diff --git a/.cspell.json b/.cspell.json index 716885cf..6591db5b 100644 --- a/.cspell.json +++ b/.cspell.json @@ -36,6 +36,7 @@ "Auslander", "Axiomatising", "axiomatization", + "Baer", "bijection", "bijections", "bijective", @@ -91,7 +92,10 @@ "cogenerating", "cogenerator", "cogenerators", + "cogroup", "Cohomology", + "coidentity", + "coinverse", "cokernel", "cokernels", "colimit", @@ -101,6 +105,7 @@ "comonadicity", "compactification", "compactifications", + "comultiplication", "conormal", "copower", "copowers", @@ -129,6 +134,7 @@ "cospans", "cosymmetric", "cosymmetry", + "cototal", "cotransitive", "cotransitivity", "counit", @@ -157,6 +163,8 @@ "extensivity", "extremal", "Faddeev", + "Farb", + "fibration", "fieldification", "filtrations", "finitary", @@ -180,6 +188,7 @@ "hausdorff", "Hertweck", "Heyting", + "homeomorphic", "homotopic", "homotopy", "Hušek", @@ -240,6 +249,7 @@ "Noetherian", "Noncommutative", "objectwise", + "opfibration", "pointwise", "Pontryagin", "poset", @@ -275,6 +285,7 @@ "saft", "Schapira", "Schepler", + "Schreier", "semigroup", "semigroups", "semisimple", @@ -318,6 +329,7 @@ "Turso", "Tychonoff", "Ulmer", + "ultrafilters", "uncountably", "unital", "unitalization", diff --git a/content/Grp_total_explicit_proof.md b/content/Grp_total_explicit_proof.md new file mode 100644 index 00000000..a107dc13 --- /dev/null +++ b/content/Grp_total_explicit_proof.md @@ -0,0 +1,68 @@ +--- +title: Explicit Proof that the Category of Groups is Total +description: An explicit construction of the left adjoint to the covariant Yoneda embedding on the category of groups +--- + +## Explicit Proof that the Category of Groups is Total + +The definition of a total category is very abstract; furthermore, it is not immediately clear how it is possible for _any_ category which is not essentially small to satisfy the definition, much less a wide variety of the algebraic and topological categories which are considered in practice. Thus, to illustrate the definition, we give an explicit construction of the functor +$$L : [\Grp^{\op},\Set] \to \Grp$$ +that is left adjoint to the Yoneda embedding $y : \Grp \hookrightarrow [\Grp^{\op},\Set]$. + +Fix a functor $T : \Grp^{\op} \to \Set$. To construct the group $L(T)$, we will make use of the usual cogroup structure on $\IZ$ in $\Grp$, which includes + +- the comultiplication homomorphism $\mu : \IZ \to \IZ * \IZ'$, $1 \mapsto 1 \cdot 1'$ (where $\IZ'$ denotes a copy of $\IZ$), +- the coidentity homomorphism $\varepsilon : \IZ \to 0$, +- the coinverse homomorphism $\iota : \IZ \to \IZ$. + +Also, let $i_1,i_2 : \IZ \rightrightarrows \IZ * \IZ'$ denote the coprojections. We define the group $L(T)$ as the group generated by elements $e(x)$, one for each element $x \in T(\IZ)$, subject to the following relations: + +- $e(T\mu(x)) = e(Ti_1(x)) \cdot e(Ti_2(x))$ for each $x \in T(\IZ * \IZ')$, +- $e(T\varepsilon(x)) = 1$ for each $x \in T0$, +- $e(T\iota(x)) = e(x)^{-1}$ for each $x \in T\IZ$, + +We first need to define a natural transformation $\eta_T : T \to \Hom({-}, L(T))$. For each group $H$ we define the function $\eta_T(H) : TH \to \Hom(H, L(T))$ by sending $x \in TH$ to $h \mapsto e(Th(x))$, where we abuse notation to identify $h \in H$ with the corresponding morphism $\IZ \to H$ mapping $1 \mapsto h$, so that $Th : TH \to T\IZ$. To see that this defines a group homomorphism from $H$ to $L(T)$, note that for $h, h' \in H$ we have three commutative diagrams of the form + +$$ +\begin{CD} +T(H) @> = >> T(H)\\ +@V T(hh') VV @VVV\\ +T(\IZ * \IZ') @>>> T(\IZ) +\end{CD} +$$ + +where on the bottom we use $T\mu, Ti_1, Ti_2$, and on the right we use $h h', h, h'$. Applying this to $x\in TH$, we get $Th(x)$, $Th'(x)$, and $T(h h')(x)$, respectively. Thus, the relation $e(T\mu(y)) = e(Ti_1(y)) \cdot e(Ti_2(y))$ with $y \coloneqq T(h h')(x)$ implies +$$e(T(hh')(x)) = e(Th(x)) \cdot e(Th'(x)),$$ +as required. Similar proofs show that the map $H \to L(T)$ respects inverses and the identity. We leave it as an exercise for the reader to show this is natural in $H$. + +We now need to show that for each group $G$ and natural transformation $\alpha : T \to y_G$, there exists a unique group homomorphism $\varphi : L(T) \to G$ such that +$$\alpha = y_{\varphi} \circ \eta_T : T \to \Hom({-}, L(T)) \to \Hom({-}, G).$$ +We start with uniqueness: suppose $x \in T\IZ$. Then by hypothesis, +$$\alpha_{\IZ} = (y_{\varphi})_{\IZ} \circ (\eta_T)_{\IZ} : T\IZ \to \Hom(\IZ, L(T)) \to \Hom(\IZ, G).$$ +For each $x \in T\IZ$, the first step on the right hand side maps $x \mapsto (1 \mapsto e(x))$, and the second step then maps this to $1 \mapsto \varphi(e(x))$. Therefore, +$$\varphi(e(x)) = \alpha_{\IZ}(x)(1)$$ +for each $x$, which establishes the uniqueness of $\varphi$. + +For the existence part, the first step is to show there is a group homomorphism $L(T) \to G$ with the images of $e(x)$ required by the previous part, i.e. $e(x) \mapsto \alpha_{\IZ}(x)(1)$. To prove this, we need to check that the relations in $L(T)$ are satisfied in $G$. Now, for each $x \in T(\IZ * \IZ')$, we have three commutative diagrams of the form + +$$ +\begin{CD} +T(\IZ * \IZ') @> \alpha_{\IZ * \IZ'} >> \Hom(\IZ * \IZ', G) @> \simeq >> UG \times UG\\ +@VVV @VVV @VVV\\ +T(\IZ) @> \alpha_{\IZ} >> \Hom(\IZ, G) @> \simeq >> UG +\end{CD} +$$ + +applying naturality to $\mu, i_1, i_2 : \IZ \to \IZ * \IZ'$. On the right hand side, we get multiplication, first projection, and second projection respectively. From this, we conclude that the images of $e(T\mu(x))$ and $e(Ti_1(x)) \cdot e(Ti_2(x))$ in $UG$ agree for any element $x \in T(\IZ * \IZ')$. Similar proofs show that the other relations are also satisfied. + +Finally, we need to show $\alpha = y_{\varphi} \circ \eta_T$, i.e. $\alpha_H = (y_{\varphi})_H \circ (\eta_T)_H$ for each group $H$. By definition, for each $x \in TH$, the first step gives the homomorphism $h \mapsto e(Th(x))$; then the second step is formed by composition with $\varphi$. By the specification of $\varphi$, this gives the homomorphism $h \mapsto \alpha_{\IZ}(Th(x))(1)$. However, by the assumption that $\alpha$ is a natural transformation, for each $h \in H$ we have a commutative diagram + +$$ +\begin{CD} +TH @> \alpha_H >> \Hom(H, G) \\ +@V Th VV @VV {-} \circ h V \\ +T\IZ @> \alpha_{\IZ} >> \Hom(\IZ, G). +\end{CD} +$$ + +Applying this to $x \in TH$ gives exactly that $\alpha_{\IZ}(Th(x))(1) = \alpha_H(x)(h)$. $\square$ diff --git a/content/missing_cogenerator.md b/content/missing_cogenerator.md index 1d071fc4..36b52e8f 100644 --- a/content/missing_cogenerator.md +++ b/content/missing_cogenerator.md @@ -12,8 +12,10 @@ Let $\C$ be a pointed category with a faithful functor $U: \C \to \Set$. Assume 1. For any $X \in \F$ and any $Y \in \C$, every non-zero morphism $f: X \to Y$ is injective on underlying sets. 2. For every $Y \in \C$ there is some object $X \in \F$ such that $\card(U(X)) > \card(U(Y))$. -Then $\C$ does not have a cogenerator. +Then $\C$ does not have a cogenerator. Moreover, if $\C$ is locally essentially small, then $\C$ is not cototal. ::: _Proof._ -Assume that there is a cogenerator $Y$. By assumption (2) there is an object $X \in \F$ such that $U(X)$ is larger than $U(Y)$ (w.r.t. cardinalities). Since $0,\id_X : X \rightrightarrows X$ are distinct, there is a morphism $f : X \to Y$ with $f \neq 0$. But then $U(f) : U(X) \to U(Y)$ is injective by assumption (1), which contradicts our choice of $X$. $\square$ +Assume that there is a cogenerator $Y$. By assumption (2) there is an object $X \in \F$ such that $U(X)$ is larger than $U(Y)$ (w.r.t. cardinalities). Since $0,\id_X : X \rightrightarrows X$ are distinct, there is a morphism $f : X \to Y$ with $f \neq 0$. But then $U(f) : U(X) \to U(Y)$ is injective by assumption (1), which contradicts our choice of $X$. + +Now assume that $\C$ is locally essentially small and cototal. Using the axiom of choice, we may assume that for each small cardinal $\kappa$, there is at most one element $X \in \F$ such that $\card(U(X)) = \kappa$. Treating $\F$ as a discrete diagram in $\C$, assumption (1) implies that for any object $Y$ of $\C$, the collection of cocones $\F \to Y$ is bijective with a set, since the maps $X \to Y$ with $\card(U(X)) > \card(U(Y))$ must all be zero in such a cocone. Therefore, by G. M. Kelly, A survey of totality for enriched and ordinary categories, Thm. 5.6, $\C$ must have a coproduct $Y$ of all elements of $\F$. But then by assumption (2), there exists $X \in \F$ such that $\card(U(X)) > \card(U(Y))$; and since $\C$ is pointed, the coprojection $X \to Y$ must be split monic and therefore non-zero. Using assumption (1), we get a contradiction. $\square$ diff --git a/database/data/categories/Alg(R).yaml b/database/data/categories/Alg(R).yaml index ab5ecf39..6f3b1ffa 100644 --- a/database/data/categories/Alg(R).yaml +++ b/database/data/categories/Alg(R).yaml @@ -41,8 +41,8 @@ unsatisfied_properties: - property: semi-strongly connected proof: This is because already the full subcategory $\CAlg(R)$ of commutative algebras is not semi-strongly connected. - - property: cogenerating set - proof: 'We apply this lemma to the collection of $R$-algebras which are fields: If $F$ is an $R$-algebra that is also a field and $A$ is a non-trivial $R$-algebra, any algebra homomorphism $F \to A$ is injective. For every infinite cardinal $\kappa$ the field of rational functions in $\kappa$ variables over some residue field of $R$ has cardinality $\geq \kappa$ and a non-trivial automorphism (swap two variables).' + - property: cototal + proof: Essentially the same proof as for $\CAlg(R)$ works here. - property: codistributive proof: 'If $\sqcup$ denotes the coproduct of $R$-algebras (see MSE/625874 for their description) and $A$ is an $R$-algebra, the canonical morphism $A \sqcup R^2 \to (A \sqcup R)^2 = A^2$ is usually no isomorphism. For example, for $A = R[X]$ the coproduct on the LHS is not commutative, it has the algebra presentation $\langle X,E : E^2=E \rangle$.' diff --git a/database/data/categories/CAlg(R).yaml b/database/data/categories/CAlg(R).yaml index d9eecee9..5ec38ed3 100644 --- a/database/data/categories/CAlg(R).yaml +++ b/database/data/categories/CAlg(R).yaml @@ -18,7 +18,7 @@ satisfied_properties: proof: There is a forgetful functor $\CAlg(R) \to \Set$ and $\Set$ is locally small. - property: finitary algebraic - proof: Take the algebraic theory of a commutative algebra. + proof: Take the algebraic theory of a commutative $R$-algebra. - property: strict terminal object proof: 'If $f : 0 \to A$ is a homomorphism of $R$-algebras, then $A$ satisfies $1=f(1)=f(0)=0$, so that $A=0$.' @@ -55,8 +55,8 @@ unsatisfied_properties: - property: balanced proof: Take a prime ideal $P \subseteq R$ and consider the commutative $R$-algebra $A \coloneqq R/P$ (which is an integral domain). Then the inclusion $A \hookrightarrow Q(A)$ is a counterexample. - - property: cogenerating set - proof: 'We apply this lemma to the collection of commutative $R$-algebras which are fields: If $F$ is a commutative $R$-algebra that is also a field and $A$ is a non-trivial commutative $R$-algebra, any algebra homomorphism $F \to A$ is injective. For every infinite cardinal $\kappa$ the field of rational functions in $\kappa$ variables over some residue field of $R$ has cardinality $\geq \kappa$ and a non-trivial automorphism (swap two variables).' + - property: cototal + proof: 'Let $\F$ be the family of commutative $R$-algebras of the form $R \times k$ where $k$ is an infinite field including a quotient field of $R$. Then for any commutative $R$-algebra $A$, we have a distinguished morphism $R \times k \to A$ consisting of the projection to $R$ followed by the unique morphism $R \to A$. Moreover, if we have any morphism $\varphi : R \times k \to A$ which is not equal to the distinguished morphism, that implies that $\varphi(0, 1) \ne 0$, so the rng homomorphism $k \to R \times k \to A$ is injective, implying $\card(U(A)) \ge \card(U(k))$. From here, an argument similar to the one here gives a contradiction, using the distinguished morphisms in place of zero morphisms.' - property: countably codistributive proof: 'The canonical homomorphism $A \otimes_R R^{\IN} \to A^{\IN}$ is given by $a \otimes (r_n)_n \mapsto (r_n a)_n$ and does not have to be surjective: Since $R \neq 0$, there is a commutative $R$-algebra $K$ which is a field. Now take $A \coloneqq K[X]$ and consider the sequence $(X^n)_{n} \in A^{\IN}$.' diff --git a/database/data/categories/Cat.yaml b/database/data/categories/Cat.yaml index 11fd7d2c..12b4aa2d 100644 --- a/database/data/categories/Cat.yaml +++ b/database/data/categories/Cat.yaml @@ -44,9 +44,6 @@ unsatisfied_properties: - property: balanced proof: Since we know that $\Mon$ 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: 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 $\Mon$, which we know does not exist.' - - property: regular proof: See Example 3.14 at the nLab. @@ -93,6 +90,12 @@ unsatisfied_properties: $$\Sub_{\reg}(\{ 0 \to 1 \to 2 \}) \to \Sub_{\reg}(\{ 0 \to 1 \}) \times_{\Sub_{\reg}(\{1\})} \Sub_{\reg}(\{ 1 \to 2 \})$$ is not injective. Therefore, $\Sub_{\reg} : \Cat^{\op} \to \Set^+$ does not preserve pullbacks, so it cannot be representable. + - property: cototal + proof: >- + For each infinite cardinal $\kappa$, choose a simple group $S_\kappa$ of cardinality $\kappa$ (for example the group of permutations of $\kappa$ of finite support which are even). Now consider the ultra-wide pushout diagram $1 \rightrightarrows B S_\kappa$. Then for any small category $\C$, the collection of cocones $1 \rightrightarrows B S_\kappa \to \C$ is bijective with a set: to form any such cocone, we must first choose the object $X$ of $\C$ which is the image of the object of $1$. Then, we must choose the morphisms $S_\kappa \to \End_{\C}(X)$; but for $\kappa > \card(\End_{\C}(X))$, the corresponding morphism must be zero. + + On the other hand, we claim that $1 \rightrightarrows B S_\kappa$ does not have a pushout in $\Cat$; by G. M. Kelly, A survey of totality for enriched and ordinary categories, Thm. 5.6, this will imply that $\Cat$ is not cototal. To see this, suppose we have a pushout $\C$ of $1 \rightrightarrows B S_\kappa$, and choose a cardinal $\lambda > \card(\Mor(\C))$. Then the coprojection $i_\lambda : B S_\lambda \to \C$ must be split monic, since we can construct a cocone $1 \rightrightarrows B S_\kappa \to B S_\lambda$ in which $B S_\kappa \to B S_\lambda$ corresponds to the zero map for $\kappa \ne \lambda$, and in which $B S_\lambda \to B S_\lambda$ is the identity. It follows that if $X$ is the image in $\C$ of the object of $B S_\lambda$ under $i_\lambda$, then $i_\lambda$ induces an injective map $S_\lambda \to \End_{\C}(X)$. This gives a contradiction since $\lambda > \card(\End_{\C}(X))$ and $S_\lambda$ is a simple group. + special_objects: initial object: description: empty category diff --git a/database/data/categories/Grp.yaml b/database/data/categories/Grp.yaml index 180e139f..01deab90 100644 --- a/database/data/categories/Grp.yaml +++ b/database/data/categories/Grp.yaml @@ -42,6 +42,10 @@ satisfied_properties: - property: effective cocongruences proof: A proof can be found here. + - property: total + proof: This follows formally from the fact that $\Grp$ is finitary algebraic and therefore locally presentable. For a more explicit proof, see here. + check_redundancy: false + - property: extremal generator proof: The group $\IZ$ is an extremal generator since it represents the forgetful functor $\Grp \to \Set$, which is faithful and conservative. check_redundancy: false @@ -54,7 +58,7 @@ unsatisfied_properties: - property: normal proof: Every non-normal subgroup (such as $C_2 \hookrightarrow S_3$) provides a counterexample. - - property: cogenerator + - property: cototal proof: 'We apply this lemma 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 diff --git a/database/data/categories/Haus.yaml b/database/data/categories/Haus.yaml index 328fb032..592a600e 100644 --- a/database/data/categories/Haus.yaml +++ b/database/data/categories/Haus.yaml @@ -27,9 +27,11 @@ satisfied_properties: - property: equalizers proof: This follows from the corresponding fact for $\Top$ since subspaces of Hausdorff spaces are again Hausdorff. + check_redundancy: false - property: products proof: This follows from the corresponding fact for $\Top$ since products of Hausdorff spaces are again Hausdorff. + check_redundancy: false - property: cocomplete proof: This follows since $\Haus$ is a reflective subcategory of $\Top$, which is cocomplete. For the reflector, see e.g. the nLab. Explicitly, we construct the colimit of Hausdorff spaces by applying the reflector to the colimit of the underlying topological spaces. @@ -76,9 +78,6 @@ unsatisfied_properties: references: - met_no_filtered_colimit_stable_monos - - property: cogenerator - proof: 'Assume that $Q$ is a cogenerator. Since $Q$ is Hausdorff, $Q$ is $T_1$. By a theorem of Herrlich (Wann sind alle stetigen Abbildungen in Y konstant. Math. Z. 90 (1965): 152-154. EUMDL), there is a regular Hausdorff space $X$ with $\geq 2$ points such that every continuous map $X \to Q$ is constant. (The author only states that $X$ is regular, but actually, $X$ is regular and $T_1$, hence Hausdorff.) But since $Q$ is a cogenerator, this implies that all maps $1 \rightrightarrows X$ are equal, i.e. that $X$ has just one point. This is a contradiction.' - - 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 (MSE/1907972).' @@ -89,6 +88,13 @@ unsatisfied_properties: Let $C \coloneqq \{1,2\}$ be the discrete two-point space. The map $f : A \to C$ defined by $f(a)=1$ for $a \in A_1$ and $f(a)=2$ for $a \in A_2$ is continuous, since $A$ is discrete. The pushout $C \sqcup_A \Gamma$ in $\Haus$ is the Hausdorff reflection of the pushout $Q$ in $\Top$. Notice that $Q$ is the quotient space of $\Gamma$ in which $A_1$ and $A_2$ are each collapsed to a point, denoted by $[A_1]$ and $[A_2]$. The canonical map $C \to Q$ is given by $i \mapsto [A_i]$. Now, $[A_1]$ and $[A_2]$ cannot be separated by disjoint open neighborhoods in $Q$, since such neighborhoods would pull back to disjoint open neighborhoods of $A_1$ and $A_2$ in $\Gamma$. Thus, they are identified in the Hausdorff reflection. This shows that the canonical map $C \to C \sqcup_A \Gamma$ is not injective and hence not a regular monomorphism. + - property: cototal + # cspell: disable-next-line + proof: >- + For each small cardinal $\kappa$, let $Q_\kappa$ be the product of all Hausdorff topological spaces whose underlying set is a non-empty subset of $\kappa$. By a theorem of Herrlich (Wann sind alle stetigen Abbildungen in Y konstant. Math. Z. 90 (1965): 152-154. EUMDL), there is a regular Hausdorff space $X_\kappa$ with at least two points such that every continuous map $X_\kappa \to Q_\kappa$ is constant. (The author only states that $X_\kappa$ is regular, but actually, $X_\kappa$ is regular and $T_1$, hence Hausdorff.) Choose a base point $x_\kappa \in X_\kappa$ for each $\kappa$. We can form an ultra-wide pushout diagram $1 \rightrightarrows X_\kappa$ where each morphism $1 \to X_\kappa$ corresponds to $x_\kappa$. Then for any Hausdorff space $Y$, the collection of cocones $1 \rightrightarrows X_\kappa \to Y$ is bijective to a set: if $Y$ is empty, then the collection of cocones is obviously empty. Otherwise, in order to form a cocone, we must first choose $y \in Y$ corresponding to the morphism $1 \to Y$. Then for each $\kappa \ge \card(U(Y))$, $Y$ is homeomorphic to one of the spaces in the product forming $Q_\kappa$. Therefore, there is a morphism $Y \to Q_\kappa$ splitting the projection map $Q_\kappa \to Y$. It follows that the map $X_\kappa \to Y$ is constant, and in fact it must be the constant map with image $y$. + + On the other hand, we claim that $1 \rightrightarrows X_\kappa$ does not have a pushout in $\Haus$; by G. M. Kelly, A survey of totality for enriched and ordinary categories, Thm. 5.6, this will imply that $\Cat$ is not cototal. To see this, suppose we had a pushout $Y$, and let $\lambda \coloneqq \card(U(Y))$. Then the coprojection $X_\lambda \to Y$ is split monic, since we can construct a cocone $1 \rightrightarrows X_\kappa \to X_\lambda$ where the map $X_\kappa \to X_\lambda$ is the constant map with image $x_\lambda$ when $\kappa \ne \lambda$, and the map $X_\lambda \to X_\lambda$ is the identity. But similarly to the previous paragraph, we can show any morphism $X_\lambda \to Y$ must be constant, giving a contradiction since $X_\lambda$ has at least two points. + - property: extremal generating set proof: The proof is the same as the one for $\Top$; there the test spaces we use are of the form $\kappa \sqcup \{ \kappa \}$ and $\kappa + 1$, which are both Hausdorff spaces. references: diff --git a/database/data/categories/LRS_R.yaml b/database/data/categories/LRS_R.yaml index 7b30bedb..25564de8 100644 --- a/database/data/categories/LRS_R.yaml +++ b/database/data/categories/LRS_R.yaml @@ -64,11 +64,8 @@ unsatisfied_properties: references: - top_not_co-malcev - - property: generating set - proof: >- - Out of any small set $S$ of locally ringed spaces, there is only a small set of residue fields at their points. Therefore, if $K$ is a field over $R$ with a strictly larger cardinality than any of these residue fields, then the only possible morphism from an element of $S$ to $\Spec K(X,Y)$ is one with an empty domain. However, that makes it impossible for $S$ to distinguish the two canonical automorphisms of $\Spec K(X,Y)$. - - Alternatively, using the usual adjunction between affine schemes and locally ringed spaces (EGA I (1971), Ch. 1, Prop. 1.6.3), a generating set in $\LRS_R$ would induce a generating set in the category of affine $R$-schemes, which contradicts the fact that $\CAlg(R)$ does not have a cogenerating set. + - property: total + proof: 'The adjunction between the global sections functor and the $\Spec$ functor (EGA I (1971), Ch. 1, Prop. 1.6.3) makes $\CAlg(R)^{\op}$ into a reflective subcategory of $\LRS_R$. Therefore, if $\LRS_R$ were total, then $\CAlg(R)$ would be cototal by G. M. Kelly, A survey of totality for enriched and ordinary categories, Cor. 6.2, which we know is not the case.' - property: cartesian closed proof: This is Corollary 4(a) here. diff --git a/database/data/categories/Meas.yaml b/database/data/categories/Meas.yaml index e5ca2c5d..c2c1e105 100644 --- a/database/data/categories/Meas.yaml +++ b/database/data/categories/Meas.yaml @@ -34,6 +34,7 @@ satisfied_properties: - property: complete proof: Take the limit of the underlying sets and take the smallest $\sigma$-algebra making all projections measurable. + check_redundancy: false - property: cocomplete 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. diff --git a/database/data/categories/Mon.yaml b/database/data/categories/Mon.yaml index 9f58e138..52741665 100644 --- a/database/data/categories/Mon.yaml +++ b/database/data/categories/Mon.yaml @@ -40,7 +40,7 @@ unsatisfied_properties: - property: Malcev proof: 'Consider the submonoid $\{(a,b) : a \leq b \}$ of $\IN^2$.' - - property: cogenerator + - property: cototal proof: 'We apply this lemma to the collection of simple groups: Any non-trivial homomorphism $G \to M$ from a simple group $G$ to a monoid $M$ must be injective (as it corestricts to a homomorphism of groups $G \to M^{\times}$), and for every infinite cardinal $\kappa$ there is a simple group of size $\geq \kappa$ (for example, the alternating group on $\kappa$ elements).' - property: counital diff --git a/database/data/categories/N.yaml b/database/data/categories/N.yaml index d4ee93cb..40806823 100644 --- a/database/data/categories/N.yaml +++ b/database/data/categories/N.yaml @@ -7,7 +7,6 @@ description: >- This category can also be seen as the path category of the infinite linear graph $$\bullet \longrightarrow \bullet \longrightarrow \bullet \longrightarrow \cdots.$$ nlab_link: null - tags: - number theory diff --git a/database/data/categories/On.yaml b/database/data/categories/On.yaml index 45fe331b..495785fa 100644 --- a/database/data/categories/On.yaml +++ b/database/data/categories/On.yaml @@ -42,9 +42,6 @@ unsatisfied_properties: - property: terminal object proof: There is no largest ordinal $\alpha$ since $\alpha + 1$ will always be larger. - - property: well-copowered - proof: The "quotients" of $0$ are all ordinals. - - property: inverse proof: Consider the strictly increasing sequence $0 < 1 < 2 < \cdots$. diff --git a/database/data/categories/Rng.yaml b/database/data/categories/Rng.yaml index 37744cd0..6e552a75 100644 --- a/database/data/categories/Rng.yaml +++ b/database/data/categories/Rng.yaml @@ -40,7 +40,7 @@ unsatisfied_properties: - property: balanced proof: The inclusion $\IZ \hookrightarrow \IQ$ is a counterexample; the proof can be reduced to the unital case. - - property: cogenerator + - property: cototal proof: 'We apply this lemma to the collection of fields: Any non-zero rng homomorphism from a field to a rng must be injective, and for every infinite cardinal $\kappa$ the field of rational functions in $\kappa$ variables has cardinality $\geq \kappa$.' - property: counital diff --git a/database/data/categories/SemiGrp.yaml b/database/data/categories/SemiGrp.yaml index 5719dfee..b64ce2c9 100644 --- a/database/data/categories/SemiGrp.yaml +++ b/database/data/categories/SemiGrp.yaml @@ -55,16 +55,6 @@ unsatisfied_properties: Let us first remark that every non-empty finite semigroup $A$ has an idempotent element $e$, and then $B \to A$, $x \mapsto e$ does define a semigroup homomorphism for any $B$. Therefore, counterexamples need to be infinite and also without idempotent elements. Let $A$ be the set of positive rational numbers of the form $m/2^n$ (with $m > 0$, $n \geq 0$), and let $B$ be the set of positive rational numbers of the form $m/3^n$ (with $m > 0$, $n \geq 0$). Both are semigroups under addition. The element $1 \in A$ is $2^\infty$-divisible, meaning that for every $n \geq 0$ there is some $a \in A$ with $1 = 2^n \cdot a$. But $B$ has no $2^\infty$-divisible element. Hence, there is no semigroup homomorphism $A \to B$. Likewise, there is no semigroup homomorphism $B \to A$. - - property: cogenerator - # TODO: find a variant of the lemma missing_cogenerating_sets - # (or missing_cogenerator) which handles this. - proof: >- - The proof is similar to the proof for $\Grp$. Assume that there is a cogenerator $Q$. There is an infinite simple group $G$ larger than $Q$ (such as an alternating group). Since $\id_G, 1 : G \rightrightarrows G$ are different, there is a homomorphism of semigroups $f : G \to Q$ with $f \neq f \circ 1$. Then - $$N \coloneqq \{g \in G : f(g) = f(1)\}$$ - is a normal subgroup of $G$. It is proper, and hence trivial. But then $f$ is injective, which is a contradiction. - references: - - grp_no_cogenerator - - 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 Lemma 2 here to the functor $\Set \to \SemiGrp$ that equips a set with the multiplication $a \cdot b \coloneqq a$. @@ -76,6 +66,14 @@ unsatisfied_properties: references: - mon_no_effective_cocongruences + - property: cototal + proof: >- + The proof is similar to the proof for $\Cat$. For each infinite cardinal $\kappa$, let $S_\kappa$ be a simple group of cardinality $\kappa$ (such as the alternating group on $\kappa$). We can then form the ultra-wide pushout diagram $1 \rightrightarrows S_\kappa$ in $\SemiGrp$. For every semigroup $A$, the collection of cocones $1 \rightrightarrows S_\kappa \to A$ is bijective to a set: for every such cocone, we must first choose an idempotent $e$ of $A$ corresponding to the map $1 \to A$. Then, whenever $\kappa > \card(U(A))$, then for $f_\kappa : S_\kappa \to A$ in the cocone, we see + $$N \coloneqq \{g \in S_\kappa : f_\kappa(g) = e\}$$ + is a normal subgroup of $G$. It must be non-trivial since otherwise $f_\kappa$ would induce an injective group homomorphism from $G$ to a group contained in $A$. Therefore, $N$ is all of $G$, so $f_\kappa$ is the constant map with image $e$. + + We now claim that $1 \rightrightarrows S_\kappa$ does not have a pushout in $\SemiGrp$; by G. M. Kelly, A survey of totality for enriched and ordinary categories, Thm. 5.6, this will imply that $\SemiGrp$ is not cototal. To see this, suppose we had a pushout $A$, and let $\lambda$ be a cardinal strictly greater than $\card(U(A))$. Then the coprojection $S_\lambda\to A$ must be split monic, since we can construct a cocone $1 \rightrightarrows S_\kappa \to S_\lambda$ such that the map $S_\kappa \to S_\lambda$ is the constant map with image 1 if $\kappa \ne \lambda$, while the map $S_\lambda \to S_\lambda$ is the identity. But this contradicts the choice of $\lambda$. + - property: natural numbers object proof: >- Assume that a natural numbers object exists. Then by this result, for every semigroup $A$ the natural homomorphism diff --git a/database/data/categories/Top.yaml b/database/data/categories/Top.yaml index d86852e3..0b65f75e 100644 --- a/database/data/categories/Top.yaml +++ b/database/data/categories/Top.yaml @@ -24,6 +24,7 @@ satisfied_properties: - property: complete proof: Take the limit of the underlying sets and endow it with the coarsest topology making all projections continuous. + check_redundancy: false - property: cocomplete proof: Take the colimit of the underlying sets and endow it with the finest topology making all inclusions continuous. diff --git a/database/data/categories/Unif.yaml b/database/data/categories/Unif.yaml index 4af96f07..f74033c2 100644 --- a/database/data/categories/Unif.yaml +++ b/database/data/categories/Unif.yaml @@ -19,6 +19,7 @@ satisfied_properties: - property: complete proof: 'Take the limit of the underlying sets and endow it with the coarsest uniform structure making all projections uniform; cf. Bourbaki, General Topology (Part 1), Chapter II, § 3, no. 3 on initial uniformities. More concretely, products are described below on this page, and the equalizer of two uniform maps $f,g : (X,\Phi) \rightrightarrows (Y,\Psi)$ is the subset $E := \{x \in X : f(x) = g(x)\}$ equipped with the uniform structure $\{U \cap (E \times E) : U \in \Phi\}$.' + check_redundancy: false - property: cocomplete proof: 'Take the colimit of the underlying sets and endow it with the finest uniform structure making all inclusions uniform. More concretely, coproducts are described below on this page, and the coequalizer of two uniform maps $f,g : (X,\Phi) \rightrightarrows (Y,\Psi)$ is the $\Set$-based coequalizer $Q = Y / (f(x) \sim g(x))$ equipped with the following uniform structure, which makes the projection $p : Y \to Q$ uniform. Let $\Theta$ be the set of all subsets $U \subseteq Q \times Q$ such that $(p \times p)^*(U) \in \Psi$. It satisfies all the axioms of a uniform structure except one, namely the composition axiom. To fix this (and this construction works in complete generality), let $\Sigma \subseteq \Theta$ be the set of all $U \in \Theta$ for which there exists a sequence $U_1,U_2,\dotsc$ in $\Theta$ such that $U_1 \subseteq U$ and $U_{k+1} \circ U_{k+1} \subseteq U_k$ for all $k$. It is then straightforward to check that $\Sigma$ is a uniform structure on $Q$. Moreover, by construction, the map $p : (Y,\Psi) \to (Q,\Sigma)$ is uniform, and one verifies that it satisfies the required universal property.' diff --git a/database/data/category-implications/total.yaml b/database/data/category-implications/total.yaml new file mode 100644 index 00000000..6f32e88e --- /dev/null +++ b/database/data/category-implications/total.yaml @@ -0,0 +1,37 @@ +# results on total and cototal categories + +- id: total_loc_ess_small + assumptions: + - total + conclusions: + - locally essentially small + proof: This is trivial. + is_equivalence: false + +- id: total_cocomplete + assumptions: + - total + conclusions: + - cocomplete + proof: 'If the category $\C$ is locally small and total, then the Yoneda embedding $y : \C^{\op} \to [\C, \Set]$ makes $\C^{\op}$ into a reflective subcategory of the presheaf category $[\C, \Set]$, where the latter is complete. For a general total category $\C$, use the equivalence to a locally small, total category.' + is_equivalence: false + +- id: total_complete + assumptions: + - total + conclusions: + - complete + proof: This is proven in G. M. Kelly, A survey of totality for enriched and ordinary categories, Thm. 5.6. + is_equivalence: false + +# TODO: replace "well-copowered" with "epi-cocomplete" if adding the latter property +- id: cocomplete_well-copowered_generator_implies_total + assumptions: + - cocomplete + - locally essentially small + - well-copowered + - generating set + conclusions: + - total + proof: Since the category is cocomplete and well-copowered, it is epi-cocomplete (meaning that it has wide pushouts of epimorphisms, even of non-small families of epimorphisms). The result then follows from B. J. Day, Further criteria for totality, Thm. 1. + is_equivalence: false diff --git a/database/data/category-properties/cototal.yaml b/database/data/category-properties/cototal.yaml new file mode 100644 index 00000000..a9e58b50 --- /dev/null +++ b/database/data/category-properties/cototal.yaml @@ -0,0 +1,27 @@ +id: cototal +relation: is +description: >- + A locally small category $\C$ is called cototal if it satisfies one of the following equivalent conditions: +