Pullback-stable regular epimorphisms

content/free-cocompletion.md

@@ -78,11 +78,11 @@ The direction $\impliedby$ is trivial in each case. For the direction $\implies$
 :::
 
 ::: Lemma 6
-If $\C$ is a locally small category, then $\widehat{\C}$ is mono-regular. Actually, every monomorphism is an effective monomorphism. Moreover, monomorphisms are stable under filtered colimits.
+If $\C$ is a locally small category, then $\widehat{\C}$ is mono-regular. Actually, every monomorphism is an effective monomorphism. Moreover, monomorphisms are stable under filtered colimits and pushouts.
 :::
 
 ::: Proof
-The first statement is a formal consequence of the fact that every monomorphism in $\Set$ is effective and the already established facts that monomorphisms and pushouts can be understood objectwise. For similar reasons, the second statement is a formal consequence of the corresponding fact for $\Set$.
+The first statement is a formal consequence of the fact that every monomorphism in $\Set$ is effective and the already established facts that monomorphisms and pushouts can be understood objectwise. For similar reasons, the second statement is a formal consequence of the corresponding facts for $\Set$.
 :::
 
 ::: Lemma 7
@@ -181,7 +181,7 @@ is a weakly terminal essentially small collection in $\int E$, which is equivale
 $$\textstyle \coprod_{A \in K,\, a \in E(A)} \Hom(-,A) \to E$$
 is an epimorphism of presheaves, as required.
 
-Let $(X,x)$ be an object of $\int E$, i.e. $X \in \Ob(\C)$ and $x \in E(X)$. In particular, $x \in P(X)$. Thus the comma category $(X,x) \downarrow \K$ is connected, and therefore non-empty. Choose an object $f : (X,x) \to (A,a)$. Thus, $A \in K$, $a \in P(A)$, and $f : X \to A$ satisfies $f^*(a)=x$. If $a \in E(A)$, we are done. Assume otherwise and, without loss of generality, $a \in F_1(A)$. Let $b \in F_2(A) \subseteq P(A)$ be the corresponding element in the other copy of $F$. Using the flip automorphism of $P$ that fixes $E$ and exchanges $F_1$ and $F_2$, we see that $f^*(b)=x$. Thus, we also have a morphism $f' : (X,x) \to (A,b)$ with the same underlying morphism $f : X \to A$.
+Let $(X,x)$ be an object of $\int E$, i.e. $X \in \Ob(\C)$ and $x \in E(X)$. In particular, $x \in P(X)$. Thus the comma category $(X,x) \downarrow \K$ is connected, and therefore non-empty. Choose an object $f : (X,x) \to (A,a)$. Thus, $A \in K$, $a \in P(A)$, and $f : X \to A$ satisfies $f^_(a)=x$. If $a \in E(A)$, we are done. Assume otherwise and, without loss of generality, $a \in F_1(A)$. Let $b \in F_2(A) \subseteq P(A)$ be the corresponding element in the other copy of $F$. Using the flip automorphism of $P$ that fixes $E$ and exchanges $F_1$ and $F_2$, we see that $f^_(b)=x$. Thus, we also have a morphism $f' : (X,x) \to (A,b)$ with the same underlying morphism $f : X \to A$.
 
 Since $(X,x) \downarrow \K$ is connected, the two morphisms $f$ and $f'$ are connected to each other. Thus, there are morphisms $(X,x) \to (A_i,a_i)$ with $A_i \in K$, $a_i \in P(A_i)$, starting with $f$ and ending with $f'$, such that for each pair of adjacent indices $i,i+1$, there is a morphism $(A_i,a_i) \to (A_{i+1},a_{i+1})$ or a morphism $(A_{i+1},a_{i+1}) \to (A_i,a_i)$. If any $a_i$ is contained in $E(A_i)$, we would be done. Assume, for a contradiction, that this is not the case.
 

database/data/categories/Alg(R).yaml

@@ -52,7 +52,7 @@ unsatisfied_properties:
   - property: co-Malcev
     proof: 'See <a href="https://mathoverflow.net/questions/509552">MO/509552</a>: Consider the forgetful functor $U : \Alg(R) \to \Set$ and the relation $S \subseteq U^2$ defined by $S(A) \coloneqq \{(a,b) \in U(A)^2 : ab = a^2\}$. Both are representable: $U$ by $R[X]$ and $S$ by $R \langle X,Y \rangle / \langle XY-X^2 \rangle$. It is clear that $S$ is reflexive, but not symmetric.'
 
-  - property: coregular
+  - property: pushout-stable regular monomorphisms
     proof: '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.'
     label: alg_not_coregular
 

database/data/categories/Ban.yaml

@@ -37,16 +37,16 @@ satisfied_properties:
     proof: >-
       The Hahn-Banach theorem implies that $\IC$ is a cogenerator. We claim that it is in fact an extremal cogenerator. Thus, suppose $f : X \to Y$ is a morphism such that ${-} \circ f : \Hom(Y, \IC) \to \Hom(X, \IC)$ is bijective on the underlying sets. Then for any non-zero $x \in X$, by the Hahn-Banach theorem, there exists $\varphi \in X^*$ such that $|\varphi| = 1$ and $\varphi(x) = |x|$. Since $|\varphi| = 1$, we see that $\varphi$ is a morphism $X \to \IC$ in $\Ban$; so by the assumption, there exists a morphism $\psi : Y \to \IC$ such that $\varphi = \psi \circ f$. Therefore, $|x| = |\psi(f(x))| \le |f(x)|$; and conversely, since $f$ is a morphism, $|f(x)| \le |x|$. On the other hand, if $x = 0$, then certainly $|f(x)| = |x| = 0$. This shows that $f$ is isometric and therefore a regular monomorphism (see below). On the other hand, since $\IC$ is a cogenerator and ${-} \circ f$ is injective, we have $f$ is also an epimorphism. Hence, $f$ is an isomorphism.
 
-  - property: regular
+  - property: pullback-stable regular epimorphisms
     proof: >-
-      It suffices to prove that regular epimorphisms are stable under pullbacks. We will use their classification via open unit balls below.
+      We will use the classification of regular epimorphisms via open unit balls below.
       So let $f : X \to Y$ be a regular epimorphism and let $g : T \to Y$ be any morphism. We need to show that the projection $X \times_Y T \to X$ is a regular epimorphism.
       Let $t \in T$ be an element of norm $<1$. Since $g$ is a linear contraction, $g(t)$ has norm $<1$. Since $f$ is a regular epimorphism, there is some $x \in X$ with norm $<1$ and $f(x) = g(t)$.
       Then $(x,t) \in X \times_Y T$ is a preimage of $t$ with norm $\max(|x|,|t|) < 1$.
 
-  - property: coregular
+  - property: pushout-stable regular monomorphisms
     proof: >-
-      It suffices to prove that regular monomorphisms are stable under pushouts. We will use their classification as isometric linear maps below.
+      We will use the classification of regular monomorphisms as isometric linear maps below.
       So let $i : X \to Y$ be an isometric linear map and let $f : X \to T$ be any morphism. We need to show that the linear contraction $\iota : T \to T \sqcup_X Y$ is isometric as well. The pushout can be constructed as the quotient of the direct sum $T \oplus Y$, equipped with the $1$-norm, modulo the closure of the subspace containing all $(-f(x),i(x))$ for $x \in X$. Using that $i$ is an isometry, it is easily checked that this subspace is already closed.
       For $t \in T$ the norm of $\iota(t) = [(t,0)]$ is the infimum of the norms of $(t,0) + (-f(x),i(x)) = (t - f(x), i(x))$ for $x \in X$.
       By taking $x=0$ we see that the infimum is $\leq |t|$.

database/data/categories/Cat.yaml

@@ -45,11 +45,11 @@ unsatisfied_properties:
   - property: balanced
     proof: Since we know that <a href="/category/Mon">$\Mon$</a> is not balanced, there is a monoid map $M \to N$ which is a monomorphism and an epimorphism which is not an isomorphism. Then $B(M) \to B(N)$ has the corresponding properties.
 
-  - property: regular
+  - property: pullback-stable regular epimorphisms
     proof: See Example 3.14 at the <a href="https://ncatlab.org/nlab/show/regular+category" target="_blank">nLab</a>.
 
-  - property: coregular
-    proof: 'We already know that <a href="/category/Mon">$\Mon$</a> is not coregular; in fact we have shown that there is a regular monomorphism $M \to N$ of monoids and a morphism $M \to K$ such that $K \to K \sqcup_M N$ is not a monomorphism. The delooping functor $B : \Mon \to \Cat$ has a left adjoint (<a href="https://math.stackexchange.com/questions/574745" target="_blank">MSE/574745</a>), hence it preserves regular monomorphisms. It also preserves pushouts (<a href="https://math.stackexchange.com/questions/5130854" target="_blank">MSE/5130854</a>), and it reflects monomorphisms since it is faithful. Therefore, $B(M) \to B(N)$ provides the desired counterexample of a non-stable regular monomorphism of categories.'
+  - property: pushout-stable regular monomorphisms
+    proof: 'We already know that <a href="/category/Mon">$\Mon$</a> has a regular monomorphism $M \to N$ and a morphism $M \to K$ such that $K \to K \sqcup_M N$ is not a monomorphism. The delooping functor $B : \Mon \to \Cat$ has a left adjoint (<a href="https://math.stackexchange.com/questions/574745" target="_blank">MSE/574745</a>), hence it preserves regular monomorphisms. It also preserves pushouts (<a href="https://math.stackexchange.com/questions/5130854" target="_blank">MSE/5130854</a>), and it reflects monomorphisms since it is faithful. Therefore, $B(M) \to B(N)$ provides the desired counterexample of a non-stable regular monomorphism of categories.'
     references:
       - mon_not_coregular
 

database/data/categories/CompHaus.yaml

@@ -43,9 +43,9 @@ satisfied_properties:
   - property: Barr-exact
     proof: The forgetful functor from $\CompHaus$ to $\Set$ is monadic; see for example <a href="https://ncatlab.org/nlab/show/compact+Hausdorff+space#compact_hausdorff_spaces_are_monadic_over_sets">nLab</a>. Therefore, by <a href="https://ncatlab.org/nlab/show/colimits+in+categories+of+algebras#exact">this result</a>, $\CompHaus$ is Barr-exact.
 
-  - property: coregular
+  - property: pushout-stable regular monomorphisms
     proof: >-
-      It suffices to show that pushouts preserve (regular) monomorphisms in $\CompHaus$. Thus, suppose we have a pushout square
+      Suppose we have a pushout square
       $$\begin{CD}
       A @> i >> B \\
       @V f VV @VV g V \\

database/data/categories/FiltVect.yaml

@@ -78,18 +78,18 @@ satisfied_properties:
       $$F_{< N}^n(V) \coloneqq \begin{cases} F^n(V) & n < N \\ 0 & n \geq N. \end{cases}$$
       Indeed, we have $F_{< N} \subseteq F_{< N+1}$, so that $\id_V : (V,F_{<N}) \to (V,F_{<N+1})$ is a filtered linear map. If $(W,F)$ is a filtered vector space together with a linear map $f : V \to W$ such that each map $f : (V,F_{<N}) \to (W,F)$ is filtered, then $f : (V,F) \to (W,F)$ is filtered as well. Indeed, for every $n \in \IZ$ we can choose some $N \in \IN$ with $n < N$, so that $F_{<N}^n(V) = F^n(V)$ is mapped into $F^n(W)$.
 
-  - property: regular
+  - property: pullback-stable regular epimorphisms
     proof: >-
-      It remains to prove that regular epimorphisms are stable under pullbacks. This follows immediately from their classification below, from the fact that $F^n$ preserves limits, and from the regularity of $\Vect$.
+      This follows immediately from the classification of regular epimorphisms below, from the fact that $F^n$ preserves limits, and from the corresponding property of $\Vect$.
 
 
       In more detail, if $(V,F) \to (W,F)$ is a regular epimorphism and $(U,F) \to (W,F)$ is any morphism, then $(V,F) \times_{(W,F)} (U,F) \to (U,F)$ is a regular epimorphism, since $V \times_W U \to U$ is surjective and, for every $n \in \IZ$, the restricted map
       $$F^n(V \times_W U) = F^n(V) \times_{F^n(W)} F^n(U) \to F^n(U)$$
       is surjective.
 
-  - property: coregular
+  - property: pushout-stable regular monomorphisms
     proof: >-
-      It remains to prove that regular monomorphisms (as classified below) are stable under pushouts. Let $i : (U,F) \to (V,F)$ be a regular monomorphism, i.e. $i$ is injective and $F^n(U) = i^*(F^n(V))$. Let $f : (U,F) \to (W,F)$ be any morphism. We must prove that the canonical morphism $(W,F) \to (V,F) \oplus_{(U,F)} (W,F)$ is a regular monomorphism. It is certainly injective, since the forgetful functor to $\Vect$ preserves colimits and $\Vect$ is abelian, and <a href="/category-implication/dual_abelian_implies_regular">hence coregular</a>. Now suppose that $w \in W$ is an element whose image $[0,w] \in V \oplus_U W$ lies in $F^n(V \oplus_U W)$; we must show that $w \in F^n(W)$.
+      Let $i : (U,F) \to (V,F)$ be a regular monomorphism, i.e. $i$ is injective and $F^n(U) = i^*(F^n(V))$. Let $f : (U,F) \to (W,F)$ be any morphism. We must prove that the canonical morphism $(W,F) \to (V,F) \oplus_{(U,F)} (W,F)$ is a regular monomorphism. It is certainly injective since the forgetful functor to $\Vect$ preserves colimits and $\Vect$ has the claimed property. Now suppose that $w \in W$ is an element whose image $[0,w] \in V \oplus_U W$ lies in $F^n(V \oplus_U W)$; we must show that $w \in F^n(W)$.
       Since, by the construction of colimits in $\FiltVect$, the subspace $F^n(V \oplus_U W)$ is the sum of the images of $F^n(V)$ and $F^n(W)$, there exist $v \in F^n(V)$ and $w' \in F^n(W)$ such that $[0,w] = [v,w']$. This means that there exists some $u \in U$ with $v = i(u)$ and $w = f(u) + w'$. Then $u \in F^n(U)$ because $i(u) \in F^n(V)$. Hence $f(u) \in F^n(W)$, and therefore $w = f(u) + w' \in F^n(W)$.
 
 unsatisfied_properties:

database/data/categories/FreeAb.yaml

@@ -46,8 +46,8 @@ satisfied_properties:
     proof: 'Let $f : A \to B$ be a homomorphism of free abelian groups. The coequalizer of its kernel pair $A \times_B A \rightrightarrows A$ in $\Ab$ is the image $\im(f)$. As a subgroup of $B$, it is also free abelian. Then it is also the coequalizer of the kernel pair in $\FreeAb$.'
     check_redundancy: false
 
-  - property: regular
-    proof: It remains to prove that regular epimorphisms are stable under pullback. This is clear since they coincide with the surjective homomorphisms (see below).
+  - property: pullback-stable regular epimorphisms
+    proof: This is clear since regular epimorphisms coincide with the surjective homomorphisms (see below).
 
 unsatisfied_properties:
   - property: balanced

database/data/categories/FreeAb_fg.yaml

@@ -40,8 +40,8 @@ satisfied_properties:
     references:
       - proj_fg_self-dual
 
-  - property: regular
-    proof: We already know that the category is finitely complete and self-dual, hence also finitely cocomplete. It remains to prove that regular epimorphisms are stable under pullback. This is clear since they coincide with the surjective homomorphisms (see below).
+  - property: pullback-stable regular epimorphisms
+    proof: This is clear since regular epimorphisms coincide with the surjective homomorphisms (see below).
 
   - property: ℵ₁-accessible
     proof: >-

database/data/categories/Grp.yaml

@@ -62,8 +62,8 @@ unsatisfied_properties:
     proof: 'We apply <a href="/content/missing_cogenerator">this lemma</a> to the collection of simple groups: Any non-trivial homomorphism from a simple group to a group must be injective, and for every infinite cardinal $\kappa$ there is a simple group of size $\geq \kappa$ (for example, the alternating group on $\kappa$ elements).'
     label: grp_no_cogenerator
 
-  - property: coregular
-    proof: This is because injective group homomorphisms are not stable under pushouts, see e.g. <a href="https://math.stackexchange.com/questions/601463/" target="_blank">MSE/601463</a> or <a href="https://math.stackexchange.com/questions/5088032" target="_blank">MSE/5088032</a>.
+  - property: pushout-stable regular monomorphisms
+    proof: See <a href="https://math.stackexchange.com/questions/601463/" target="_blank">MSE/601463</a> or <a href="https://math.stackexchange.com/questions/5088032" target="_blank">MSE/5088032</a>.
 
   - property: counital
     proof: The canonical morphism $F_2 = \IZ \sqcup \IZ \to \IZ \times \IZ$ is not a monomorphism since $F_2$ is not abelian.

database/data/categories/Grp_c.yaml

@@ -91,8 +91,8 @@ unsatisfied_properties:
     references:
       - grp_no_regular_quotient_object_classifier
 
-  - property: coregular
-    proof: Pushouts of injective homomorphisms between countable groups do not need to be injective, see <a href="https://math.stackexchange.com/questions/5088032" target="_blank">MSE/5088032</a>.
+  - property: pushout-stable regular monomorphisms
+    proof: See <a href="https://math.stackexchange.com/questions/5088032" target="_blank">MSE/5088032</a>.
 
   - property: cogenerator
     proof: 'Assume that a cogenerator $Q$ exists in $\Grp_\c$. There are only countably many finitely generated subgroups of $Q$. But there are continuum many finitely generated simple groups; this follows from Corollary 1.5 in <a href="https://arxiv.org/abs/1807.06478" target="_blank">Finitely generated infinite simple groups of homeomorphisms of the real line</a> by J. Hyde and Y. Lodha. Hence, there is a finitely generated (and hence countable) simple group $H$ which does not embed into $Q$. Since $H$ is simple, any homomorphism $H \to Q$ must be trivial then. But then $\id_H, 1 : H \rightrightarrows H$ are not separated by a homomorphism $H \to Q$.'

database/data/categories/Haus.yaml

@@ -78,10 +78,10 @@ unsatisfied_properties:
     references:
       - met_no_filtered_colimit_stable_monos
 
-  - property: regular
-    proof: 'The regular epimorphisms are precisely the surjective quotient maps of Hausdorff spaces (see below). In a regular category, for every regular epimorphism $X \to Y$ and every object $Z$, the induced morphism $X \times Z \to Y \times Z$ is again a regular epimorphism. This is not the case in $\Haus$ (or $\Top$, for that matter). The standard example is the quotient map $\IR \to \IR / \IZ^+$, for which the induced map $\IR \times \IQ \to \IR/\IZ^+ \times \IQ$ is not a quotient map (<a href="https://math.stackexchange.com/questions/1907972/">MSE/1907972</a>).'
+  - property: pullback-stable regular epimorphisms
+    proof: In a category with pullback-regular epimorphisms and products, for every regular epimorphism $X \to Y$ and every object $Z$, the induced morphism $X \times Z \to Y \times Z$ is again a regular epimorphism, since it can be seen as $X \times_Y (Y \times Z) \to Y \times Z$. This is not the case in $\Haus$ (or $\Top$, for that matter). The standard example is the quotient map $\IR \to \IR / \IZ^+$, for which the induced map $\IR \times \IQ \to \IR/\IZ^+ \times \IQ$ is not a quotient map (<a href="https://math.stackexchange.com/questions/1907972/">MSE/1907972</a>).
 
-  - property: coregular
+  - property: pushout-stable regular monomorphisms
     proof: >-
       Let $\Gamma$ be the <a href="https://en.wikipedia.org/wiki/Moore_plane" target="_blank">Moore plane</a>. Its underlying set is $\{(x,y) \in \IR^2 : y \geq 0 \}$. The open neighborhoods of points $(x,y)$ with $y > 0$ are those of $\IR^2$ (intersected with $\Gamma$), and the basic open neighborhoods of a point $(x,0)$ are open disks centered at $(x,\varepsilon)$ with radius $\varepsilon$ for some $\varepsilon > 0$. Then $\Gamma$ is Hausdorff, and the $x$-axis $A \coloneqq \{(x,0) : x \in \IR\}$ is a closed discrete subspace of $\Gamma$. In particular, by the classification of regular monomorphisms below, the inclusion map $i : A \to \Gamma$ is a regular monomorphism.
       Consider the two subsets $A_1 \coloneqq \{(x,0) : x \in \IQ \}$ and $A_2 \coloneqq \{(x,0) : x \in \IR \setminus \IQ \}$ of $A$. They are closed in $A$ (since $A$ is closed and discrete), disjoint, but cannot be separated by disjoint open neighborhoods in $\Gamma$; this is part of the proof of the well-known fact that $\Gamma$ is not normal (<a href="https://math.stackexchange.com/questions/2528435" target="_blank">MSE/2528435</a>).