Add total/cototal category properties (WIP)

content/Grp_total_explicit_proof.md

@@ -0,0 +1,37 @@
+---
+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
+author: Daniel Schepler
+---
+
+## Explicit Proof that the Category of Groups is Total
+
+We will fix a functor $T : \Grp^{\op} \to \Set$. Based on this, we need to construct a group $L(T)$. The construction will work as follows: we start with a set of generators $e_x$, one for each element $x \in T\IZ$. Now let $\mu : \IZ \to \IZ * \IZ'$ be the comultiplication homomorphism, $1 \mapsto 1 \cdot 1'$, and $i_1, i_2 : \IZ \rightrightarrows \IZ * \IZ'$ the two coprojections. We now add a relation $e_{T\mu(x)} = e_{Ti_1(x)} \cdot e_{Ti_2(x)}$ for each element $x \in T(\IZ * \IZ')$. Similarly, letting $\iota : \IZ \to \IZ$ be the coinverse homomorphism, we add a relation $e_{T\iota(x)} = e_x^{-1}$ for each element $x \in T\IZ$; and letting $\varepsilon : \IZ \to 0$ be the coidentity homomorphism, we add a relation $e_{T\varepsilon(x)} = 1$ for each element $x \in T0$.
+
+We first need to define the natural transformation $\eta_T : T \to \Hom({-}, L(T))$. To start, for each group $H$ we need a function $TH \to \Hom(H, L(T))$. We will define this function to send $x \in TH$ to $h \mapsto e_{Th(x)}$, where we abuse notation to identify $h \in H$ with the corresponding morphism $\IZ \to H$, 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
+$$\begin{CD}
+T(H) @> = >> T(H)\\
+@V T(h * h') VV @VVV\\
+T(\IZ * \IZ') @>>> T(\IZ)
+\end{CD}$$
+where on the bottom we use $T\mu, Ti_1, Ti_2$. 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
+$$e_{T(hh')(x)} = e_{Th(x)} e_{Th'(x)},$$
+as required. Similar proofs show that the map $H \to G$ 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, $\alpha_{\IZ} = y_{\varphi} \circ (\eta_T)_{\alpha} : T\IZ \to \Hom(\IZ, L(T)) \to \Hom(\IZ, G)$. 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 gruop 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 collapsed 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 collapsed.
+
+Finally, we need to show $\alpha = y_{\varphi} \circ \eta_T$. It suffices to show $\alpha_H = (y_{\varphi})_H \circ (\eta_T)_H : TH \to \Hom(H, L(T)) \to \Hom(H, G)$ 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 homorphism $h \mapsto \alpha_{\IZ}(Th(x))(1)$. 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 $\alpha_{\IZ}(Th(x))(1) = \alpha_H(x)(h)$. <span class="qed">$\square$</span>

content/missing_cogenerator.md

@@ -13,8 +13,18 @@ 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, $\C$ 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 $X$. <span class="qed">$\square$</span>
+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 cocotal. 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 $\card(U(X)) = \kappa$. We define a functor $T : \C \to \Set$ which is morally defined by $T(Y) \coloneqq \prod_{X \in \F} \Hom(X, Y)$. In this product, we know that $\Hom(X, Y) = \{ 0 \}$ whenever $\card(U(X)) > \card(U(Y))$, so every term except for a small number is a singleton and does not change the product. Thus, we may define
+$$T(Y) \coloneqq \prod_{X \in \F : \card(U(X)) \le \card(U(Y))} \Hom(X, Y),$$
+and define the functor structure by extending tuples with zero morphisms as necessary.
+
+Now, if $L : [\C, \Set] \to \C^{\op}$ is the left adjoint of the contravariant Yoneda embedding $y : \C^{op} \to [\C, \Set]$, then we have a bijection
+$$\Hom_{\C}(X, L(T)) \simeq \Hom(T, \Hom(X, {-}))$$
+for each object $X$ of $\C$. Now let $X_0$ be an element of $\F$ such that $\card(U(X_0)) > \card(U(L(T)))$. Then we can define two natural transformations $T \to \Hom({-}, X_0)$: one is the constant zero map; and the other is selection of the $X$ component, i.e. given an element of $TY = \prod_{X\in F : \card(U(X)) \le \card(U(Y))} \Hom(X, Y)$ we select the $X$ component if $\card(U(X)) \le \card(U(Y))$, or set the result to the zero morphism if $\card(U(X)) > \card(U(Y))$. Note these two natural transformations are unequal, since for example we can define an element of $T(X_0)$ which is $\id_{X_0}$ on the $X_0$ component and zero on the other components, and the two transformations map this element to $\id_{X_0} \ne 0$.
+
+Therefore, by the above bijection, $\Hom_{\C}(X_0, L(T))$ has at least two elements, out of which at least one must be non-zero and therefore injective on underlying sets. This contradicts the assumption that $\card(U(X_0)) > \card(U(L(T))$. <span class="qed">$\square$</span>

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

@@ -43,6 +43,9 @@ unsatisfied_properties:
   - property: cogenerating set
     proof: 'We apply <a href="/content/missing_cogenerating_sets">this lemma</a> 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 <a href="/category/CAlg(R)">$\CAlg(R)$</a> works here.
+
   - property: codistributive
     proof: 'If $\sqcup$ denotes the coproduct of $R$-algebras (see <a href="https://math.stackexchange.com/questions/625874" target="_blank">MSE/625874</a> 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 $\langle X,E : E^2=E \rangle$.'
 

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

@@ -19,7 +19,7 @@ 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 ring.
+    proof: Take the algebraic theory of a commutative $R$-algebra.
 
   - property: strict terminal object
     proof: 'If $f : 0 \to R$ is a homomorphism, then $R$ satisfies $1=f(1)=f(0)=0$, so that $R=0$.'
@@ -41,6 +41,9 @@ unsatisfied_properties:
   - property: cogenerating set
     proof: 'We apply <a href="/content/missing_cogenerating_sets">this lemma</a> 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 <a href="/content/missing_cogenerator">here</a> 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 $(X^n)_{n} \in A^{\IN}$.'
 

databases/catdat/data/categories/CRing.yaml

@@ -47,6 +47,9 @@ unsatisfied_properties:
   - property: cogenerating set
     proof: 'We apply <a href="/content/missing_cogenerating_sets">this lemma</a> to the collection of fields: If $F$ is a field and $R$ is a non-trivial commutative ring, any ring homomorphism $F \to R$ is injective. For every infinite cardinal $\kappa$ the field of rational functions in $\kappa$ variables has cardinality $\geq \kappa$ and a non-trivial automorphism (swap two variables).'
 
+  - property: cototal
+    proof: 'This is a special case of the proof for <a href="/category/CAlg(R)">$\CAlg(R)$</a> with $R = \IZ$.'
+
   - property: countably codistributive
     proof: 'The canonical homomorphism $\IQ \otimes \IZ^{\IN} \to (\IQ \otimes \IZ)^{\IN} = \IQ^{\IN}$ is not an isomorphism: its image consists of those sequences of rational numbers whose denominators can be bounded.'
 

databases/catdat/data/categories/Cat.yaml

@@ -40,9 +40,6 @@ 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: 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
     proof: See Example 3.14 at the <a href="https://ncatlab.org/nlab/show/regular+category" target="_blank">nLab</a>.
 
@@ -85,6 +82,16 @@ 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 $\kappa$ with at least $\kappa$ distinct automorphisms (for example the group of permutations of $\kappa$ of finite support which are even). We will define a functor $T : \Cat \to \Set$ which morally will send a category $\C$ to the collection-sized wide pullback of $[B S_\kappa, \C] \to [1, \C]$. Namely, if $\kappa$ is greater than the cardinality of any endomorphism monoid of an object of $\C$, then the image in $[1, \C]$ uniquely determines a functor in $[B S_\kappa, \C]$. Thus, we can restrict to the corresponding pullback with $\kappa$ restricted to be at most $\max(\aleph_0, \lambda)$ where $\lambda$ is the supremum of the cardinalities of endomorphism monoids; and this also forms a pullback of the full collection-sized diagram.
+
+      Now assume $\Cat$ is cototal with $L : [\Cat, \Set] \to \Cat^{\op}$ a left adjoint to the contravariant Yoneda embedding. Then for each infinite cardinal $\kappa$, we would have a bijection
+      $$[B S_\kappa, L(T)] \simeq \Hom(T, [B S_\kappa, -]).$$
+      On the other hand, we have at least $\kappa$ distinct morphisms $T \to [B S_\kappa, -]$: for each automorphism $\pi$ of $S_\kappa$, we can take the natural transformation which selects the $\kappa$ component of the limit, and then composes with $B \pi$. To see these are distinct, apply them to the element of $T(B S_\kappa)$ which is the identity on the $\kappa$ component and $B(0)$ on the $\lambda$ component for $\lambda \ne \kappa$.
+
+      However, if $\kappa$ is greater than the cardinality of any endomorphism monoid of $L(T)$, then any functor $B S_\kappa \to L(T)$ must act on the morphisms of $B S_\kappa$ by sending each to the identity morphism on the image object. Therefore, $\card([B S_\kappa, L(T)]) \le \card(\Ob(L(T)))$. Thus, if we also choose $\kappa > \card(\Ob(L(T)))$, then we get a contradiction.
+
 special_objects:
   initial object:
     description: empty category

databases/catdat/data/categories/Grp.yaml

@@ -39,14 +39,18 @@ satisfied_properties:
   - property: effective cocongruences
     proof: A proof can be found <a href="/content/cocongruences_of_groups">here</a>.
 
+  - property: total
+    proof: This follows formally from the fact that $\Grp$ is finitary algebraic and therefore locally presentable. For a more explicit proof, see <a href="/content/Grp_total_explicit_proof.md">here</a>.
+    check_redundancy: false
+
 unsatisfied_properties:
   - property: skeletal
     proof: This is trivial.
 
   - property: normal
     proof: Every non-normal subgroup (such as $C_2 \hookrightarrow S_3$) provides a counterexample.
 
-  - property: cogenerator
+  - property: cototal
     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).'
 
   - property: coregular

databases/catdat/data/categories/Haus.yaml

@@ -26,9 +26,11 @@ satisfied_properties:
 
   - property: equalizers
     proof: This follows from the corresponding fact for <a href="/category/Top">$\Top$</a> since subspaces of Hausdorff spaces are again Hausdorff.
+    check_redundancy: false
 
   - property: products
     proof: This follows from the corresponding fact for <a href="/category/Top">$\Top$</a> since products of Hausdorff spaces are again Hausdorff.
+    check_redundancy: false
 
   - property: cocomplete
     proof: This follows since $\Haus$ is a reflective subcategory of <a href="/category/Top">$\Top$</a>, which is cocomplete. For the reflector, see e.g. the <a href="https://ncatlab.org/nlab/show/Hausdorff+space#HausdorffReflection" target="_blank">nLab</a>. Explicitly, we construct the colimit of Hausdorff spaces by applying the reflector to the colimit of the underlying topological spaces.

databases/catdat/data/categories/LRS_R.yaml

@@ -53,6 +53,9 @@ unsatisfied_properties:
 
       Alternatively, using the usual adjunction between affine schemes and locally ringed spaces (<a href="https://en.wikipedia.org/wiki/%C3%89l%C3%A9ments_de_g%C3%A9om%C3%A9trie_alg%C3%A9brique" target="_blank">EGA I</a> (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 <a href="/category/CAlg(R)">$\CAlg(R)$</a> does not have a cogenerating set.
 
+  - property: total
+    proof: 'The adjunction between the global sections functor and the $\Spec$ functor (<a href="https://en.wikipedia.org/wiki/%C3%89l%C3%A9ments_de_g%C3%A9om%C3%A9trie_alg%C3%A9brique" target="_blank">EGA I</a> (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 <a href="/category/CAlg(R)">\CAlg(R)</a> would be cototal, which we know is not the case.'
+
   - property: cartesian closed
     proof: This is Corollary 4(a) <a href="/content/Top-embeds-in-LRS">here</a>.
     check_redundancy: false

databases/catdat/data/categories/Meas.yaml

@@ -36,9 +36,11 @@ 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.
+    check_redundancy: false
 
   - property: infinitary extensive
     proof: '[Sketch] Since <a href="/category/Set">$\Set$</a> is infinitary extensive, a map $f : Y \to \coprod_i X_i \eqqcolon X$ corresponds to a decomposition $Y = \coprod_i Y_i$ (as sets) with maps $f_i : Y_i \to X_i$. Endow the measurable subset $Y_i \subseteq Y$ with the restricted $\sigma$-algebra. If $f$ is measurable, each $f_i$ is measurable, and $Y = \coprod_i Y_i$ holds as measurable spaces.'

databases/catdat/data/categories/Mon.yaml

@@ -39,7 +39,7 @@ unsatisfied_properties:
   - property: Malcev
     proof: 'Consider the submonoid $\{(a,b) : a \leq b \}$ of $\IN^2$.'
 
-  - property: cogenerator
+  - property: cototal
     proof: 'We apply <a href="/content/missing_cogenerator">this lemma</a> 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

databases/catdat/data/categories/On.yaml

@@ -45,6 +45,7 @@ unsatisfied_properties:
 
   - property: well-copowered
     proof: The "quotients" of $0$ are all ordinals.
+    check_redundancy: false
 
   - property: inverse
     proof: Consider the strictly increasing sequence $0 < 1 < 2 < \cdots$.