Combine content pages about subcategories

content/subcategories.md

@@ -0,0 +1,50 @@
+---
+title: Results on subcategories
+description: We prove that several properties of categories descend to suitable subcategories.
+author: Martin Brandenburg
+---
+
+## Results on subcategories
+
+This page collects several useful results of the following form: if $U : \C \to \D$ is a faithful functor (perhaps even fully faithful, or satisfying additional assumptions) and $\D$ has a certain property, then $\C$ has this property as well.
+
+::: Lemma 1
+Let $\D$ be a category with a (regular) subobject classifier $\Omega$. Assume that $U : \C \to \D$ is a fully faithful functor such that (1) $U$ is coreflective, i.e. there is a functor $R : \D \to \C$ right adjoint to $U$, and (2) every (regular) monomorphism $Y \to U(X)$ in $\D$ is the image of a (regular) monomorphism $X' \to X$ in $\C$. Then $R(\Omega)$ is a (regular) subobject classifier in $\C$.
+:::
+
+_Proof._
+If $X \in \C$, then
+$$\Hom(X,R(\Omega)) \cong \Hom(U(X),\Omega) \cong \Sub(U(X)) \cong \Sub(X).$$
+The same proof works for regular subobjects. <span class="qed">$\square$</span>
+
+::: Lemma 2
+Let $\C$ be a category with filtered colimits. Assume that $U : \C \to \D$ is a faithful functor that preserves monomorphisms and filtered colimits. If monomorphisms in $\D$ are stable under filtered colimits, then the same is true in $\C$.
+:::
+
+For the record, here is the dual statement: let $\C$ be a category with cofiltered limits. Assume that $U : \C \to \D$ is a faithful functor that preserves epimorphisms and cofiltered limits. If epimorphisms in $\D$ are stable under cofiltered limits, then the same is true in $\C$.
+
+_Proof._
+If $(f_i : X_i \to Y_i)$ is a filtered diagram of monomorphisms in $\C$, it induces a filtered diagram $(U(f_i) : U(X_i) \to U(Y_i))$ of monomorphisms in $\D$. Hence, its colimit
+$\colim_i U(f_i) : \colim_i U(X_i) \to \colim_i U(Y_i)$
+is a monomorphism in $\D$. This morphism is isomorphic to
+$U(\colim_i f_i) : U(\colim_i X_i) \to U(\colim_i Y_i)$.
+Since $U(\colim_i f_i)$ is a monomorphism in $\D$ and $U$ is faithful, it follows that $\colim_i f_i$ is a monomorphism in $\C$. <span class="qed">$\square$</span>
+
+::: Lemma 3
+Let $U : \C \to \D$ be a fully faithful functor with a left adjoint $L : \D \to \C$ (i.e. $\C$ is equivalent to a reflective subcategory of $\D$). Assume that $\D$ has exact filtered colimits, that $\C$ has finite limits, and that $L$ preserves finite limits. Then $\C$ also has exact filtered colimits.
+:::
+
+_Proof._
+It is well known (and easy to prove) that the colimit of a diagram $(X_j)$ in $\C$ is given by $L(\colim_j U(X_j))$, provided that the colimit in $\D$ exists. In particular, $\C$ has filtered colimits. By assumption, it also has finite limits, and $U$ preserves them since it is a right adjoint. Now let $X : \I \times \J \to \C$ be a diagram, where $\I$ is finite and $\J$ is filtered. We compute:
+
+$$
+\begin{align*}
+\colim_j {\lim}_i X(i,j) & \cong L(\colim_j U({\lim}_i X(i,j))) \\
+& \cong L(\colim_j {\lim}_i U(X(i,j))) \\
+& \cong L({\lim}_i \colim_j U(X(i,j))) \\
+& \cong {\lim}_i L(\colim_j U(X(i,j))) \\
+& \cong {\lim}_i \colim_j X(i,j)
+\end{align*}
+$$
+
+<span class="qed">$\square$</span>

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

@@ -53,7 +53,7 @@ unsatisfied_properties:
     proof: '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 $B \coloneqq M_2(K)$ and $A \coloneqq K \times K$. 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 $M \coloneqq \bigl(\begin{smallmatrix} 1 & 0 \\ 0 & \lambda \end{smallmatrix}\bigr)$, 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: regular quotient object classifier
-    proof: 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="/content/subobject_classifiers_coreflection">this lemma</a> (dualized).
+    proof: 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 the reflective subcategory $\CAlg(R)$ by Lemma 1 <a href="/content/subcategories">here</a> (dualized).
 
   - property: cocartesian cofiltered limits
     proof: >-
@@ -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: cofiltered-limit-stable epimorphisms
-    proof: 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="/content/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>.
+    proof: We already know that <a href="/category/CRing">$\CAlg(R)$</a> does not have this property. Now apply the contrapositive of the dual of Lemma 2 <a href="/content/subcategories">here</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: effective cocongruences
     proof: 'The counterexample is similar to the one for <a href="/category/Ring">$\Ring$</a>: Let $X \coloneqq R[p] / (p^2-p)$ with cocongruence $E \coloneqq R \langle p, q \rangle / (p^2-p, q^2-q, pq-q, qp-p)$.'

databases/catdat/data/categories/Cat.yaml

@@ -56,7 +56,7 @@ unsatisfied_properties:
     proof: 'We can adapt the proof from <a href="/category/Mon">$\Mon$</a> as follows: Consider the functor $U : \Cat \to \Set^+$ sending a category $\C$ to the (large) set $\{(x,u) : x \in \Ob(\C) ,\, u \in \End(x) \}$. It is represented by $B \IN$, the one-object category associated to the free monoid in one generator. Consider the relation $R \subseteq U^2$ consisting of those pairs $((x,u),(y,v))$ where $x = y$ and $uv = u^2$. This also representable, namely be the one-object category associated to the monoid with the presentation $\langle u,v : uv = u^2 \rangle$. Clearly, $R$ is reflexive, but not symmetric.'
 
   - property: cofiltered-limit-stable epimorphisms
-    proof: We already know that <a href="/category/Set">$\Set$</a> does not have this property. Now apply the contrapositive of the dual of <a href="/content/filtered-monos">this lemma</a> to the functor $\Set \to \Cat$ that maps a set to its discrete category.
+    proof: We already know that <a href="/category/Set">$\Set$</a> does not have this property. Now apply the contrapositive of the dual of Lemma 2 <a href="/content/subcategories">here</a> to the functor $\Set \to \Cat$ that maps a set to its discrete category.
 
   - property: effective cocongruences
     proof: >-

databases/catdat/data/categories/Grp.yaml

@@ -66,7 +66,7 @@ unsatisfied_properties:
     proof: 'Assume that $\Grp$ 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 $G$.'
 
   - property: cofiltered-limit-stable epimorphisms
-    proof: We already know that <a href="/category/Ab">$\Ab$</a> does not have this property. Now apply the contrapositive of the dual of <a href="/content/filtered-monos">this lemma</a> to the forgetful functor $\Ab \to \Grp$ which indeed preserves epimorphisms.
+    proof: We already know that <a href="/category/Ab">$\Ab$</a> does not have this property. Now apply the contrapositive of the dual of Lemma 2 <a href="/content/subcategories">here</a> to the forgetful functor $\Ab \to \Grp$ which indeed preserves epimorphisms.
 
   - property: cocartesian cofiltered limits
     proof: >-

databases/catdat/data/categories/Haus.yaml

@@ -66,7 +66,7 @@ unsatisfied_properties:
     proof: 'It is shown in <a href="https://math.stackexchange.com/questions/1255678">MSE/1255678</a> that $\IQ \times - : \Top \to \Top$ does not preserve sequential colimits (so that it cannot be a left adjoint). The same example also works in $\Haus$: Surely $\IQ$ is Hausdorff, $X_n$ is Hausdorff, as is their colimit $X$, and the colimit (taken in $\Top$) of the $X_n \times \IQ$ admits a bijective continuous map to a Hausdorff space, therefore is also Hausdorff, meaning it is also the colimit taken in $\Haus$.'
 
   - property: cofiltered-limit-stable epimorphisms
-    proof: 'Recall the counterexample for sets: The unique maps $\IN_{\geq n} \to 1$ are surjective, but their limit $0 = \bigcap_{n \geq 0} \IN_{\geq n} \to 1$ is not. This also works in $\Haus$ by using discrete topologies. We could also apply a variant of (the dual of) <a href="/content/filtered-monos">this lemma</a> to the discrete topology functor $\Set \to \Haus$, which does not preserve all cofiltered limits, but does preserve intersections.'
+    proof: 'Recall the counterexample for sets: The unique maps $\IN_{\geq n} \to 1$ are surjective, but their limit $0 = \bigcap_{n \geq 0} \IN_{\geq n} \to 1$ is not. This also works in $\Haus$ by using discrete topologies. We could also apply a variant of (the dual of) Lemma 2 <a href="/content/subcategories">here</a> to the discrete topology functor $\Set \to \Haus$, which does not preserve all cofiltered limits, but does preserve intersections.'
 
   - property: filtered-colimit-stable monomorphisms
     proof: |-

databases/catdat/data/categories/Meas.yaml

@@ -44,7 +44,7 @@ satisfied_properties:
     proof: '[Sketch] Since <a href="/category/Set">$\Set$</a> is infinitary extensive, a map $f : Y \to \coprod_i X_i \eqqcolon X$ corresponds to a decomposition $Y = \coprod_i Y_i$ (as sets) with maps $f_i : Y_i \to X_i$. Endow the measurable subset $Y_i \subseteq Y$ with the restricted $\sigma$-algebra. If $f$ is measurable, each $f_i$ is measurable, and $Y = \coprod_i Y_i$ holds as measurable spaces.'
 
   - property: filtered-colimit-stable monomorphisms
-    proof: This follows from <a href="/content/filtered-monos">this lemma</a> applied to the forgetful functor to $\Set$.
+    proof: This follows from Lemma 2 <a href="/content/subcategories">here</a> applied to the forgetful functor to $\Set$.
 
   - property: regular subobject classifier
     proof: The set $\{0,1\}$ with the trivial $\sigma$-algebra is a regular subobject classifier since measurable maps $X \to \{0,1\}$ correspond to subsets of $X$.
@@ -63,7 +63,7 @@ unsatisfied_properties:
     proof: See <a href="https://math.stackexchange.com/questions/5027218" target="_blank">MSE/5027218</a>.
 
   - property: cofiltered-limit-stable epimorphisms
-    proof: We already know that <a href="/category/Set">$\Set$</a> does not have this property. Now apply the contrapositive of the dual of <a href="/content/filtered-monos">this lemma</a> to the functor $\Set \to \Meas$ which equips a set with the trivial $\sigma$-algebra.
+    proof: We already know that <a href="/category/Set">$\Set$</a> does not have this property. Now apply the contrapositive of the dual of Lemma 2 <a href="/content/subcategories">here</a> to the functor $\Set \to \Meas$ which equips a set with the trivial $\sigma$-algebra.
 
   - property: effective cocongruences
     proof: '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/categories/Met_oo.yaml

@@ -55,7 +55,7 @@ unsatisfied_properties:
     proof: We can copy the proof from <a href="/category/Met">$\Met$</a>.
 
   - property: cofiltered-limit-stable epimorphisms
-    proof: We already know that <a href="/category/Set">$\Set$</a> does not have this property. Now apply the contrapositive of the dual of <a href="/content/filtered-monos">this lemma</a> to the functor $\Set \to \Met_{\infty}$ that equips a set with the discrete topology.
+    proof: We already know that <a href="/category/Set">$\Set$</a> does not have this property. Now apply the contrapositive of the dual of Lemma 2 <a href="/content/subcategories">here</a> to the functor $\Set \to \Met_{\infty}$ that equips a set with the discrete topology.
 
   - property: effective cocongruences
     proof: 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 $(-1,0) \cup (0,1)$.

databases/catdat/data/categories/Mon.yaml

@@ -59,10 +59,10 @@ unsatisfied_properties:
     proof: 'Assume that $\Omega$ is a regular subobject classifier. Since the trivial monoid is a zero object, every regular submonoid $U \subseteq M$ of any monoid $M$ would have the form $\{m \in M : h(m) = 1 \}$ for some homomorphism $M \to \Omega$. Now take any monoid $M$ with zero that has two different homomorphisms with zero $f,g : M \rightrightarrows N$ (for example, let $M = N = \{0\} \cup \{x^n : n \geq 0\}$ be the free monoid with zero on one generator, $f(x) = 0$,and $g(x) = x$). Take their equalizer $U \subseteq M$, and choose a homomorphism $h : M \to \Omega$ with $U = \{m \in M : h(m) = 1\}$. Since $0 \in U$, we have $h(0)=1$. But then for all $m \in M$ we have $h(m) = h(m) h(0) = h(m 0) = h(0) = 1$, i.e. $U = M$, which yields the contradiction $f = g$.'
 
   - property: regular quotient object classifier
-    proof: We can just copy the proof for <a href="/category/CMon">$\CMon$</a>. Alternatively, we may use <a href="/content/subobject_classifiers_coreflection">this lemma</a> (dualized).
+    proof: We can just copy the proof for <a href="/category/CMon">$\CMon$</a>. Alternatively, we may use Lemma 1 <a href="/content/subcategories">here</a> (dualized) applied to the forgetful functor $\CMon \to \Mon$.
 
   - property: cofiltered-limit-stable epimorphisms
-    proof: We already know that <a href="/category/Grp">$\Grp$</a> does not have this property. Now apply the contrapositive of the dual of <a href="/content/filtered-monos">this lemma</a> to the forgetful functor $\Grp \to \Mon$. It preserves epimorphisms since it has a right adjoint, the unit group functor.
+    proof: We already know that <a href="/category/Grp">$\Grp$</a> does not have this property. Now apply the contrapositive of the dual of Lemma 2 <a href="/content/subcategories">here</a> to the forgetful functor $\Grp \to \Mon$. It preserves epimorphisms since it has a right adjoint, the unit group functor.
 
   - property: cocartesian cofiltered limits
     proof: 'We know that <a href="/category/Grp">$\Grp$</a> fails to satisfy this property. The same counterexample works here since the inclusion $\Grp \hookrightarrow \Mon$ 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.'