use LaTeX macros in results about effective (co)congruences

Author: ScriptRaccoon

databases/catdat/data/003_category-property-assignments/Alg(R).sql

@@ -101,5 +101,5 @@ VALUES
 	'Alg(R)',
 	'effective cocongruences',
 	FALSE,
-	'The counterexample is similar to the one for <a href="/category/Ring">$\mathbf{Ring}$</a>: Let $X := R[p] / (p^2-p)$ with cocongruence $E := R \langle p, q \rangle / (p^2-p, q^2-q, pq-q, qp-p)$.'
+	'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/003_category-property-assignments/Cat.sql

@@ -93,7 +93,7 @@ VALUES
 	'Cat',
 	'effective cocongruences',
 	FALSE,
-	'The counterexample is similar to the one for <a href="/category/Mon">$\mathbf{Mon}$</a>: Let $X$ be the <a href="/category/walking_idempotent">walking idempotent</a>, and let $E$ be the delooping of the monoid with presentation
+	'The counterexample is similar to the one for <a href="/category/Mon">$\Mon$</a>: Let $X$ be the <a href="/category/walking_idempotent">walking idempotent</a>, and let $E$ be the delooping of the monoid with presentation
 	$$\langle p, q \mid p^2=p,\, q^2=q,\, pq=q,\, qp=p \rangle.$$
-	The induced relation on functors in $[X, \mathcal{C}]$ is that $F \sim G$ if and only if $F$ and $G$ send the object of $X$ to the same object of $\mathcal{C}$, and they send the idempotent of $X$ to idempotent morphisms $a, b$ in $\mathcal{C}$ satisfying $ab=b$, $ba=a$. From here, the proof that this gives a cocongruence on $\mathbf{Cat}$ which is not effective is similar to the one in $\mathbf{Mon}$.'
+	The induced relation on functors in $[X, \C]$ is that $F \sim G$ if and only if $F$ and $G$ send the object of $X$ to the same object of $\C$, and they send the idempotent of $X$ to idempotent morphisms $a, b$ in $\C$ satisfying $ab=b$, $ba=a$. From here, the proof that this gives a cocongruence on $\Cat$ which is not effective is similar to the one in $\Mon$.'
 );

databases/catdat/data/003_category-property-assignments/FinGrp.sql

@@ -69,7 +69,7 @@ VALUES
 	'FinGrp',
 	'effective congruences',
 	TRUE,
-	'Suppose we have a congruence $f, g : E \rightrightarrows X$ in $\mathbf{FinGrp}$. Since the embedding $\mathbf{FinGrp} \hookrightarrow \mathbf{Grp}$ preserves finite limits, it is also a congruence in $\mathbf{Grp}$. We already know that $\mathbf{Grp}$ has effective congruences since it is algebraic. Using <a href="/lemma/effective-congruence-quotients">this result</a>, we see that $E$ is the kernel pair of $X \to (X/E)_{\mathbf{Grp}}$ in $\mathbf{Grp}$. Also, the quotient $(X/E)_{\mathbf{Grp}}$ is finite; and the forgetful functor $\mathbf{FinGrp} \to \mathbf{Grp}$ is fully faithful <a href="https://ncatlab.org/nlab/show/reflected+limit#FullSubcategoryInclusionReflectCoLimits" target="_blank">and therefore reflects limits</a>. Thus, we conclude that $E$ is the kernel pair of $X \to (X/E)_{\mathbf{Grp}}$ in $\mathbf{FinGrp}$ as well.'
+	'Suppose we have a congruence $f, g : E \rightrightarrows X$ in $\FinGrp$. Since the embedding $\FinGrp \hookrightarrow \Grp$ preserves finite limits, it is also a congruence in $\Grp$. We already know that $\Grp$ has effective congruences since it is algebraic. Using <a href="/lemma/effective-congruence-quotients">this result</a>, we see that $E$ is the kernel pair of $X \to (X/E)_{\Grp}$ in $\Grp$. Also, the quotient $(X/E)_{\Grp}$ is finite; and the forgetful functor $\FinGrp \to \Grp$ is fully faithful <a href="https://ncatlab.org/nlab/show/reflected+limit#FullSubcategoryInclusionReflectCoLimits" target="_blank">and therefore reflects limits</a>. Thus, we conclude that $E$ is the kernel pair of $X \to (X/E)_{\Grp}$ in $\FinGrp$ as well.'
 ),
 (
 	'FinGrp',

databases/catdat/data/003_category-property-assignments/FreeAb.sql

@@ -89,8 +89,8 @@ VALUES
 	'FreeAb',
 	'effective cocongruences',
 	FALSE,
-	'We will let $E$ be the abelian group with presentation $\langle a, b, c \mid a - b = 2c \rangle$, with two morphisms $\mathbb{Z} \rightrightarrows E$ given by $1\mapsto a$, $1\mapsto b$. Note that $E$ is free with basis $\{ b, c \}$. Then $\operatorname{Hom}(E, G) \cong \{ (x, y, z) \in G^3 \mid x - y = 2z \}$. 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 $\mathbb{Z}$.<br>
-	On the other hand, if $E$ were the cokernel pair of $h : H \to \mathbb{Z}$, that would mean that for $x, y : \mathbb{Z} \to G$, $x \equiv y \pmod{2G}$ if and only if $x \circ h = y \circ h$. In particular, from the case $G := \mathbb{Z}$, $x := 2 \operatorname{id}$, $y := 0$, we would have $2h = 0$. That implies $h = 0$, but then that would give $\operatorname{id}_{\mathbb{Z}} \equiv 0 \pmod{2}$, resulting in a contradiction.'
+	'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.'
 );
 
 INSERT INTO category_property_comments (category_id, property_id, comment)

databases/catdat/data/003_category-property-assignments/Haus.sql

@@ -69,7 +69,7 @@ VALUES
 	'Haus',
 	'effective cocongruences',
 	TRUE,
-	'As the proof at <a href="https://mathoverflow.net/a/509582/2841" target="_blank">MO/509548</a> shows, in fact any coreflexive corelation on $X$ in $\mathbf{Haus}$ is of the form $X +_S X$ for a closed subset $S$ of $X$. Such a cocongruence is clearly effective.'
+	'As the proof at <a href="https://mathoverflow.net/a/509582/2841" target="_blank">MO/509548</a> shows, in fact any coreflexive corelation on $X$ in $\Haus$ is of the form $X +_S X$ for a closed subset $S$ of $X$. Such a cocongruence is clearly effective.'
 ),
 (
 	'Haus',

databases/catdat/data/003_category-property-assignments/Man.sql

@@ -84,7 +84,7 @@ VALUES
 	'Man',
 	'effective cocongruences',
 	TRUE,
-	'From the proof that $\mathbf{Man}$ 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 $U \hookrightarrow X$.'
+	'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 $U \hookrightarrow X$.'
 ),
 (
 	'Man',

databases/catdat/data/003_category-property-assignments/Meas.sql

@@ -105,5 +105,5 @@ VALUES
 	'Meas',
 	'effective cocongruences',
 	FALSE,
-	'The proof is similar to the one for <a href="/category/Top">$\mathbf{Top}$</a>: Use the trivial $\sigma$-algebra on a two-point set.'
+	'The proof is similar to the one for <a href="/category/Top">$\Top$</a>: Use the trivial $\sigma$-algebra on a two-point set.'
 );

databases/catdat/data/003_category-property-assignments/Met.sql

@@ -149,17 +149,17 @@ VALUES
 	'Met',
 	'effective congruences',
 	FALSE,
-	'Any kernel pair of $h : X \to Z$ in $\mathbf{Met}$ corresponds to a closed subset of $X\times X$. However, there are plenty of non-closed congruences, such as $\Delta \cup (\mathbb{Q} \times \mathbb{Q}) \subseteq \mathbb{R} \times \mathbb{R}$ with the subspace metric.'
+	'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.'
 ),
 (
 	'Met',
 	'effective cocongruences',
 	FALSE,
-	'We will define a cocongruence on the interval $(0,1) \subseteq \mathbb{R}$ where $E := (-1, 0) \cup (0, 1) \subseteq \mathbb{R}$, 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
+	'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 $\mathbf{Met}$.'
+	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 $\Met$.'
 );
 

databases/catdat/data/003_category-property-assignments/Met_c.sql

@@ -69,7 +69,7 @@ VALUES
 	'Met_c',
 	'effective cocongruences',
 	TRUE,
-	'Suppose we have a cocongruence $f, g : X \rightrightarrows E$ in $\mathbf{Met}_\mathrm{c}$. Then the image in $\mathbf{Haus}$ is a coreflexive corelation (since epimorphisms in both categories are continuous maps with dense image). By <a href="https://mathoverflow.net/a/509582/2841" target="_blank">MO/509548</a>, that implies that image is of the form $X +_S X$ for a closed subset $S$ of $X$. Since $S$ is metrizable, and the functor $\mathbf{Met}_\mathrm{c} \to \mathbf{Haus}$ is fully faithful <a href="https://ncatlab.org/nlab/show/reflected+limit#FullSubcategoryInclusionReflectCoLimits" target="_blank">and therefore reflects colimits</a>, we conclude that $E$ is effective in $\mathbf{Met}_\mathrm{c}$.'
+	'Suppose we have a cocongruence $f, g : X \rightrightarrows E$ in $\Met_\c$. Then the image in $\Haus$ is a coreflexive corelation (since epimorphisms in both categories are continuous maps with dense image). By <a href="https://mathoverflow.net/a/509582/2841" target="_blank">MO/509548</a>, that implies that image is of the form $X +_S X$ for a closed subset $S$ of $X$. Since $S$ is metrizable, and the functor $\Met_\c \to \Haus$ is fully faithful <a href="https://ncatlab.org/nlab/show/reflected+limit#FullSubcategoryInclusionReflectCoLimits" target="_blank">and therefore reflects colimits</a>, we conclude that $E$ is effective in $\Met_\c$.'
 ),
 (
 	'Met_c',

databases/catdat/data/003_category-property-assignments/Met_oo.sql

@@ -93,5 +93,5 @@ VALUES
 	'Met_oo',
 	'effective cocongruences',
 	FALSE,
-	'The same counterexample as for <a href="/category/Met">$\mathbf{Met}$</a> works here. The difference in this case is that a binary copower of two copies of $(0,1)$ does exist in $\mathbf{Met}_\infty$. However, this would assign a distance of $\infty$ between points in $(-1,0)$ and points in $(0,1)$, which does not agree with the chosen subspace metric on $(-1,0) \cup (0,1)$.'
+	'The same counterexample as for <a href="/category/Met">$\Met$</a> works here. The difference in this case is that a binary copower of two copies of $(0,1)$ does exist in $\Met_\infty$. However, this would assign a distance of $\infty$ between points in $(-1,0)$ and points in $(0,1)$, which does not agree with the chosen subspace metric on $(-1,0) \cup (0,1)$.'
 );