle p \mid p^2 = p \rangle \cong (\{ 0, 1 \}, \cdot),\\ E & := \langle p, q \mid p^2 = p,\, q^2 = q,\, pq = q,\, qp = p \rangle. \end{align*}$$ Then $X$ represents the functor sending a monoid $M$ to its idempotents, and $E$ represents the relation on idempotents $a, b$ of $M$ that $ab = b$, $ba = a$. The equations are equivalent to $aM = bM$, showing that the relation is indeed an equivalence relation.
+      On the other hand, using the multiplicative map $$E \to M_{2\times 2}(\IZ), \quad p \mapsto \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix},\quad q \mapsto \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix},$$ we can see that $p \ne q$ in $E$, so the equalizer of the two maps $X \rightrightarrows E$ is the trivial submonoid $\{ 1 \}$. Therefore, if $E$ were effective, it would be isomorphic to the coproduct $X \sqcup X$, whose underlying set consists of words in $p,q$ with $p,q$ strictly alternating. In particular, in this coproduct, $pq \ne q$.
 
 special_objects:
   initial object:

databases/catdat/data/categories/Pos.yaml

@@ -61,7 +61,9 @@ unsatisfied_properties:
     reason: Pick any poset $X$ which has a decreasing sequence of non-empty sets $X = X_0 \supseteq X_1 \supseteq \cdots$ with empty intersection, such as $X_n = \IN_{\geq n}$ with the natural ordering. The unique map $X_n \to 1$ is surjective, but their limit $\varnothing \to 1$ is not surjective.
 
   - property_id: effective cocongruences
-    reason: 'Let $X$ be $\IR$ with the standard (total) order, and let $E$ be the poset with underlying set $\IR \times \{ 0, 1 \}$ and partial order such that $(x, m) \le (y, n)$ if and only if $x < y$ or $(x, m) = (y, n)$. The two maps $\IR \rightrightarrows E$ will be $x \mapsto (x, 0)$ and $x \mapsto (x, 1)$ respectively. For any partial order $(\IP, \le)$, the induced equivalence relation on the set of order-preserving functions $\IR \to \IP$ is that $f \sim g$ if and only if $f(x) \le g(y)$ and $g(x) \le f(y)$ whenever $x < y$. This relation is clearly reflexive and symmetric; for transitivity, if $f \sim g$ and $g \sim h$, then whenever $x < y$, we have $f(x) \le g(\frac{x+y}{2}) \le h(y)$ and similarly $h(x) \le g(\frac{x+y}{2}) \le f(y)$, showing that $f \sim h$.<br> On the other hand, if this cocongruence on $\IR$ were effective, then by the dual of <a href="/lemma/effective-congruence-quotients">this result</a>, $E$ would be the cokernel pair of the equalizer of the two maps $\IR \rightrightarrows E$. However, that equalizer is the empty poset, so $E$ would have to be the coproduct poset $\IR + \IR$, giving a contradiction.'
+    reason: |
+      Let $X$ be $\IR$ with the standard (total) order, and let $E$ be the poset with underlying set $\IR \times \{ 0, 1 \}$ and partial order such that $(x, m) \le (y, n)$ if and only if $x < y$ or $(x, m) = (y, n)$. The two maps $\IR \rightrightarrows E$ will be $x \mapsto (x, 0)$ and $x \mapsto (x, 1)$ respectively. For any partial order $(\IP, \le)$, the induced equivalence relation on the set of order-preserving functions $\IR \to \IP$ is that $f \sim g$ if and only if $f(x) \le g(y)$ and $g(x) \le f(y)$ whenever $x < y$. This relation is clearly reflexive and symmetric; for transitivity, if $f \sim g$ and $g \sim h$, then whenever $x < y$, we have $f(x) \le g(\frac{x+y}{2}) \le h(y)$ and similarly $h(x) \le g(\frac{x+y}{2}) \le f(y)$, showing that $f \sim h$.
+      On the other hand, if this cocongruence on $\IR$ were effective, then by the dual of <a href="/lemma/effective-congruence-quotients">this result</a>, $E$ would be the cokernel pair of the equalizer of the two maps $\IR \rightrightarrows E$. However, that equalizer is the empty poset, so $E$ would have to be the coproduct poset $\IR + \IR$, giving a contradiction.
 
 special_objects:
   initial object:

databases/catdat/data/categories/R-Mod.yaml

@@ -3,7 +3,9 @@ name: category of left modules over a ring
 notation: $R{-}\Mod$
 objects: left $R$-modules
 morphisms: $R$-linear maps
-description: 'This is the prototype of an abelian category. The category of right modules is the same with the opposite ring $R^{\op}$, hence not listed here.<br> To settle the unsatisfied properties, we make the assumption that $R$ is <i>not</i> semisimple: If $R$ is semisimple, then by the Artin-Wedderburn theorem, the category is equivalent to a finite direct product of categories $D{-}\Mod$ for division rings $D$, and the case of division rings is in a separate entry. In particular, $R \neq 0$ and $R$ is not a field.'
+description: |
+  This is the prototype of an abelian category. The category of right modules is the same with the opposite ring $R^{\op}$, hence not listed here.
+  To settle the unsatisfied properties, we make the assumption that $R$ is <i>not</i> semisimple: If $R$ is semisimple, then by the Artin-Wedderburn theorem, the category is equivalent to a finite direct product of categories $D{-}\Mod$ for division rings $D$, and the case of division rings is in a separate entry. In particular, $R \neq 0$ and $R$ is not a field.
 nlab_link: https://ncatlab.org/nlab/show/module
 tags:
   - algebra