Add total/cototal category properties

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 +13,10 @@ 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 $X$. <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 locally essentially small and 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 $\card(U(X)) = \kappa$. 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 <a href="https://www.numdam.org/item/?id=CTGDC_1986__27_2_109_0" target="_blank">Kelly</a>, 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. <span class="qed">$\square$</span>

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

@@ -40,8 +40,8 @@ unsatisfied_properties:
   - property: semi-strongly connected
     proof: This is because already the full subcategory <a href="/category/CAlg(R)">$\CAlg(R)$</a> of commutative algebras is not semi-strongly connected.
 
-  - 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 $\langle X,E : E^2=E \rangle$.'

database/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 $R=0$.'
@@ -38,8 +38,8 @@ 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 <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 $(X^n)_{n} \in A^{\IN}$.'

database/data/categories/CRing.yaml

@@ -44,8 +44,8 @@ unsatisfied_properties:
   - property: balanced
     proof: The inclusion $\IZ \hookrightarrow \IQ$ is a counterexample.
 
-  - 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.'

database/data/categories/Cat.yaml

@@ -43,9 +43,6 @@ 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>.
 
@@ -88,6 +85,12 @@ 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 <a href="https://www.numdam.org/item/?id=CTGDC_1986__27_2_109_0" target="_blank">Kelly</a>, 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

database/data/categories/Grp.yaml

@@ -39,6 +39,10 @@ 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">here</a>.
+    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
@@ -50,7 +54,7 @@ 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 <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