Add extremal (co)generating sets and single extremal (co)generators

content/generator_construction.md

@@ -1,13 +1,19 @@
 ---
 title: Construction of Generators
 description: How to construct a generator from a generating set
-author: Martin Brandenburg
+authors:
+    - Martin Brandenburg
+    - Daniel Schepler
 ---
 
 ## Construction of Generators
 
 ::: Lemma
-In a category let $S$ be a generating set which is [strongly connected](/category-property/strongly_connected) (between any two objects in $S$ there is a morphism). If the coproduct $U \coloneqq \coprod_{G \in S} G$ exists, then it is a generator.
+In a category let $S$ be a generating set which is [strongly connected](/category-property/strongly_connected) (between any two objects in $S$ there is a morphism). If the coproduct $U \coloneqq \coprod_{G \in S} G$ exists, then it is a generator. Moreover, if $S$ is an extremal generating set, then $U$ is an extremal generator.
 :::
 
-_Proof._ This is a straight forward generalization of [this result](/category-implication/generator_via_coproduct). We remark that the assumption about $S$ implies that each inclusion $G \to U$ has a left inverse. Now let $f,g : A \rightrightarrows B$ be two morphisms with $f h = g h$ for all $h : U \to A$. If $G \in S$, any morphism $G \to A$ extends to $U$ by our preliminary remark. Thus, $fh = gh$ holds for all $h : G \to A$ and $G \in S$. Since $S$ is a generating set, this implies $f = g$. <span class="qed">$\square$</span>
+_Proof._ We remark that the assumption on $S$ implies that each coprojection $i_G : G \to U$ has a left inverse. Now let $f,g : A \rightrightarrows B$ be two morphisms with $f \circ \bar a = g \circ \bar a$ for all $\bar a : U \to A$. If $G \in S$, any morphism $G \to A$ extends to $U$ by our preliminary remark. Thus, $f \circ a = g \circ a$ holds for all morphisms $a : G \to A$ with $G \in S$. Since $S$ is a generating set, this implies $f = g$.
+
+Similarly, for the case where $S$ is an extremal generating set, suppose we have a morphism $f : A \to B$ such that $f \circ {-} : \Hom(U, A) \to \Hom(U, B)$ is a bijection. In particular, because it is injective and $U$ is a generator, we can conclude that $f$ is a monomorphism, so $f \circ {-} : \Hom(G, A) \to \Hom(G, B)$ is injective for each $G \in S$. Now suppose $b \in \Hom(G, B)$ for $G \in S$. Then $b$ extends to a morphism $\bar b : U \to B$. By assumption, there exists $\bar a : U \to A$ such that $f \circ \bar a = \bar b$. Composing with the coprojection $i_G : G \to U$, we see
+$$f \circ \bar a \circ i_G = \bar b \circ i_G = b.$$
+This shows that $f \circ {-} : \Hom(G, A) \to \Hom(G, B)$ is also surjective for each $G \in S$. Since $S$ is an extremal generating set, this implies $f$ is an isomorphism. <span class="qed">$\square$</span>

content/subcategories.md

@@ -116,3 +116,9 @@ $$
 $$
 
 is a composition of faithful functors, hence faithful. <span class="qed">$\square$</span>
+
+::: Lemma 10
+Any fully faithful functor reflects extremal generating sets (and therefore, by duality, it also reflects extremal cogenerating sets). In other words, if $U : \C \to \D$ is a fully faithful functor, and $S$ is a set of objects such that $U(S)$ is an extremal generating set of $\D$, then $S$ is an extremal generating set of $\C$.
+:::
+
+_Proof:_ Under the given assumptions, we can factor $\C \to (\Set^+)^S$, $X \mapsto (\Hom_\C(G, X))_{G\in S}$, as being isomorphic to the composition of $U : \C \to D$ followed by $Y \mapsto (\Hom_\D(UG, Y))_{G\in S}$, using the assumption on $U$ to identify $\Hom_\D(UG, UX)$ with $\Hom_C(G, X)$ naturally in $X$. In this composition, the first is fully faithful and therefore also conservative; and the second is assumed to be faithful and conservative. Therefore, the composition is also faithful and conservative. <span class="qed">$\square$</span>

content/thin_extremal_generator.md

@@ -0,0 +1,23 @@
+---
+title: Thin Category with an Extremal Generator
+description: A result restricting which thin categories can have an extremal generator
+author: Daniel Schepler
+---
+
+# Thin Category with an Extremal Generator
+
+::: Lemma
+Suppose $G$ is an object of a thin category. Then $G$ is an extremal generator if and only if for every object $X$, either $X \cong G$ or every morphism with codomain $X$ is an isomorphism.
+:::
+
+_Proof._ ($\Rightarrow$) Since the category is thin, $\Hom(G, X)$ is either a singleton or empty. In the first case, let $f \in \Hom(G, X)$. Then $f \circ {-} : \Hom(G, G) \to \Hom(G, X)$ is automatically a bijection since $\Hom(G, G) = \{ \id_G \}$ is also a singleton, implying that $f$ is an isomorphism.
+
+In the second case, suppose we have a morphism $g : Y \to X$. Then $g \circ {-} : \Hom(G, Y) \to \Hom(G, X)$ is a function with empty codomain, so it is automatically a bijection, implying that $g$ is an isomorphism.
+
+($\Leftarrow$) Since the category is thin, any object is automatically a generator. Now suppose we have a morphism $f : X \to Y$ such that $f \circ {-} : \Hom(G, X) \to \Hom(G, Y)$ is a bijection. Then by assumption, either $Y \cong G$ or every morphism with codomain $Y$ is an isomorphism. In the first case, $\Hom(G, Y)$ is non-empty, so $\Hom(G, X)$ is also non-empty. We also have $\Hom(Y, G)$ is non-empty. Therefore, $\Hom(Y, X)$ is non-empty, and the (necessarily unique) morphism $Y \to X$ is automatically an inverse to $f$. In the second case, $f$ is already a morphism with codomain $Y$ so it is an isomorphism. <span class="qed">$\square$</span>
+
+::: Corollary
+For a poset $P$, the corresponding thin category has an extremal generator if and only if $P$ is non-empty and it has at most one non-minimal element. In particular, if this is the case, then either the poset is discrete, in which case any element gives an extremal generator; or otherwise, there is exactly one non-minimal element which is the unique extremal generator.
+:::
+
+_Proof._ In a thin category coming from a poset, the condition in the previous lemma that every morphism with codomain $X$ is an isomorphism is equivalent to the corresponding element of the poset being minimal. <span class="qed">$\square$</span>

database/data/categories/Ab_fg.yaml

@@ -21,8 +21,8 @@ satisfied_properties:
   - property: abelian
     proof: This follows from the fact for abelian groups and the fact that subgroups of finitely generated abelian groups are also finitely generated.
 
-  - property: generator
-    proof: The group $\IZ$ is a generator since it represents the forgetful functor to $\Set$.
+  - property: extremal generator
+    proof: The group $\IZ$ is an extremal generator since it represents the forgetful functor to $\Set$ which is faithful and conservative.
 
   - property: essentially countable
     proof: Every finitely generated abelian group is isomorphic to a group of the form $\IZ^n / U$, where $n \in \IN$ and $U$ is a subgroup of $\IZ^n$. Since $\IZ^n$ is Noetherian as a $\IZ$-module, $U$ is finitely generated, hence the category $\Ab_\fg$ has only countably many objects up to isomorphism. Furthermore, for any objects $A \cong \IZ^n / U$ and $B \cong \IZ^m / T$, the hom-set $\Hom(A,B)$ is countable. Indeed, precomposition with the quotient map yields an injection $\Hom(A,B) \hookrightarrow \Hom(\IZ^n, B) \cong B^n$, and $B^n$ is countable.

database/data/categories/Ban.yaml

@@ -20,8 +20,9 @@ satisfied_properties:
     proof: The trivial Banach space $\{0\}$ is a zero object.
     check_redundancy: false
 
-  - property: cogenerator
-    proof: The Hahn-Banach theorem implies that $\IC$ is a cogenerator.
+  - property: extremal cogenerator
+    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: CIP
     proof: This is immediate from the concrete description of coproducts and products.

database/data/categories/Cat.yaml

@@ -27,8 +27,11 @@ satisfied_properties:
   - property: semi-strongly connected
     proof: Every non-empty category is weakly terminal (by using constant functors).
 
-  - property: generator
-    proof: 'The <a href="/category/walking_morphism">walking morphism</a> $I$ is a generator: Assume that $F,G : \C \rightrightarrows \D$ are functors that agree when being precomposed with any functor from $I$. This means that $F(f) = G(f)$ for all morphisms $f : X \to Y$ in $\C$. By comparing the domains and applying this to $f = \id_X$, we see that $F(X) = G(X)$ for all objects $X$. And we just saw that $F,G$ also agree on morphisms.'
+  - property: extremal generator
+    proof: >-
+      The <a href="/category/walking_morphism">walking morphism</a> $I$ is a generator: Assume that $F,G : \C \rightrightarrows \D$ are functors that agree when being precomposed with any functor from $I$. This means that $F(f) = G(f)$ for all morphisms $f : X \to Y$ in $\C$. By comparing the domains and applying this to $f = \id_X$, we see that $F(X) = G(X)$ for all objects $X$. And we just saw that $F,G$ also agree on morphisms.
+
+      In fact, $I$ is an extremal generator: suppose $F : \C \to \D$ is a functor which induces a bijection $\Mor(\C) \to \Mor(\D)$. By considering the images of identity morphisms in $\C$, we see that $F$ is injective on objects; and then by considering the preimages of identity morphisms in $\D$, we see that $F$ is surjective on objects. By assumption, $F$ is also bijective on morphisms, so $F$ is an isomorphism of categories.
 
   - property: infinitary extensive
     proof: '[Sketch] This is straight forward from the fact that <a href="/category/Set">$\Set$</a> is infinitary extensive: A functor $\C \to \coprod_i \D_i$ yields full subcategories $\C_i \subseteq \C$ (the preimages of $\D_i)$ with $\C = \coprod_i \C_i$.'

database/data/categories/Delta.yaml

@@ -33,11 +33,11 @@ satisfied_properties:
   - property: strongly connected
     proof: For all $n,m$ there are morphisms $[n] \to [0] \to [m]$.
 
-  - property: generator
-    proof: The ordered set $[0] = \{0\}$ is a generator.
+  - property: extremal generator
+    proof: The ordered set $[1] = \{0 < 1\}$ is an extremal generator, even for <a href="/category/PreOrd">$\PreOrd$</a>. Now apply Lemma 10 <a href="/content/subcategories">here</a>.
 
-  - property: cogenerator
-    proof: The ordered set $[1] = \{0 < 1\}$ is a cogenerator, even for <a href="/category/Pos">$\Pos$</a>.
+  - property: extremal cogenerator
+    proof: The ordered set $[1] = \{0 < 1\}$ is an extremal cogenerator, even for <a href="/category/Pos">$\Pos$</a>. Now apply Lemma 10 <a href="/content/subcategories">here</a>.
 
   - property: skeletal
     proof: 'If $f : [n] \to [m]$ is an isomorphism, then $n + 1 = m + 1$ by comparing the cardinalities, hence $n = m$.'

database/data/categories/FI.yaml

@@ -26,8 +26,8 @@ satisfied_properties:
   - property: left cancellative
     proof: This is trivial.
 
-  - property: generator
-    proof: The one-point set is a generator since it represents the forgetful functor $\FI \to \Set$.
+  - property: extremal generator
+    proof: The one-point set is a generator since it represents the forgetful functor $\FI \to \Set$, which is faithful and conservative.
 
   - property: essentially countable
     proof: Every finite set is isomorphic to some $\{1,\dotsc,n\}$ for some $n \in \IN$.

database/data/categories/FS.yaml

@@ -28,8 +28,11 @@ satisfied_properties:
   - property: right cancellative
     proof: This is trivial.
 
-  - property: cogenerator
-    proof: 'We prove that $\{0,1\}$ is a cogenerator: The surjective maps $X \to \{0,1\}$ correspond to the non-empty proper subsets of $X$. If $a,b \in X$ are elements that have the same image under each surjective map $X \to \{0,1\}$, it therefore means that they lie in the same non-empty proper subsets of $X$. This implies $a=b$: If $X = \{a\}$, this is trivial. Otherwise, use the subset $\{a\}$.'
+  - property: extremal cogenerator
+    proof: >-
+      We prove that $\{0,1\}$ is an extremal cogenerator. First, to prove it is a cogenerator: The surjective maps $X \to \{0,1\}$ correspond to the non-empty proper subsets of $X$. If $a,b \in X$ are elements that have the same image under each surjective map $X \to \{0,1\}$, it therefore means that they lie in the same non-empty proper subsets of $X$. This implies $a=b$: If $X = \{a\}$, this is trivial. Otherwise, use the subset $\{a\}$.
+
+      Now, suppose we have a surjective map $f : X \to Y$ of finite sets such that ${-} \circ f : \Hom(Y, \{0,1\}) \to \Hom(X, \{0,1\})$ is bijective. That means that $f^* : P(Y) \to P(X)$ is bijective on non-empty proper subsets, and it certainly also maps $\varnothing \mapsto \varnothing$ and $Y \mapsto X$. Therefore, since the <a href="/functor/power_set_contravariant">contravariant powerset functor</a> is conservative, that implies $f$ is an isomorphism.
 
   - property: coequalizers
     proof: We construct coequalizers as in <a href="/category/FinSet">$\FinSet$</a> (or $\Set$) and observe that the universal property still holds when we restrict to surjective maps.

database/data/categories/FiltVect.yaml

@@ -47,6 +47,15 @@ satisfied_properties:
 
   - property: cogenerator
     proof: It is straightforward to check that the vector space $K$ equipped with the maximal filtration $F^n(K) \coloneqq K$ is a cogenerator.
+    check_redundancy: false
+
+  - property: extremal cogenerating set
+    proof: >-
+      Let $K_n$ denote the vector space $K$ equipped with the filtration such that $F_m(K) = K$ for $m < n$ and $F_m(K) = 0$ for $m \ge n$; and similarly, let $K_\infty$ denote the vector space denoted with the filtration such that $F_m(K) = K$ for each $m$. Then $K_n$ represents the functor mapping $(V, F)$ to $F_n(V)^\perp$, i.e. the space of functionals on $V$ whose kernels contain $F_n(V)$. Also, $K_\infty$ represents the functor sending $(V, F)$ to the dual $V^*$; and the canonical epimorphism $K_\infty \twoheadrightarrow K_n$ corresponds under the Yoneda embedding to the natural inclusion $F_n(V)^\perp \hookrightarrow V$. We claim that $\{ K_n : n \in \IZ \} \cup \{ K_\infty \}$ is an extremal cogenerating set of $\FiltVect_K$.
+
+      First, the set includes $K_\infty$, which we have already seen above is a cogenerator of $\FiltVect_K$. Now, suppose we have a morphism $f : (V, F) \to (W, F')$ such that ${-} \circ f : \Hom((W, F'), K_\infty) \to \Hom((V, F), K_\infty)$ is a bijection. This implies that $f^* : W^* \to V^*$ is an isomorphism of the dual vector spaces. But since $K$ is a cogenerator of $\Vect_K$ and $\Vect_K$ is balanced, in fact $K$ is an extremal cogenerator of $\Vect_K$; so from $f^*$ being an isomorphism, we conclude $f : V \to W$ is an isomorphism of vector spaces.
+
+      It remains to show that if $f$ also induces a bijection ${-} \circ f : \Hom((W, F'), K_n) \to \Hom(V, F), K_n)$ for each $n$, then $f$ induces an isomorphism of the filtrations. For this step, we may assume without loss of generality that $f = \id_V : V \to V$. By the observations above, this then reduces to showing that if $S_1^\perp = S_2^\perp$ where $S_1 \coloneqq F_n(V), S_2 \coloneqq F_n'(V)$ are subspaces of $V$ such that $S_1^\perp = S_2^\perp$, then $S_1 = S_2$. However, if we had $x \in S_1 \setminus S_2$, then starting with a basis of $S_2$, adjoining $x$, and then extending to a basis of $V$, we see that there is a functional $\varphi \in V^*$ such that $\varphi |_{S_2} = 0$ and $\varphi(x) = 1$. But then $\varphi \in S_2^* \setminus S_1^*$, giving a contradiction. This shows $S_1 \subseteq S_2$; and similarly $S_2 \subseteq S_1$.
 
   - property: finitely accessible
     proof: >-

database/data/categories/FinOrd.yaml

@@ -34,11 +34,11 @@ satisfied_properties:
   - property: semi-strongly connected
     proof: Every non-empty totally ordered set is weakly terminal (by using constant maps).
 
-  - property: generator
-    proof: The one-point finite ordered set is a generator since it represents the forgetful functor $\FinOrd \to \Set$.
+  - property: extremal generator
+    proof: The ordered set $\{0 < 1\}$ is an extremal generator, even for <a href="/category/PreOrd">$\PreOrd$</a>. Now apply Lemma 10 <a href="/content/subcategories">here</a>.
 
-  - property: cogenerator
-    proof: The ordered set $\{0 < 1\}$ is a cogenerator, even for <a href="/category/Pos">$\Pos$</a>.
+  - property: extremal cogenerator
+    proof: The ordered set $\{0 < 1\}$ is an extremal cogenerator, even for <a href="/category/Pos">$\Pos$</a>. Now apply Lemma 10 <a href="/content/subcategories">here</a>.
 
   - property: equalizers
     proof: Take the equalizer in $\FinSet$ and restrict the order.

database/data/categories/FinSet.yaml

@@ -26,11 +26,11 @@ satisfied_properties:
   - property: essentially countable
     proof: Every finite set is isomorphic to some $\{1,\dotsc,n\}$ for some $n \in \IN$.
 
-  - property: generator
-    proof: The one-point set is a generator since it represents the forgetful functor $\FinSet \to \Set$.
+  - property: extremal generator
+    proof: The one-point set is an extremal generator even in <a href="/category/Set">$\Set$</a>. Now apply Lemma 10 <a href="/content/subcategories">here</a>.
 
-  - property: cogenerator
-    proof: The two-element set is a cogenerator.
+  - property: extremal cogenerator
+    proof: The two-element set is an extremal cogenerator even in <a href="/category/Set">$\Set$</a>. Now apply Lemma 10 <a href="/content/subcategories">here</a>.
 
   - property: semi-strongly connected
     proof: Every non-empty finite set is weakly terminal (by using constant maps).

database/data/categories/FinVect_c.yaml

@@ -23,8 +23,8 @@ satisfied_properties:
   - property: essentially countable
     proof: Every object is isomorphic to $K^n$ for some $n \in \IN$, and $\Hom(K^n,K^m) \cong M_{m \times n}(K)$ is a countable set.
 
-  - property: generator
-    proof: The forgetful functor $\FinVect_K \to \Set$ is faithful and represented by $K$. Hence, $K$ is a generator.
+  - property: extremal generator
+    proof: The one-dimensional vector space $K$ is an extremal generator even in <a href="/category/Vect">$\Vect_K$</a>. Now apply Lemma 10 <a href="/content/subcategories">here</a>.
 
   - property: split abelian
     proof: This follows directly from the corresponding fact for <a href="/category/Vect">$\Vect_K$</a>.

database/data/categories/FinVect_f.yaml

@@ -26,8 +26,8 @@ satisfied_properties:
   - property: locally finite
     proof: Each hom-set $\Hom(K^n,K^m) \cong M_{m \times n}(K)$ is finite by assumption.
 
-  - property: generator
-    proof: The forgetful functor $\FinVect_K \to \Set$ is faithful and represented by $K$. Hence, $K$ is a generator.
+  - property: extremal generator
+    proof: The one-dimensional vector space $K$ is an extremal generator even in <a href="/category/Vect">$\Vect_K$</a>. Now apply Lemma 10 <a href="/content/subcategories">here</a>.
 
   - property: split abelian
     proof: This follows directly from the corresponding fact for <a href="/category/Vect">$\Vect_K$</a>.

database/data/categories/FinVect_u.yaml

@@ -23,8 +23,8 @@ satisfied_properties:
   - property: essentially small
     proof: Every object is isomorphic to $K^n$ for some $n \in \IN$, and $\Hom(K^n,K^m) \cong M_{m \times n}(K)$ is a set.
 
-  - property: generator
-    proof: The forgetful functor $\FinVect_K \to \Set$ is faithful and represented by $K$. Hence, $K$ is a generator.
+  - property: extremal generator
+    proof: The one-dimensional vector space $K$ is an extremal generator even in <a href="/category/Vect">$\Vect_K$</a>. Now apply Lemma 10 <a href="/content/subcategories">here</a>.
 
   - property: split abelian
     proof: This follows directly from the corresponding fact for <a href="/category/Vect">$\Vect_K$</a>.