Pullback-stable regular epimorphisms

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

@@ -50,7 +50,7 @@ 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

@@ -47,11 +47,11 @@ unsatisfied_properties:
   - property: cogenerating set
     proof: 'Assume that $S$ is a cogenerating set in $\Cat$. Then one checks that the set of monoids $\{\End(X) : X \in \C \in S\}$ is a cogenerating set in <a href="/category/Mon">$\Mon$</a>, which we know does not exist.'
 
-  - 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

@@ -42,15 +42,15 @@ 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
-    proof:
-      'It suffices to show that pushouts preserve (regular) monomorphisms in $\CompHaus$. Thus, suppose we have a pushout square
+  - property: pushout-stable regular monomorphisms
+    proof: >-
+      Suppose we have a pushout square
       $$\begin{CD}
       A @> i >> B \\
       @V f VV @VV g V \\
       C @>> j > D,
       \end{CD}$$
-      with $i : A \hookrightarrow B$ a monomorphism. Then for any pair of distinct elements $c, c'' \in C$, by Urysohn''s lemma there exists $\gamma : C \to [0, 1]$ with $\gamma(c) = 0$ and $\gamma(c'') = 1$. Also, by Tietze''s extension theorem, there exists $\beta : B \to [0, 1]$ such that $\beta \circ i = \gamma \circ f$. By the pushout property, there is a unique $\delta : D \to [0, 1]$ such that $\delta \circ g = \beta$ and $\delta \circ j = \gamma$. Since $\delta(j(c)) \ne \delta(j(c''))$, we conclude that $j(c) \ne j(c'')$. This shows that $j$ is injective, so it is a regular monomorphism.'
+      with $i : A \hookrightarrow B$ a monomorphism. Then for any pair of distinct elements $c, c' \in C$, by Urysohn's lemma there exists $\gamma : C \to [0, 1]$ with $\gamma(c) = 0$ and $\gamma(c') = 1$. Also, by Tietze's extension theorem, there exists $\beta : B \to [0, 1]$ such that $\beta \circ i = \gamma \circ f$. By the pushout property, there is a unique $\delta : D \to [0, 1]$ such that $\delta \circ g = \beta$ and $\delta \circ j = \gamma$. Since $\delta(j(c)) \ne \delta(j(c'))$, we conclude that $j(c) \ne j(c')$. This shows that $j$ is injective, so it is a regular monomorphism.
 
   - property: extensive
     proof: This follows from Lemma 11 <a href="/content/subcategories">here</a> since <a href="/category/Top">$\Top$</a> is infinitary extensive and its full subcategory $\CompHaus$ is closed under pullbacks and finite coproducts in $\Top$.

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/Grp.yaml

@@ -58,8 +58,8 @@ 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

@@ -88,8 +88,8 @@ 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

@@ -82,7 +82,7 @@ unsatisfied_properties:
   - 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: 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>).

database/data/categories/Meas.yaml

@@ -39,11 +39,11 @@ satisfied_properties:
     proof: Take the colimit of the underlying sets and take the largest $\sigma$-algebra making all inclusions measurable. That is, a set is measurable iff its preimage under each inclusion is measurable.
     check_redundancy: false
 
-  - property: coregular
+  - property: pushout-stable regular monomorphisms
     proof: >-
-      The proof is similar to the proof for <a href="/category/Top">$\Top$</a>. We already know that (finite) colimits and equalizers exist, and that they are preserved by the forgetful functor to $\Set$. It remains to show that regular monomorphisms, i.e. embeddings, are stable under pushouts. Thus, let $i : A \to X$ be an embedding and let $f : A \to Y$ be any measurable map. We claim that the induced measurable map
+      The proof is similar to the proof for <a href="/category/Top">$\Top$</a>. Let $i : A \to X$ be an embedding and let $f : A \to Y$ be any measurable map. We claim that the induced measurable map
       $$j : Y \to Y \sqcup_A X$$
-      is again an embedding. It is certainly injective, since <a href="/category/Set">$\Set$</a> is coregular. More precisely, the underlying set of $Y \sqcup_A X$ can be identified with $Y \sqcup (X \setminus \im(i))$. Now let $T \subseteq Y$ be a measurable subset. Then its preimage $f^*(T) \subseteq A$ is measurable. Since $i$ is an embedding, there exists a measurable subset $S \subseteq X$ such that $i^*(S) = f^*(T)$. Let $u : X \to Y \sqcup_A X$ denote the canonical map, so that $u \circ i = j \circ f$, and consider the subset
+      is again an embedding. It is certainly injective, since <a href="/category/Set">$\Set$</a> has the claimed property. More precisely, the underlying set of $Y \sqcup_A X$ can be identified with $Y \sqcup (X \setminus \im(i))$. Now let $T \subseteq Y$ be a measurable subset. Then its preimage $f^*(T) \subseteq A$ is measurable. Since $i$ is an embedding, there exists a measurable subset $S \subseteq X$ such that $i^*(S) = f^*(T)$. Let $u : X \to Y \sqcup_A X$ denote the canonical map, so that $u \circ i = j \circ f$, and consider the subset
       $$M \coloneqq j_*(T) \cup u_*(S \setminus \im(i))$$
       of the pushout. It is straightforward to verify that $j^*(M) = T$ and $u^*(M) = S$. Since both $T$ and $S$ are measurable, it follows that $M$ is measurable. Finally, the equality $j^*(M) = T$ shows that every measurable subset of $Y$ is the preimage of a measurable subset of the pushout. Hence $j$ is an embedding, as claimed.
     references:

database/data/categories/Met.yaml

@@ -160,7 +160,7 @@ unsatisfied_properties:
       On the other hand, if this cocongruence were effective, then by the dual of <a href="/content/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 $\Met$.
     label: met_no_effective_cocongruences
 
-  - property: regular
+  - property: pullback-stable regular epimorphisms
     proof: We can take the same counterexample as for <a href="/category/PMet">$\PMet$</a>.
     references:
       - pmet_not_regular