diff --git a/content/exact_filtered_colimits_descend.md b/content/exact_filtered_colimits_descend.md deleted file mode 100644 index 002fcdc0..00000000 --- a/content/exact_filtered_colimits_descend.md +++ /dev/null @@ -1,26 +0,0 @@ ---- -title: Exact filtered colimits descend to nice subcategories -description: Exact filtered colimits are inherited by reflective subcategories with a finite-limit preserving reflector. -author: Martin Brandenburg ---- - -## Exact filtered colimits descend to nice subcategories - -::: Lemma -Let $G : \C \to \D$ be a fully faithful functor with a left adjoint $F : \D \to \C$ that preserves finite limits. Assume that $\D$ has exact filtered colimits and that $\C$ has finite limits. Then $\C$ has exact filtered colimits as well. -::: - -_Proof._ -It is well-known (and easy to prove) that the colimit of a diagram $(X_j)$ in $\C$ is constructed as $F(\colim_j G(X_j))$, provided that colimit in $\D$ exists. In particular, $\C$ has filtered colimits. By assumption, it also has finite limits, and $G$ preserves these 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: - -$$ -\begin{align*} -\colim_j {\lim}_i X(i,j) & \cong F(\colim_j G({\lim}_i X(i,j))) \\ -& \cong F(\colim_j {\lim}_i G(X(i,j))) \\ -& \cong F({\lim}_i \colim_j G(X(i,j))) \\ -& \cong {\lim}_i F(\colim_j G(X(i,j))) \\ -& \cong {\lim}_i \colim_j X(i,j) -\end{align*} -$$ - -$\square$ diff --git a/content/filtered-monos.md b/content/filtered-monos.md deleted file mode 100644 index 5e839a23..00000000 --- a/content/filtered-monos.md +++ /dev/null @@ -1,16 +0,0 @@ ---- -title: Detection of filtered-colimit-stable monomorphisms -description: A useful result to prove (and disprove) filtered-colimit-stable monomorphisms for several categories based on other categories -author: Martin Brandenburg ---- - -## Detection of filtered-colimit-stable monomorphisms - -::: Lemma -Let $\C$ be a category with filtered colimits. Assume that $U : \C \to \D$ is a faithful functor which preserves monomorphisms and filtered colimits. If monomorphisms in $\D$ are stable under filtered colimits, then the same is true for $\C$. -::: - -For the record, here is the dual statement: Let $\C$ be a category with cofiltered limits. Assume that $U : \C \to \D$ is faithful functor which preserves epimorphisms and cofiltered limits. If epimorphisms in $\D$ are stable under cofiltered limits, then the same is true for $\C$. - -_Proof._ -Since $U$ is faithful, it reflects monomorphisms. From here the proof is straight forward. $\square$ diff --git a/content/subcategories.md b/content/subcategories.md new file mode 100644 index 00000000..afb257c0 --- /dev/null +++ b/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. $\square$ + +::: 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 $\C$. $\square$ + +::: 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: + +$$ +\begin{align*} +\colim_j {\lim}_i X(i,j) & \cong L(\colim_j U({\lim}_i X(i,j))) \\ +& \cong L(\colim_j {\lim}_i U(X(i,j))) \\ +& \cong L({\lim}_i \colim_j U(X(i,j))) \\ +& \cong {\lim}_i L(\colim_j U(X(i,j))) \\ +& \cong {\lim}_i \colim_j X(i,j) +\end{align*} +$$ + +$\square$ diff --git a/content/subobject_classifiers_coreflection.md b/content/subobject_classifiers_coreflection.md deleted file mode 100644 index 00c8795d..00000000 --- a/content/subobject_classifiers_coreflection.md +++ /dev/null @@ -1,14 +0,0 @@ ---- -title: Coreflection of subobject classifiers -description: How to construct subobject classifiers in a coreflective subcategory. -author: Martin Brandenburg ---- - -## Coreflection of subobject classifiers - -::: Lemma -Let $\D$ be a category with a (regular) subobject classifier $\Omega$. Assume that $\C \to \D$ is a full subcategory such that (1) any (regular) $\D$-subobject of an object in $\C$ already lies in $\C$, (2) it is coreflective, i.e. there is a functor $R : \D \to \C$ right adjoint to the inclusion. Then $R(\Omega)$ is a (regular) subobject classifier in $\C$. -::: - -_Proof._ -If $X \in \C$, then $\Hom(X,R(\Omega)) \cong \Hom(X,\Omega)$ is isomorphic to the collection of $\D$-subobjects of $X$, which by assumption coincide with the $\C$-subobjects of $X$. $\square$ diff --git a/databases/catdat/data/categories/Alg(R).yaml b/databases/catdat/data/categories/Alg(R).yaml index d6632b4e..96fc09ad 100644 --- a/databases/catdat/data/categories/Alg(R).yaml +++ b/databases/catdat/data/categories/Alg(R).yaml @@ -53,7 +53,7 @@ unsatisfied_properties: proof: 'We just need to tweak the proof for $\Ring$. 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 (proof) or via a direct calculation with elementary matrices.' - property: regular quotient object classifier - proof: We may copy the proof for $\CRing$ (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 this lemma (dualized). + proof: We may copy the proof for $\CRing$ (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 here (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 $\CAlg(R)$ does not have this property. Now apply the contrapositive of the dual of this lemma to the forgetful functor $\CAlg(R) \to \Alg(R)$. It preserves epimorphisms by MSE/5133488. + proof: We already know that $\CAlg(R)$ does not have this property. Now apply the contrapositive of the dual of Lemma 2 here to the forgetful functor $\CAlg(R) \to \Alg(R)$. It preserves epimorphisms by MSE/5133488. - property: effective cocongruences proof: 'The counterexample is similar to the one for $\Ring$: Let $X \coloneqq R[p] / (p^2-p)$ with cocongruence $E \coloneqq R \langle p, q \rangle / (p^2-p, q^2-q, pq-q, qp-p)$.' diff --git a/databases/catdat/data/categories/Cat.yaml b/databases/catdat/data/categories/Cat.yaml index 3d1f40ef..80e08f45 100644 --- a/databases/catdat/data/categories/Cat.yaml +++ b/databases/catdat/data/categories/Cat.yaml @@ -56,7 +56,7 @@ unsatisfied_properties: proof: 'We can adapt the proof from $\Mon$ 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 $\Set$ does not have this property. Now apply the contrapositive of the dual of this lemma to the functor $\Set \to \Cat$ that maps a set to its discrete category. + proof: We already know that $\Set$ does not have this property. Now apply the contrapositive of the dual of Lemma 2 here to the functor $\Set \to \Cat$ that maps a set to its discrete category. - property: effective cocongruences proof: >- diff --git a/databases/catdat/data/categories/Grp.yaml b/databases/catdat/data/categories/Grp.yaml index 453dacf8..1b541e60 100644 --- a/databases/catdat/data/categories/Grp.yaml +++ b/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 $\Ab$ does not have this property. Now apply the contrapositive of the dual of this lemma to the forgetful functor $\Ab \to \Grp$ which indeed preserves epimorphisms. + proof: We already know that $\Ab$ does not have this property. Now apply the contrapositive of the dual of Lemma 2 here to the forgetful functor $\Ab \to \Grp$ which indeed preserves epimorphisms. - property: cocartesian cofiltered limits proof: >- diff --git a/databases/catdat/data/categories/Haus.yaml b/databases/catdat/data/categories/Haus.yaml index c71cfcda..81bcd23e 100644 --- a/databases/catdat/data/categories/Haus.yaml +++ b/databases/catdat/data/categories/Haus.yaml @@ -66,7 +66,7 @@ unsatisfied_properties: proof: 'It is shown in MSE/1255678 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) this lemma 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 here 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: |- diff --git a/databases/catdat/data/categories/Meas.yaml b/databases/catdat/data/categories/Meas.yaml index c9710ff5..d7ce5a65 100644 --- a/databases/catdat/data/categories/Meas.yaml +++ b/databases/catdat/data/categories/Meas.yaml @@ -44,7 +44,7 @@ satisfied_properties: proof: '[Sketch] Since $\Set$ 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 this lemma applied to the forgetful functor to $\Set$. + proof: This follows from Lemma 2 here 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 MSE/5027218. - property: cofiltered-limit-stable epimorphisms - proof: We already know that $\Set$ does not have this property. Now apply the contrapositive of the dual of this lemma to the functor $\Set \to \Meas$ which equips a set with the trivial $\sigma$-algebra. + proof: We already know that $\Set$ does not have this property. Now apply the contrapositive of the dual of Lemma 2 here 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 $\Top$: Use the trivial $\sigma$-algebra on a two-point set.' diff --git a/databases/catdat/data/categories/Met_oo.yaml b/databases/catdat/data/categories/Met_oo.yaml index 74fa5c9b..ee2171c2 100644 --- a/databases/catdat/data/categories/Met_oo.yaml +++ b/databases/catdat/data/categories/Met_oo.yaml @@ -55,7 +55,7 @@ unsatisfied_properties: proof: We can copy the proof from $\Met$. - property: cofiltered-limit-stable epimorphisms - proof: We already know that $\Set$ does not have this property. Now apply the contrapositive of the dual of this lemma to the functor $\Set \to \Met_{\infty}$ that equips a set with the discrete topology. + proof: We already know that $\Set$ does not have this property. Now apply the contrapositive of the dual of Lemma 2 here to the functor $\Set \to \Met_{\infty}$ that equips a set with the discrete topology. - property: effective cocongruences proof: The same counterexample as for $\Met$ 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 $(-1,0) \cup (0,1)$. diff --git a/databases/catdat/data/categories/Mon.yaml b/databases/catdat/data/categories/Mon.yaml index 4647b42c..10474de5 100644 --- a/databases/catdat/data/categories/Mon.yaml +++ b/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 $\CMon$. Alternatively, we may use this lemma (dualized). + proof: We can just copy the proof for $\CMon$. Alternatively, we may use Lemma 1 here (dualized) applied to the forgetful functor $\CMon \to \Mon$. - property: cofiltered-limit-stable epimorphisms - proof: We already know that $\Grp$ does not have this property. Now apply the contrapositive of the dual of this lemma 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 $\Grp$ does not have this property. Now apply the contrapositive of the dual of Lemma 2 here 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 $\Grp$ 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.' diff --git a/databases/catdat/data/categories/Prost.yaml b/databases/catdat/data/categories/Prost.yaml index 5db0c545..c120b60d 100644 --- a/databases/catdat/data/categories/Prost.yaml +++ b/databases/catdat/data/categories/Prost.yaml @@ -57,7 +57,7 @@ unsatisfied_properties: proof: 'See MO/509552: Consider the forgetful functor $U : \Prost \to \Set$ and the relation $R \subseteq U^2$ defined by $R(A) \coloneqq \{(a,b) \in U(A)^2 : a \leq b\}$. Both are representable: $U$ by the singleton preordered set and $R$ by $\{0 \leq 1 \}$. It is clear that $R$ is reflexive, but not symmetric.' - property: cofiltered-limit-stable epimorphisms - proof: We know that $\Set$ does not have this property. Now use the contrapositive of the dual of this lemma applied to the functor $\Set \to \Prost$ that equips a set with the chaotic preorder. + proof: We know that $\Set$ does not have this property. Now use the contrapositive of the dual of Lemma 2 here applied to the functor $\Set \to \Prost$ that equips a set with the chaotic preorder. - property: effective cocongruences proof: 'Consider the proset $E \coloneqq \{ a, b \}$ with the chaotic preorder. This represents the functor which sends a proset to the pairs of elements $x,y$ with $x \le y$ and $y \le x$. Therefore, it defines a cocongruence $1 \rightrightarrows E$, where the maps are the two possible functions. However, this cannot be effective: for any map $h : Z \to 1$ which equalizes the two functions, $Z$ must be empty. But that means the cokernel pair of $h$ is the two-element proset with the trivial preorder.' diff --git a/databases/catdat/data/categories/Ring.yaml b/databases/catdat/data/categories/Ring.yaml index 2798a58b..0e2a6342 100644 --- a/databases/catdat/data/categories/Ring.yaml +++ b/databases/catdat/data/categories/Ring.yaml @@ -56,7 +56,7 @@ unsatisfied_properties: proof: 'Let $B \coloneqq M_2(\IQ)$ and $A \coloneqq \IQ^2$. 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 $M \coloneqq \bigl(\begin{smallmatrix} 1 & 0 \\ 0 & 2 \end{smallmatrix}\bigr)$, 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 (proof) or via a direct calculation with elementary matrices.' - property: regular quotient object classifier - proof: We may copy the proof for $\CRing$ (since the proof there did not use that $P$ is commutative). Alternatively, any regular quotient object classifier in $\Ring$ would produce one in $\CRing$ by this lemma (dualized). + proof: We may copy the proof for $\CRing$ (since the proof there did not use that $P$ is commutative). Alternatively, any regular quotient object classifier in $\Ring$ would produce one in $\CRing$ by Lemma 1 here (dualized). - property: cocartesian cofiltered limits proof: >- @@ -65,7 +65,7 @@ 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 know that $\CRing$ does not have this property. Now use the contrapositive of the dual of this lemma applied to the forgetful functor $\CRing \to \Ring$. It preserves epimorphisms by MSE/5133488. + proof: We know that $\CRing$ does not have this property. Now use the contrapositive of the dual of Lemma 2 here applied to the forgetful functor $\CRing \to \Ring$. It preserves epimorphisms by MSE/5133488. - property: effective cocongruences proof: See MO/510744. diff --git a/databases/catdat/data/categories/Rng.yaml b/databases/catdat/data/categories/Rng.yaml index b1ad2b70..d491880c 100644 --- a/databases/catdat/data/categories/Rng.yaml +++ b/databases/catdat/data/categories/Rng.yaml @@ -62,7 +62,7 @@ unsatisfied_properties: proof: 'We can copy the proof for $\Ring$. In short, the inclusion of diagonal matrices $\IQ^2 \hookrightarrow M_2(\IQ)$ is a regular monomorphism, but becomes zero after taking the pushout with $p_1 : \IQ^2 \twoheadrightarrow \IQ$ because $M_2(\IQ)$ is simple.' - property: regular quotient object classifier - proof: 'Assume that $\Rng$ has a regular quotient object classifier $P$. Consider the functor $N : \Ab \to \Rng$ 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 this lemma (dualized) apply and we conclude that $P/P^2$ is a regular quotient object classifier of $\Ab$. But we already know that $\Ab$ has no such object (in fact, the only additive categories with such an object are trivial by MSE/4086192).' + proof: 'Assume that $\Rng$ has a regular quotient object classifier $P$. Consider the functor $N : \Ab \to \Rng$ 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 Lemma 1 here (dualized) apply and we conclude that $P/P^2$ is a regular quotient object classifier of $\Ab$. But we already know that $\Ab$ has no such object (in fact, the only additive categories with such an object are trivial by MSE/4086192).' - property: cocartesian cofiltered limits proof: >- @@ -71,7 +71,7 @@ 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 know that $\Ring$ does not have this property. Now use the contrapositive of the dual of this lemma applied to the forgetful functor $\Ring \to \Rng$. We only need to verify that it preserves epimorphisms: Let $f : R \to S$ be an epimorphism in $\Ring$ and let $g,h : S \rightrightarrows T$ be two homomorphisms of rngs with $gf = hf$. The element $e = g(1) = h(1) \in T$ is idempotent, and $g,h$ become homomorphisms of rings $S \rightrightarrows eTe$. Hence, $g=h$.' + proof: 'We know that $\Ring$ does not have this property. Now use the contrapositive of the dual of Lemma 2 here applied to the forgetful functor $\Ring \to \Rng$. We only need to verify that it preserves epimorphisms: Let $f : R \to S$ be an epimorphism in $\Ring$ and let $g,h : S \rightrightarrows T$ be two homomorphisms of rngs with $gf = hf$. The element $e = g(1) = h(1) \in T$ is idempotent, and $g,h$ become homomorphisms of rings $S \rightrightarrows eTe$. Hence, $g=h$.' - property: effective cocongruences proof: >- diff --git a/databases/catdat/data/categories/SemiGrp.yaml b/databases/catdat/data/categories/SemiGrp.yaml index 1f699617..06b1659d 100644 --- a/databases/catdat/data/categories/SemiGrp.yaml +++ b/databases/catdat/data/categories/SemiGrp.yaml @@ -63,7 +63,7 @@ unsatisfied_properties: is a normal subgroup of $G$. It is proper, and hence trivial. But then $f$ is injective, which is a contradiction. - property: cofiltered-limit-stable epimorphisms - proof: We already know that $\Set$ does not have this property (by this result). Now apply the contrapositive of the dual of this lemma to the functor $\Set \to \SemiGrp$ that equips a set with the multiplication $a \cdot b \coloneqq a$. + proof: We already know that $\Set$ does not have this property (by this result). Now apply the contrapositive of the dual of Lemma 2 here to the functor $\Set \to \SemiGrp$ that equips a set with the multiplication $a \cdot b \coloneqq a$. - property: effective cocongruences proof: >- diff --git a/databases/catdat/data/categories/Set_pointed.yaml b/databases/catdat/data/categories/Set_pointed.yaml index 23a04e53..e14cfef7 100644 --- a/databases/catdat/data/categories/Set_pointed.yaml +++ b/databases/catdat/data/categories/Set_pointed.yaml @@ -67,7 +67,7 @@ unsatisfied_properties: proof: 'Every cokernel is "injective away from the base point". Formally, if $p : A \to B$ is a cokernel in $\Set_*$, it has the property that $p(x)=p(y) \neq 0$ implies $x=y$ (where $0$ denotes the base point). Clearly this is not satisfied for every surjective pointed map, consider $(\IN,0) \to (\{0,1\},0)$ defined by $0 \mapsto 0$ and $x \mapsto 1$ for $x > 0$.' - property: cofiltered-limit-stable epimorphisms - proof: We already know that $\Set$ does not have this property. Now apply the contrapositive of the dual of this lemma to the functor $\Set \to \Set_*$ that freely adds a base point. + proof: We already know that $\Set$ does not have this property. Now apply the contrapositive of the dual of Lemma 2 here to the functor $\Set \to \Set_*$ that freely adds a base point. special_objects: initial object: diff --git a/databases/catdat/data/categories/Setne.yaml b/databases/catdat/data/categories/Setne.yaml index 373584a0..e4c3c5ef 100644 --- a/databases/catdat/data/categories/Setne.yaml +++ b/databases/catdat/data/categories/Setne.yaml @@ -51,7 +51,7 @@ satisfied_properties: proof: Any natural numbers object in $\Set$, such as $(\IN,0,n \mapsto n+1)$, is clearly also one in $\Setne$. - property: filtered-colimit-stable monomorphisms - proof: This follows from this lemma applied to the forgetful functor to $\Set$. + proof: This follows from Lemma 2 here applied to the forgetful functor to $\Set$. - property: multi-complete proof: Let $D$ be a diagram in $\Setne$, and let $L$ be a limit of $D$ in $\Set$. If $L$ is non-empty, it gives a limit in $\Setne$ as well. If $L$ is the empty set, there is no cone over $D$ in $\Setne$; hence the empty set of cones gives a multi-limit of $D$ in $\Setne$. diff --git a/databases/catdat/data/categories/Top.yaml b/databases/catdat/data/categories/Top.yaml index 56f8cd0f..608ae1da 100644 --- a/databases/catdat/data/categories/Top.yaml +++ b/databases/catdat/data/categories/Top.yaml @@ -51,7 +51,7 @@ satisfied_properties: proof: The category has all limits and colimits, and the regular monomorphisms are the subspace inclusions. Thus, it suffices to prove that subspace inclusions are stable under pushouts. For a proof see e.g. Lemma 3.6 at the nLab. - property: filtered-colimit-stable monomorphisms - proof: This follows from this lemma applied to the forgetful functor to $\Set$. + proof: This follows from Lemma 2 here applied to the forgetful functor to $\Set$. unsatisfied_properties: - property: skeletal @@ -79,7 +79,7 @@ unsatisfied_properties: proof: 'Assume $\Top$ is coaccessible. Let $p : S \to I$ be the identity map from the Sierpinski space to the two-element indiscrete space. Then, a topological space is discrete if and only if it is projective to the morphism $p$. This implies that the full subcategory spanned by all discrete spaces, which is equivalent to $\Set$, is coaccessible by Prop. 4.7 in Adamek-Rosicky. However, since $\Set$ is not coaccessible, this is a contradiction.' - property: cofiltered-limit-stable epimorphisms - proof: We already know that $\Set$ does not have this property. Now apply the contrapositive of the dual of this lemma to the functor $\Set \to \Top$ which equips a set with the indiscrete topology. + proof: We already know that $\Set$ does not have this property. Now apply the contrapositive of the dual of Lemma 2 here to the functor $\Set \to \Top$ which equips a set with the indiscrete topology. - property: effective cocongruences proof: 'Consider the indiscrete topological space $I$ on two points. This represents the functor which takes a topological space $X$ to the pairs of indistinguishable points of $X$. Therefore, we get a cocongruence $1 \rightrightarrows I$, where the maps are the two possible functions. However, this cannot be effective: if we have $h : Z\to 1$ which equalizes the two maps, then $Z$ must be empty. But that means the cokernel pair of $h$ is the discrete space on two points.' diff --git a/databases/catdat/data/categories/Top_pointed.yaml b/databases/catdat/data/categories/Top_pointed.yaml index 966e2ed7..73930b88 100644 --- a/databases/catdat/data/categories/Top_pointed.yaml +++ b/databases/catdat/data/categories/Top_pointed.yaml @@ -61,7 +61,7 @@ satisfied_properties: 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. - property: filtered-colimit-stable monomorphisms - proof: This follows from this lemma applied to the forgetful functor to $\Set$. + proof: This follows from Lemma 2 here applied to the forgetful functor to $\Set$. - property: coregular proof: Regular monomorphisms coincide with the embeddings (see below). Since $\Top$ is coregular, they are stable under pushouts, and pushouts in $\Top_*$ are the same. @@ -99,7 +99,7 @@ unsatisfied_properties: proof: 'We can adjust the proof for $\Top$ as follows: Assume $\Top_*$ is coaccessible. Let $S_0=\{x,*\}$ be the pointed topological space such that $\{*\}$ is the only non-trivial open set, and let $S_1=\{x,*\}$ be the pointed space such that $\{x\}$ is the only non-trivial open set. Let $p_i : S_i \to \{x,*\}$ be the identity function to the two-element indiscrete pointed space. Then, a pointed topological space is discrete if and only if it is projective to the morphisms $p_0$ and $p_1$. This implies that the full subcategory spanned by all discrete pointed spaces, which is equivalent to $\Set_*$, is coaccessible by Prop. 4.7 in Adamek-Rosicky. However, since $\Set_*$ is not coaccessible, this is a contradiction.' - property: cofiltered-limit-stable epimorphisms - proof: We already know that $\Set_*$ does not have this property. Now apply the contrapositive of the dual of this lemma to the functor $\Set_* \to \Top_*$ that equips a pointed set with the indiscrete topology. + proof: We already know that $\Set_*$ does not have this property. Now apply the contrapositive of the dual of Lemma 2 here to the functor $\Set_* \to \Top_*$ that equips a pointed set with the indiscrete topology. - property: effective congruences proof: Suppose that $\Top_*$ had effective congruences. Then by this result, $\Top$ would also have effective congruences, which we know is not the case. diff --git a/databases/catdat/data/categories/Z.yaml b/databases/catdat/data/categories/Z.yaml index 1be7b099..850de6c6 100644 --- a/databases/catdat/data/categories/Z.yaml +++ b/databases/catdat/data/categories/Z.yaml @@ -72,7 +72,7 @@ unsatisfied_properties: proof: 'Consider the functor $F$ from MO/390611 for example. The collection of subobjects of $F$ is not isomorphic to a set: for each infinite cardinal $\kappa$, simply cut off the construction of $F$ at $\kappa$. This yields a different subobject for each $\kappa$.' - property: cofiltered-limit-stable epimorphisms - proof: We already know that $\Set$ does not have this property. Now apply the contrapositive of the dual of this lemma to the functor $\Set \to [\CRing, \Set]$ that maps a set to its constant functor. + proof: We already know that $\Set$ does not have this property. Now apply the contrapositive of the dual of Lemma 2 here to the functor $\Set \to [\CRing, \Set]$ that maps a set to its constant functor. special_objects: initial object: diff --git a/databases/catdat/data/category-implications/topos.yaml b/databases/catdat/data/category-implications/topos.yaml index 1076f87b..80d56add 100644 --- a/databases/catdat/data/category-implications/topos.yaml +++ b/databases/catdat/data/category-implications/topos.yaml @@ -57,7 +57,7 @@ - exact filtered colimits - infinitary extensive - locally presentable - proof: A Grothendieck topos is locally presentable by Prop. 3.4.16 in Handbook of Categorical Algebra Vol. 3, has a cogenerator (see nLab) and is infinitary extensive by Giraud's Theorem. To show that it has exact filtered colimits, first observe that this is clearly true in every presheaf topos (since $\Set$ has the property). Every Grothendieck topos is a full reflective subcategory of a presheaf topos such that the reflector preserves finite limits (nLab), so we conclude with this lemma. + proof: A Grothendieck topos is locally presentable by Prop. 3.4.16 in Handbook of Categorical Algebra Vol. 3, has a cogenerator (see nLab) and is infinitary extensive by Giraud's Theorem. To show that it has exact filtered colimits, first observe that this is clearly true in every presheaf topos (since $\Set$ has the property). Every Grothendieck topos is a full reflective subcategory of a presheaf topos such that the reflector preserves finite limits (nLab), so we conclude with Lemma 3 here. is_equivalence: false - id: topos_is_locally_cartesian_closed