diff --git a/.cspell.json b/.cspell.json index 6566b4d8a..887f72204 100644 --- a/.cspell.json +++ b/.cspell.json @@ -197,6 +197,7 @@ "épimorphismes", "epimorphisms", "epis", + "equivariant", "EUMDL", "exponentials", "extensivity", diff --git a/content/deloopings-aleph1-accessible.md b/content/deloopings-aleph1-accessible.md new file mode 100644 index 000000000..185e0aff5 --- /dev/null +++ b/content/deloopings-aleph1-accessible.md @@ -0,0 +1,111 @@ +--- +title: ℵ₁-accessibility of deloopings of monoids +description: We show that the delooping of a left cancellative or countable monoid is ℵ₁-accessible when its idempotents split and its principal right ideals satisfy the ascending chain condition. +--- + +# $\aleph_1$-accessibility of deloopings of monoids + +Every monoid $M$ induces a category $BM$ with a single object $*$, morphisms given by the elements of $M$, and composition given by $a \circ b \coloneqq ab$. Since $BM$ is small, [this result](/category-implication/small_accessible_characterization) shows that $BM$ is accessible when it is [Cauchy complete](/category-property/Cauchy_complete), and the converse holds by [this result](/category-implication/accessible_require_Cauchy_complete). Explicitly, $BM$ is Cauchy complete if and only if every idempotent $e \in M$ can be written as $e = sr$ with $s,r \in M$ and $rs = 1$. However, this does not tell us for which regular cardinals $\kappa$ the category $BM$ is $\kappa$-accessible. We show that $BM$ is [$\aleph_1$-accessible](/category-property/ℵ₁-accessible) when, in addition, $M$ is [left cancellative](/category-property/left_cancellative) (i.e. $ab = ac$ implies $b = c$ for all $a,b,c \in M$) or countable, and $M$ satisfies the following finiteness condition. + +A _principal right ideal_ of $M$ is a subset of the form $aM = \{am : m \in M\}$ for some $a \in M$. We say that $M$ satisfies the _ascending chain condition on principal right ideals_ if every ascending chain +$$a_0 M \subseteq a_1 M \subseteq a_2 M \subseteq \cdots$$ +of principal right ideals of $M$ becomes stationary. For example, this holds when $M$ is finite or a group, and also when $M$ is a free monoid, such as $\IN$, since then $aM \subseteq bM$ means that $b$ is a prefix of $a$. + +We identify presheaves on $BM$ with right $M$-sets, i.e. sets $X$ with a map $X \times M \to X$, $(x,a) \mapsto xa$, satisfying $x1 = x$ and $(xa)b = x(ab)$. The representable presheaf $\Hom(-,*)$ corresponds to $M$ with the action given by multiplication from the right. By the Yoneda lemma, the $M$-equivariant maps $M \to X$ correspond to the elements $x \in X$ via $a \mapsto xa$, and a morphism $a : * \to *$ of $BM$ corresponds to the left multiplication $M \to M$, $b \mapsto ab$. + +## Idea + +Let $D : \I \to BM$ be a diagram indexed by a small $\aleph_1$-filtered category $\I$. Its colimit $W$ in the category of right $M$-sets is the union of the cyclic sub-$M$-sets $e_i M$, where $e_i \in W$ is the image of $1 \in M$ under the colimit map at $i \in \I$. The diagram $D$ has a colimit in $BM$ which is preserved by $\Hom(*,-)$ if and only if $W$ is isomorphic to $M$. + +Each $e_i M$ is a quotient of $M$. Along the diagram, more and more elements of $e_i M$ get identified, and the sub-$M$-sets $e_i M$ grow. We show that both processes stop at some stage. Since $\I$ is $\aleph_1$-filtered, any countably many stages are dominated by a single one, and this is where the hypotheses come in. When $M$ is left cancellative, no identifications are made at all. When $M$ is countable, only countably many identifications have to be made in $e_i M$, so they are all made at a single stage. In both cases, $e_i M$ is isomorphic to a principal right ideal of $M$ (Lemma 1). Then the ascending chain condition implies that the sub-$M$-sets $e_i M$ stop growing, i.e. that $W = e_k M$ for some $k \in \I$ (Lemma 2). Once both processes have stopped, the next stage provides a section of the projection $M \to W$, $a \mapsto e_k a$, so that $W$ is a retract of $M$ (Proposition 3). Finally, since $BM$ is Cauchy complete, $W$ is isomorphic to $M$ (Theorem 4). + +## Proofs + +Throughout, let $D : \I \to BM$ be a diagram, where $\I$ is a small $\aleph_1$-filtered category. For a morphism $u : i \to j$ in $\I$, we write $D(u) \in M$ for the corresponding element. Let $W$ be the colimit of the composite of $D$ with the Yoneda embedding, i.e. of the diagram of right $M$-sets which maps every $i \in \I$ to $M$ and every morphism $u : i \to j$ to the left multiplication $M \to M$, $a \mapsto D(u) a$. Since colimits of presheaves are computed pointwise, the underlying set of $W$ is the colimit of the underlying diagram of sets. Since $\I$ is filtered, the elements of $W$ are classes $[i,a]$ with $i \in \I$ and $a \in M$, where $[i,a] = [i',a']$ if and only if there are morphisms $u : i \to j$ and $u' : i' \to j$ with $D(u) a = D(u') a'$. The action of $M$ on $W$ is given by $[i,a] c = [i,ac]$, and the colimit maps are $M \to W$, $a \mapsto [i,a]$. + +Let $e_i \coloneqq [i,1]$, so that $[i,a] = e_i a$. For every morphism $u : i \to j$ we have $e_i = [j,D(u)] = e_j D(u)$, so that $e_i M \subseteq e_j M$. Moreover, the sub-$M$-sets $e_i M$ cover $W$. For $a,b \in M$ we have $e_i a = e_i b$ if and only if there is a morphism $u : i \to j$ with $D(u) a = D(u) b$. For the non-trivial direction, assume that $D(u) a = D(u') b$ for morphisms $u,u' : i \rightrightarrows j$. Since $\I$ is filtered, there is a morphism $v : j \to k$ with $v u = v u'$. Then $D(vu) a = D(v) D(u) a = D(v) D(u') b = D(vu') b = D(vu) b$. + +::: Lemma 1 +Assume that $M$ is left cancellative or countable. Then for every $i \in \I$ there is a morphism $u : i \to j$ such that for all $a,b \in M$ we have +$$e_i a = e_i b \iff D(u) a = D(u) b.$$ +In particular, the map $e_i M \to D(u) M$, $e_i a \mapsto D(u) a$ is a well-defined isomorphism of right $M$-sets. +::: + +::: Proof +If $M$ is left cancellative, let $u \coloneqq \id_i$, so that $D(u) = 1$. By the description of $W$ above, $e_i a = e_i b$ holds if and only if $D(v) a = D(v) b$ for some morphism $v : i \to j$, which is equivalent to $a = b$ since $M$ is left cancellative. + +If $M$ is countable, for every pair $(a,b) \in M \times M$ with $e_i a = e_i b$ we choose a morphism $u_{a,b} : i \to j_{a,b}$ with $D(u_{a,b}) a = D(u_{a,b}) b$. Since $M$ is countable, these are countably many morphisms. Since $\I$ is $\aleph_1$-filtered, the diagram consisting of them has a cocone, that is, there are an object $j$ and morphisms $u : i \to j$ and $v_{a,b} : j_{a,b} \to j$ with $v_{a,b} u_{a,b} = u$. Then +$$D(u) a = D(v_{a,b}) D(u_{a,b}) a = D(v_{a,b}) D(u_{a,b}) b = D(u) b.$$ +Conversely, if $D(u) a = D(u) b$, then $e_i a = e_i b$ by the description of $W$ above. + +In both cases, it follows that the map $e_i M \to D(u) M$, $e_i a \mapsto D(u) a$ is well-defined and injective. It is surjective and $M$-equivariant by construction. +::: + +::: Lemma 2 +Assume that $M$ is left cancellative or countable, and that $M$ satisfies the ascending chain condition on principal right ideals. Then there is some $k \in \I$ with $W = e_k M$. +::: + +::: Proof +Let $P$ be the set of all sub-$M$-sets $e_i M$ of $W$ with $i \in \I$, partially ordered by inclusion. First, we show that every ascending chain +$$e_{i_0} M \subseteq e_{i_1} M \subseteq e_{i_2} M \subseteq \cdots$$ +in $P$ becomes stationary. Since $\I$ is $\aleph_1$-filtered, there are an object $j$ and morphisms $i_n \to j$ for all $n \in \IN$. Then $e_{i_n} \in e_j M$, so that $e_{i_n} = e_j a_n$ for some $a_n \in M$. Choose a morphism $u : j \to j'$ as in Lemma 1, and let $\varphi : e_j M \to D(u) M$ be the isomorphism $e_j a \mapsto D(u) a$. Then +$$\varphi(e_{i_n} M) = \varphi(e_j a_n M) = D(u) a_n M.$$ +These sets form an ascending chain of principal right ideals of $M$, which becomes stationary by assumption. Since $\varphi$ is injective, the chain $(e_{i_n} M)_{n \in \IN}$ becomes stationary as well. + +Since $\I$ is non-empty, so is $P$. Therefore, $P$ has a maximal element $e_k M$, since otherwise we could choose a strictly ascending chain in $P$. Let $i \in \I$. Since $\I$ is filtered, there are an object $l$ and morphisms $i \to l$, $k \to l$. Then $e_i M \subseteq e_l M$ and $e_k M \subseteq e_l M$. By maximality, we have $e_l M = e_k M$, and hence $e_i M \subseteq e_k M$. Since the sub-$M$-sets $e_i M$ cover $W$, we conclude that $W = e_k M$. +::: + +::: Proposition 3 +Assume that $M$ is left cancellative or countable, and that $M$ satisfies the ascending chain condition on principal right ideals. Then $W$ is a retract of $M$ as a right $M$-set. More precisely, there are an idempotent $e \in M$ and $M$-equivariant maps $\pi : M \to W$ and $\iota : W \to M$ with $\pi \circ \iota = \id_W$ and $\iota(\pi(a)) = ea$ for all $a \in M$. +::: + +::: Proof +Choose $k \in \I$ with $W = e_k M$ as in Lemma 2, and then a morphism $u : k \to l$ as in Lemma 1. Since $e_l \in W = e_k M$, there is some $x \in M$ with $e_l = e_k x$. Let $e \coloneqq x D(u)$. We define $\pi : M \to W$ by $\pi(a) \coloneqq e_k a$ and $\iota : W \to M$ by +$$\iota(e_k a) \coloneqq e a = x D(u) a.$$ +This is well-defined: if $e_k a = e_k b$, then $D(u) a = D(u) b$ by the choice of $u$, and hence $x D(u) a = x D(u) b$. Both maps are $M$-equivariant, and $\iota(\pi(a)) = ea$ holds by definition. We compute +$$\pi(\iota(e_k a)) = e_k x D(u) a = e_l D(u) a = e_k a.$$ +Hence, $\pi \circ \iota = \id_W$. Finally, $e$ is idempotent, since +$$ee = \iota(\pi(e)) = \iota(\pi(\iota(\pi(1)))) = \iota(\pi(1)) = e,$$ +where the first two equations use that $\iota \circ \pi$ is the left multiplication with $e$. +::: + +::: Theorem 4 +Let $M$ be a monoid such that $M$ is left cancellative or countable, $M$ satisfies the ascending chain condition on principal right ideals, and $BM$ is Cauchy complete. Then every $\aleph_1$-filtered colimit of representable presheaves on $BM$ is representable. Hence, $BM$ is $\aleph_1$-accessible. +::: + +::: Proof +Let $D : \I \to BM$ and $W$ be as above. Choose $e$, $\pi$, $\iota$ as in Proposition 3. Since $BM$ is Cauchy complete, there are $s,r \in M$ with $e = sr$ and $rs = 1$. We claim that the $M$-equivariant maps + +$$ +\begin{align*} +\theta & : W \to M, \quad w \mapsto r \iota(w), \\ +\theta' & : M \to W, \quad a \mapsto \pi(sa) +\end{align*} +$$ + +are inverse to each other. For $a \in M$ we have +$$\theta(\theta'(a)) = r \iota(\pi(sa)) = r e s a = (rs)(rs) a = a.$$ +For $w \in W$ we have $e \iota(w) = \iota(\pi(\iota(w))) = \iota(w)$, and hence +$$\theta'(\theta(w)) = \pi(s r \iota(w)) = \pi(e \iota(w)) = \pi(\iota(w)) = w.$$ +Thus, $W$ is representable. By Lemma 5 below, $BM$ is $\aleph_1$-accessible. +::: + +::: Lemma 5 +Let $\kappa$ be a regular cardinal and $\C$ be an essentially small category such that every $\kappa$-filtered colimit of representable presheaves on $\C$ is representable. Then $\C$ is $\kappa$-accessible. +::: + +::: Proof +Let $y : \C \to [\C^\op,\Set]$ be the Yoneda embedding, let $D : \I \to \C$ be a diagram, where $\I$ is a small $\kappa$-filtered category, and choose $X \in \C$ with $\colim_{i \in \I} y(D(i)) \cong y(X)$. Since $y$ is fully faithful, it reflects colimits. Therefore, $X$ is a colimit of $D$, and $y$ preserves this colimit, i.e. the canonical morphism +$$\textstyle \colim_{i \in \I} \Hom(-,D(i)) \to \Hom(-,\colim_{i \in \I} D(i))$$ +is an isomorphism. Evaluating it at an object $A \in \C$ shows that $\Hom(A,-)$ preserves the colimit of $D$. Hence, $\C$ has $\kappa$-filtered colimits, and every object of $\C$ is $\kappa$-presentable. Since $\C$ is essentially small, it follows that $\C$ is $\kappa$-accessible. +::: + +When $M$ is left cancellative, the assumption in Theorem 4 that $BM$ is Cauchy complete is automatic by [this result](/category-implication/cauchy_complete_criterion). In this case, the results above extend to large monoids. + +::: Proposition 6 +Let $M$ be a large monoid, i.e. a monoid whose underlying collection is not necessarily a set (see [here](/content/foundations)). Assume that $M$ is left cancellative and satisfies the ascending chain condition on principal right ideals. Then $BM$ has $\aleph_1$-filtered colimits. +::: + +::: Proof +In general, $BM$ is not locally small. However, the proofs of Lemma 1 (in the left cancellative case), Lemma 2, Proposition 3 and Theorem 4 work verbatim when we replace sets by collections, i.e. right $M$-sets by right $M$-collections. They show that every $\aleph_1$-filtered colimit of representable presheaves with values in collections is representable, and hence, as in the proof of Lemma 5, that $BM$ has $\aleph_1$-filtered colimits. Here, $BM$ is Cauchy complete by [this result](/category-implication/cauchy_complete_criterion). +::: diff --git a/database/data/categories/BC.yaml b/database/data/categories/BC.yaml new file mode 100644 index 000000000..e80169e57 --- /dev/null +++ b/database/data/categories/BC.yaml @@ -0,0 +1,121 @@ +id: BC +name: delooping of the bicyclic monoid +notation: $BC$ +objects: a single object $*$ +morphisms: 'the elements of the bicyclic monoid $C \coloneqq \langle p,q : pq = 1 \rangle$' +description: >- + Every monoid $M$ induces a category $BM$ with a single object $*$, morphisms given by the elements of $M$, and composition given by the monoid operation. Some of the properties of this category depend on the specific monoid. + + In this entry, we take the monoid $C$ freely generated by $p,q$ subject to the relation $pq=1$, known as the bicyclic monoid and often denoted by $C(p,q)$. Every element of $C$ can be uniquely written as $q^n p^m$ for $n,m \geq 0$. + + This category is a one-object version of the walking splitting, but its properties are completely different. +nlab_link: null + +tags: + - algebra + +related: + - BD + - BN + - BG_c + +satisfied_properties: + - property: small + proof: This is trivial. + + - property: countable + proof: This is because the monoid $C$ is countable. + + - property: core-connected + proof: The category has exactly one object. + + - property: skeletal + proof: The category has exactly one object. + + - property: self-dual + proof: The monoid $C$ is isomorphic to its dual $C^{\op}$ by mapping $p \mapsto q$, $q \mapsto p$. + + - property: Cauchy complete + proof: We need to show that every idempotent of $C$ splits. Assume that $e = q^n p^m$ is idempotent. Thus, $e^2 = q^n p^m q^n p^m$. If $m \geq n$, then $p^m q^n = p^{m-n}$, so that $e^2 = q^n p^{2m-n}$. Thus, $m=2m-n$, i.e. $m=n$. If $m \leq n$, then $p^m q^n = q^{n-m}$, so that $e^2 = q^{2n-m} p^m$. Thus, $n=2n-m$, and again we get $m=n$. Every idempotent therefore has the form $e = q^n p^n$ for some $n \geq 0$. It splits because $p^n q^n = 1$. + check_redundancy: false + + - property: core-thin + proof: We need to show that $1$ is the only invertible element of $C$. Assume that $f = q^n p^m$ is invertible. If $m > 0$, then $f = f'p$ for some $f'$, so that $p$ is a split monomorphism. Because of $pq=1$, it is also a split epimorphism. Hence, it must be an isomorphism. But then $pq=1$ implies $qp=1$, which is not correct. Therefore, we must have $m=0$. Then $f=q^n$ is not an isomorphism for $n>0$ since otherwise $q$ would be an isomorphism, from which again $qp=1$ would follow. + + - property: mono-regular + proof: This follows from our descriptions of monomorphisms and regular monomorphisms below. It shows that in fact, every monomorphism splits. + + - property: reflexive coequalizers + proof: >- + It suffices to prove that every reflexive pair $f,g$ satisfies $f=g$. Algebraically, this means that left inverses are uniquely determined in $C$. So assume that $xy=1$ for some $x,y \in C$. In particular, $x$ is an epimorphism and $y$ is a monomorphism. By the classification of these morphisms below, we have $x=p^m$ for some $m \geq 0$ and $y=q^n$ for some $n \geq 0$. If $n > m$, then + $$1 = xy = p^m q^n = q^{n-m}$$ + is a contradiction. Likewise, if $n < m$, we obtain the contradiction $1 = p^{m-n}$. Therefore, $m=n$. Thus, $x$ is uniquely determined by $y$. + + - property: ℵ₁-accessible + proof: >- + We apply Theorem 4 here. We already know that $C$ is countable and that $BC$ is Cauchy complete. It remains to show that $C$ satisfies the ascending chain condition on principal right ideals. Every principal right ideal of $C$ has the form $q^n p^m C = q^n C$, since $p^m q^m = 1$ implies $p^m C = C$. If $q^n C \subseteq q^{n'} C$, then $n \geq n'$. Otherwise, write $q^n = q^{n'} x$ with $x \in C$. Since $q^n$ is a monomorphism, we get $1 = q^{n'-n} x$, so that $q^{n'-n}$ is a split epimorphism, which contradicts the classification of epimorphisms below. Therefore, every ascending chain $q^{n_0} C \subseteq q^{n_1} C \subseteq \cdots$ satisfies $n_0 \geq n_1 \geq \cdots$ and hence becomes stationary. + label: BC_aleph1-accessible + +unsatisfied_properties: + - property: locally finite + proof: This is because $C$ is infinite. + + - property: zero morphisms + proof: We need to show that $C$ has no absorbing element. Assume that $q^n p^m$ is an absorbing element. Then $q^{n+1} p^m = q (q^n p^m) = q^n p^m$ implies $n+1=n$, a contradiction. + + - property: subobject-trivial + proof: The monomorphism $q$ is not an isomorphism. + + - property: cokernel pairs + proof: >- + We show that $q$ does not have a cokernel pair, i.e. that there is no universal pair of elements $(x,y) \in C^2$ with $xq=yq$; the universal property means that any other such pair $(u,v)$ factors as $u=tx$, $v=ty$ for a unique $t \in C$. + + Assume that $(x,y)$ is a universal pair. Since $(1,1)$ is a valid pair, there is a unique $t \in C$ with $1=tx$ and $1=ty$. In particular, $x$ and $y$ are monomorphisms. By their classification below, we get $x=q^n$, $y=q^k$ for some $n,k \geq 0$. Since $xq=yq$, we actually have $k=n$, i.e. $x=y$. + + Since $(qp)q=q$, also $(qp,1)$ is a valid pair. Hence, there is a unique $t \in C$ with $qp=tx$ and $1=ty$. Since $x=y$, we obtain the contradiction $qp=1$. + + - property: sifted + proof: >- + Consider the homomorphism $\deg : C \to (\IZ,+,0)$ defined by $\deg(p)=1$ and $\deg(q)=-1$. To every cospan $(x,y)$ (over the unique pair of objects in $BC$) we associate the number $\deg(x) - \deg(y)$. If a cospan $(x,y)$ has a morphism to another cospan $(u,v)$, i.e. there is some element $t$ with $tx=u$ and $ty=v$, then + $$\deg(u) - \deg(v) = (\deg(t)+\deg(x)) - (\deg(t)+\deg(y)) = \deg(x) - \deg(y).$$ + Therefore, all cospans in a connected component have the same associated number. In particular, the cospans $(1,1)$ and $(p,1)$ do not lie in the same connected component. + label: BC_not_sifted + + - property: sequential colimits + proof: >- + We will show that the sequence + $$* \xrightarrow{q} * \xrightarrow{q} * \xrightarrow{q} \cdots$$ + has no colimit. Assume that the sequence has a colimit, i.e. a universal cocone $x=(x_n)$. Write $x_0 = q^a p^b$. Choose some $m > a$ and define the sequence of elements $(c_n)$ by + $$c_n \coloneqq \begin{cases} + q^{m-n} & n \leq m \\ + p^{n-m} & n \geq m. + \end{cases}$$ + It is easy to verify that $(c_n)$ is a cocone. By universality, there is some $h \in C$ such that + $$c_n = h x_n \tag{1}$$ + for all $n \geq 0$. Write $h = q^c p^d$. For $n=0$ the cocone equality $(1)$ is: + $$q^{m} = c_0 = h x_0 = q^c p^d q^a p^b.$$ + For $d \geq a$ the right hand side is $q^c p^{d-a+b}$, so that $m=c$ and $0=d-a+b$. For $d < a$ the right hand side is $q^{c+a-d} p^b$, so that $b=0$ and $c=m-a+d \geq m - a > 0$. In each case, we find $c > 0$. That is, $h = qh'$ for some $h' \in C$. + + Next, we evaluate the cocone equality $(1)$ at $n=m+1$. We get + $$p = c_{m+1} = h x_{m+1} = q (h' x_{m+1}).$$ + Since $p$ is an epimorphism, this implies that $q$ is also an epimorphism, which is a contradiction. + label: BC_no_sequential_colimits + +special_objects: {} + +special_morphisms: + isomorphisms: + description: only the neutral element $1$ + proof: We have shown this in the proof that the category is core-thin. + monomorphisms: + description: the elements $q^n$ for $n \geq 0$ + proof: Since $q$ is a split monomorphism, also $q^n$ is a split monomorphism. For $m > 0$ the element $q^n p^m$ is not a monomorphism, since otherwise $p$ would be a monomorphism, from which we would obtain the contradiction $qp=1$ from $pq=1$. + epimorphisms: + description: the elements $p^m$ for $m \geq 0$ + proof: The proof is identical to the proof for monomorphisms. This is also because $C$ is self-dual. + regular monomorphisms: + description: the elements $q^n$ for $n \geq 0$ + proof: We already know that these are precisely the monomorphisms. They are split monomorphisms because $q$ is a split monomorphism, and every split monomorphism is a regular monomorphism. + regular epimorphisms: + description: the elements $p^m$ for $m \geq 0$ + proof: We already know that these are precisely the epimorphisms. They are split epimorphisms because $p$ is a split epimorphism, and every split epimorphism is a regular epimorphism. diff --git a/database/data/categories/BD.yaml b/database/data/categories/BD.yaml new file mode 100644 index 000000000..d890a536f --- /dev/null +++ b/database/data/categories/BD.yaml @@ -0,0 +1,166 @@ +id: BD +name: delooping of the double bicyclic monoid +notation: $BD$ +objects: a single object $*$ +morphisms: 'the elements of the monoid $D \coloneqq \langle f,g,s : fs = gs = 1 \rangle$' +description: >- + Every monoid $M$ induces a category $BM$ with a single object $*$, morphisms given by the elements of $M$, and composition given by the monoid operation. Some of the properties of this category depend on the specific monoid. + + In this entry, we take the monoid $D$ freely generated by $f,g,s$ subject to the relations $fs = gs = 1$, which we call the double bicyclic monoid. It contains two copies of the bicyclic monoid, generated by $f,s$ and by $g,s$. Every element of $D$ can be uniquely written as $s^n w$ for $n \geq 0$ and a word $w$ in $f,g$. + + This category provides an example of a core-connected category that does not have reflexive coequalizers. +nlab_link: null + +tags: + - algebra + +related: + - BC + - BN + +satisfied_properties: + - property: small + proof: This is trivial. + + - property: countable + proof: This is because the monoid $D$ is countable. + + - property: core-connected + proof: The category has exactly one object. + + - property: skeletal + proof: The category has exactly one object. + + - property: Cauchy complete + proof: >- + We need to show that every idempotent of $D$ splits. For a word $u$ in $f,g$ of length $k$, we have $u s^k = 1$, since every letter of $u$ is a left inverse of $s$. + + Now assume that $e = s^n w$ is idempotent, where $w$ is a word of length $k$. Thus, $e^2 = s^n w s^n w$. If $k \geq n$, we write $w = w' u$ with words $w'$ of length $k-n$ and $u$ of length $n$. Then $w s^n = w' u s^n = w'$, so that $e^2 = s^n w' w$. By the uniqueness of the normal form, $e^2 = e$ implies $w' w = w$, so $w'$ is empty, i.e. $k = n$. If $k < n$, then $w s^n = w s^k s^{n-k} = s^{n-k}$, so that $e^2 = s^{2n-k} w$. By the uniqueness of the normal form, $e^2 = e$ implies $2n-k = n$, i.e. $k = n$, which contradicts $k < n$. + + Every idempotent therefore has the form $e = s^n w$ for a word $w$ of length $n$. It splits because $w s^n = 1$. + check_redundancy: false + + - property: core-thin + proof: We need to prove that $1$ is the only invertible element of $D$. Every isomorphism is a monomorphism and an epimorphism. By the classification of monomorphisms and epimorphisms (see below), the only element that is both is $s^0 = 1$. + + - property: mono-regular + proof: By our description of monomorphisms below, every monomorphism has the form $s^n$. Since $s$ is a split monomorphism, so is $s^n$. Every split monomorphism is a regular monomorphism. + + - property: epi-regular + proof: By our description of epimorphisms below, every epimorphism is a word in $f,g$. Since $f$ and $g$ are split epimorphisms, so is every word in $f,g$. Every split epimorphism is a regular epimorphism. + + - property: coreflexive equalizers + proof: >- + It suffices to prove that every coreflexive pair $x,y$ satisfies $x=y$. Algebraically, this means that right inverses are uniquely determined in $D$. So assume that $r x = r y = 1$ for some $r,x,y \in D$. In particular, $r$ is an epimorphism, so that $r$ is a word in $f,g$ of some length $m$ by the classification of epimorphisms below. We claim that $x = s^m$; by symmetry, then also $y = s^m$. + + Write $x = s^n w$, where $w$ is a word in $f,g$. Since every letter of $r$ is a left inverse of $s$, we have $r s^m = 1$. If $n \geq m$, it follows that + $$1 = r x = r s^m s^{n-m} w = s^{n-m} w,$$ + so that $n = m$ and $w$ is empty by the uniqueness of the normal form, i.e. $x = s^m$. If $n < m$, we write $r = r' u$ with words $r'$ of length $m-n$ and $u$ of length $n$. Then $u s^n = 1$, so that + $$1 = r x = r' u s^n w = r' w.$$ + This is a contradiction, since $r' w$ is a non-empty word. + + - property: ℵ₁-accessible + proof: >- + The proof is very similar to the one for $BC$. We apply Theorem 4 here. We already know that $D$ is countable and that $BD$ is Cauchy complete. It remains to show that $D$ satisfies the ascending chain condition on principal right ideals. Every principal right ideal of $D$ has the form $s^n w D = s^n D$, where $w$ is a word in $f,g$, since $w$ is a split epimorphism and hence $w D = D$. If $s^n D \subseteq s^{n'} D$, then $n \geq n'$. Otherwise, write $s^n = s^{n'} x$ with $x \in D$. Since $s^n$ is a monomorphism, we get $1 = s^{n'-n} x$, so that $s^{n'-n}$ is a split epimorphism, which contradicts the classification of epimorphisms below. Therefore, every ascending chain $s^{n_0} D \subseteq s^{n_1} D \subseteq \cdots$ satisfies $n_0 \geq n_1 \geq \cdots$ and hence becomes stationary. + references: + - BC_aleph1-accessible + + - property: ℵ₁-coaccessible + proof: >- + We apply Theorem 4 here to the opposite monoid $D^{\op}$, whose delooping is $(BD)^{\op}$. We already know that $D^{\op}$ is countable and that $(BD)^{\op}$ is Cauchy complete. It remains to show that $D^{\op}$ satisfies the ascending chain condition on principal right ideals, i.e. that $D$ satisfies the ascending chain condition on principal left ideals. + + Every principal left ideal of $D$ has the form $D s^n w = D w$, where $w$ is a word in $f,g$, since $s^n$ is a split monomorphism and hence $D s^n = D$. We claim that $D w \subseteq D w'$ implies that $w'$ is a suffix of $w$. To see this, write $w = t w'$ with $t \in D$ and write $t = s^k v$, where $v$ is a word in $f,g$. Then $w = s^k (v w')$, where $v w'$ is a word in $f,g$. By the uniqueness of the normal form, we get $k = 0$ and $w = v w'$. + + Now let $D w_0 \subseteq D w_1 \subseteq \cdots$ be an ascending chain of principal left ideals, where every $w_i$ is a word in $f,g$. By the claim, $w_{i+1}$ is a suffix of $w_i$ for all $i \geq 0$. Hence, the lengths of the words $w_i$ form a decreasing sequence of natural numbers, which becomes stationary. Once the lengths are equal, the suffix $w_{i+1}$ of $w_i$ equals $w_i$. Therefore, the chain becomes stationary. + +unsatisfied_properties: + - property: locally finite + proof: This is because $D$ is infinite. + + - property: zero morphisms + proof: We need to show that $D$ has no absorbing element. Assume that $s^n w$ is an absorbing element. Then $s^{n+1} w = s (s^n w) = s^n w$ implies $n+1=n$ by the uniqueness of the normal form, a contradiction. + + - property: sifted + proof: >- + The proof is very similar to the one for $BC$. Consider the homomorphism $\deg : D \to (\IZ,+,0)$ defined by $\deg(s)=1$ and $\deg(f)=\deg(g)=-1$. It is well-defined since $\deg(fs) = \deg(gs) = 0 = \deg(1)$. To every cospan $(x,y)$ (over the unique pair of objects in $BD$) we associate the number $\deg(x) - \deg(y)$. If a cospan $(x,y)$ has a morphism to another cospan $(u,v)$, i.e. there is some element $t$ with $tx=u$ and $ty=v$, then + $$\deg(u) - \deg(v) = (\deg(t)+\deg(x)) - (\deg(t)+\deg(y)) = \deg(x) - \deg(y).$$ + Therefore, all cospans in a connected component have the same associated number. In particular, the cospans $(1,1)$ and $(s,1)$ do not lie in the same connected component. + label: BD_not_sifted + references: + - BC_not_sifted + + - property: cosifted + proof: >- + We use the homomorphism $\deg : D \to \IZ$ from the proof that $BD$ is not sifted. To every span $(x,y)$ we associate the number $\deg(x) - \deg(y)$. If a span $(u,v)$ has a morphism to another span $(x,y)$, i.e. there is some element $t$ with $u=xt$ and $v=yt$, then + $$\deg(u) - \deg(v) = (\deg(x)+\deg(t)) - (\deg(y)+\deg(t)) = \deg(x) - \deg(y).$$ + Therefore, all spans in a connected component have the same associated number. In particular, the spans $(1,1)$ and $(s,1)$ do not lie in the same connected component. + references: + - BD_not_sifted + + - property: reflexive coequalizers + proof: The pair $f,g$ is reflexive since $fs = gs = 1$. We claim that it is not coequalized by any morphism, so that in particular it has no coequalizer. Let $x = s^n w$ be an element of $D$, where $w$ is a word in $f,g$. Then $xf = s^n (wf)$ and $xg = s^n (wg)$ are in normal form, and the words $wf$ and $wg$ are different. By the uniqueness of the normal form, we get $xf \neq xg$. + + - property: kernel pairs + proof: >- + We show that $f$ does not have a kernel pair, i.e. that there is no universal pair of elements $(x,y) \in D^2$ with $fx=fy$; the universal property means that any other such pair $(u,v)$ factors as $u=xt$, $v=yt$ for a unique $t \in D$. + + Assume that $(x,y)$ is a universal pair. Since $(1,1)$ is a valid pair, there is a unique $t \in D$ with $1=xt$ and $1=yt$. In particular, $x$ and $y$ are epimorphisms. By their classification below, $x$ and $y$ are words in $f,g$. Then $fx$ and $fy$ are words in $f,g$ as well, so that $fx = fy$ implies $x=y$ by the uniqueness of the normal form. + + Since $f(sf) = f = f \cdot 1$, also $(sf,1)$ is a valid pair. Hence, there is some $t \in D$ with $sf=xt$ and $1=yt$. Since $x=y$, we obtain the contradiction $sf=1$. + + - property: cokernel pairs + proof: >- + We show that $s$ does not have a cokernel pair, i.e. that there is no universal pair of elements $(x,y) \in D^2$ with $xs=ys$; the universal property means that any other such pair $(u,v)$ factors as $u=tx$, $v=ty$ for a unique $t \in D$. + + Assume that $(x,y)$ is a universal pair. Since $(1,1)$ is a valid pair, there is a unique $t \in D$ with $1=tx$ and $1=ty$. In particular, $x$ and $y$ are monomorphisms. By their classification below, we get $x=s^n$, $y=s^k$ for some $n,k \geq 0$. Since $xs=ys$, we have $s^{n+1} = s^{k+1}$, so that $n=k$, i.e. $x=y$. + + Since $fs = gs$, also $(f,g)$ is a valid pair. Hence, there is some $t \in D$ with $f=tx$ and $g=ty$. Since $x=y$, we obtain the contradiction $f=g$. + + - property: sequential colimits + proof: >- + The proof is very similar to the one for $BC$. We will show that the sequence + $$* \xrightarrow{s} * \xrightarrow{s} * \xrightarrow{s} \cdots$$ + has no colimit. A cocone is a sequence of elements $(c_n)$ with $c_{n+1} s = c_n$ for all $n \geq 0$. Assume that the sequence has a colimit, i.e. a universal cocone $x=(x_n)$. Write $x_0 = s^a w$, where $w$ is a word in $f,g$. Choose some $m > a$ and define the sequence of elements $(c_n)$ by + $$c_n \coloneqq \begin{cases} + s^{m-n} & n \leq m \\ + f^{n-m} & n \geq m. + \end{cases}$$ + This is a cocone: for $n < m$ we have $c_{n+1} s = s^{m-n-1} s = s^{m-n} = c_n$, and for $n \geq m$ we have $c_{n+1} s = f^{n-m} f s = f^{n-m} = c_n$. By universality, there is some $h \in D$ such that + $$c_n = h x_n \tag{1}$$ + for all $n \geq 0$. Write $h = s^c v$, where $v$ is a word in $f,g$ of length $k$. For $n=0$ the cocone equality $(1)$ is: + $$s^m = c_0 = h x_0 = s^c v s^a w.$$ + If $k \geq a$, we write $v = v' u$ with a word $u$ of length $a$. Since every letter of $u$ is a left inverse of $s$, we have $u s^a = 1$, so that the right hand side is $s^c v' w$. By the uniqueness of the normal form, $c = m > 0$. If $k < a$, then $v s^a = v s^k s^{a-k} = s^{a-k}$, so that the right hand side is $s^{c+a-k} w$. By the uniqueness of the normal form, $c = m-a+k \geq m-a > 0$. In each case, we find $c > 0$. That is, $h = sh'$ for some $h' \in D$. + + Next, we evaluate the cocone equality $(1)$ at $n=m+1$. We get + $$f = c_{m+1} = h x_{m+1} = s (h' x_{m+1}).$$ + Since $f$ is an epimorphism, this implies that $s$ is also an epimorphism, which contradicts the classification of epimorphisms below. + references: + - BC_no_sequential_colimits + + - property: sequential limits + proof: >- + We will show that the sequence + $$\cdots \xrightarrow{f} * \xrightarrow{f} * \xrightarrow{f} *$$ + has no limit. A cone is a sequence of elements $(x_n)$ with $f x_{n+1} = x_n$ for all $n \geq 0$. Assume that the sequence has a limit, i.e. a universal cone $x = (x_n)$. + + We claim that every $x_m$ is an epimorphism. To see this, let $m \geq 0$ and define the sequence of elements $(y_n)$ by + $$y_n \coloneqq \begin{cases} + f^{m-n} & n \leq m \\ + s^{n-m} & n \geq m. + \end{cases}$$ + This is a cone: for $n < m$ we have $f y_{n+1} = f f^{m-n-1} = f^{m-n} = y_n$, and for $n \geq m$ we have $f y_{n+1} = f s s^{n-m} = s^{n-m} = y_n$. By universality, there is some $h \in D$ such that $y_n = x_n h$ for all $n \geq 0$. For $n = m$, this means $1 = x_m h$, so that $x_m$ is a split epimorphism. + + By the classification of epimorphisms below, every $x_n$ is therefore a word in $f,g$. By induction, we have $x_0 = f^n x_n$ for all $n \geq 0$, so that the word $x_0$ has length at least $n$ for all $n \geq 0$, which is impossible. + +special_objects: {} + +special_morphisms: + isomorphisms: + description: only the neutral element $1$ + proof: We have shown this in the proof that the category is core-thin. + monomorphisms: + description: the elements $s^n$ for $n \geq 0$ + proof: Since $fs = 1$, the element $s$ is a split monomorphism, and hence so is $s^n$. Conversely, let $x = s^n w$ be a monomorphism and assume that the word $w$ is non-empty. Then $w$ ends with $f$ or $g$, say $w = w' f$ (the case of $g$ is analogous). Since $x = s^n w' f$, the element $f$ is also a monomorphism. From $fs = 1$ we get $f (sf) = f = f \cdot 1$, hence $sf = 1$. But $sf \neq 1$ by the uniqueness of the normal form, which is a contradiction. + epimorphisms: + description: the words in $f,g$, i.e. the elements $s^0 w$ + proof: Since $fs = gs = 1$, the elements $f$ and $g$ are split epimorphisms, and hence so is every word in $f,g$. Conversely, let $x = s^n w$ be an epimorphism and assume that $n > 0$. Since $x = s (s^{n-1} w)$, the element $s$ is also an epimorphism. From $fs = 1$ we get $(sf) s = s = 1 \cdot s$, hence $sf = 1$, which is a contradiction as before. diff --git a/database/data/categories/BN.yaml b/database/data/categories/BN.yaml index 9dc60f687..a7c5c16c9 100644 --- a/database/data/categories/BN.yaml +++ b/database/data/categories/BN.yaml @@ -13,6 +13,9 @@ tags: related: - BG_c - BOn + - BC + - BD + - Set_c_inf satisfied_properties: - property: small @@ -36,52 +39,10 @@ satisfied_properties: - property: locally cartesian closed proof: The slice category $B\IN / *$ is isomorphic to the poset $(\IN,\geq)$ (not to $(\IN,\leq)$). This category is thin and semi-strongly connected, hence cartesian closed. - - property: ℵ₁-filtered colimits - check_redundancy: false - label: BN_aleph1-filtered_colimits - proof: >- - Let $D : \I \to B\IN$ be an $\aleph_1$-filtered diagram. Every two parallel morphisms $i \rightrightarrows j$ are mapped to the same morphism in $B \IN$, because they are coequalized by some morphism and $(\IN,+)$ is cancellative. Hence, $D$ factors through the preorder reflection of $\I$, and we may therefore assume that $\I$ itself is a preordered set. Thus, the diagram consists of numbers $D(i,j) \in \IN$ for all $i \leq j$ satisfying - $$D(j,k) + D(i,j) = D(i,k)$$ - for all $i \leq j \leq k$. In particular, $D(i,j) \leq D(i,k)$. - - Let $i \in \I$. The set of natural numbers $\{D(i,k) : k \geq i\}$ is bounded above. Otherwise, for every $n \in \IN$ we could find $k_n \in \I$ with $k_n \geq i$ and $D(i,k_n) \geq n$. Since $\I$ is $\aleph_1$-filtered, the family $(k_n)_{n \in \IN}$ has an upper bound $k_\infty \in \I$. But then - $$D(i, k_\infty) = D(k_n,k_\infty) + D(i, k_n) \geq D(i, k_n) \geq n$$ - for all $n \in \IN$, contradicting the fact that $D(i,k_\infty) \in \IN$. - - Therefore, the maximum - $$u_i \coloneqq \max \{D(i,k) : k \geq i\} \in \IN$$ - is well-defined, which we regard as a morphism in $B\IN$. For $i \leq j$ we compute - $$\begin{align*} - u_i & = \max \{D(i,k) : k \geq i\} \\ - & = \max \{D(i,k) : k \geq j\} \\ - & = \max \{ D(j,k) + D(i,j) : k \geq j\} \\ - & = \max \{D(j,k) : k \geq j\} + D(i,j)\\ - & = u_j + D(i,j), - \end{align*}$$ - showing that $(u_i)$ defines a cocone. It is universal: let $(v_i)$ be another cocone, i.e. $v_i \in \IN$ and $v_i = v_j + D(i,j)$ for all $i \leq j$. Then $v_i \geq D(i,j)$ for all $i \leq j$, hence $v_i \geq u_i$. Write $v_i = w_i + u_i$ for some uniquely determined $w_i \in \IN$. For $i \leq j$ we compute - $$w_j + u_j + D(i,j) = v_j + D(i,j) = v_i = w_i + u_i = w_i + u_j + D(i,j),$$ - hence $w_j = w_i$. Therefore, the $w_i$ are constant, and the required factorization follows. - - property: ℵ₁-accessible - references: - - BN_aleph1-filtered_colimits - proof: >- - Since we have just proven that $\aleph_1$-filtered colimits exist, it remains to show that the unique object $*$ is $\aleph_1$-presentable, i.e. that for every $\aleph_1$-filtered diagram $D : \I \to B\IN$, the canonical map - $$\alpha : \colim_{i \in \I} \Hom(*,D(i)) \to \Hom(*,\colim_{i \in \I} D(i))$$ - is bijective. On objects, we necessarily have $D(i)=*$ and $\colim_{i \in \I} D(i)=*$. Hence, the codomain of $\alpha$ is simply $\IN$, while the domain consists of equivalence classes $[i,n]$ of pairs $(i,n) \in \I \times \IN$, where $(i,n) \sim (j,m)$ iff there exists some $k \geq i,j$ such that - $$D(i,k) + n = D(j,k) + m.$$ - By the construction of the colimit cocone in the previous proof, we have - $$\alpha([i,n]) = u_i + n = \max \{D(i,j) : j \geq i\} + n.$$ - (1) The map $\alpha$ is surjective: Pick some $i \in \I$. Choose $j \geq i$ such that $u_i = D(i,j)$. For all $k \geq j$ we then have - $$u_i \geq D(i,k) = D(j,k) + D(i,j) = D(j,k) + u_i,$$ - hence $D(j,k)=0$. Therefore, $u_j=0$, and thus $\alpha([j,n]) = n$ for all $n \in \IN$. - - (2) The map $\alpha$ is injective: Assume that $[i,n]$ and $[j,m]$ have the same image. Since $\I$ is filtered, we may assume $i=j$. The condition then becomes $u_i + n = u_i + m$, and therefore $n=m$. This completes the proof. + proof: We apply Theorem 4 here. We already know that $B\IN$ is left cancellative and Cauchy complete. The principal right ideals of $\IN$ are the sets $n + \IN$ for $n \in \IN$. If $n + \IN \subseteq n' + \IN$, then $n \in n' + \IN$, i.e. $n \geq n'$. Hence, every ascending chain $n_0 + \IN \subseteq n_1 + \IN \subseteq \cdots$ satisfies $n_0 \geq n_1 \geq \cdots$ and therefore becomes stationary. unsatisfied_properties: - - property: one-way - proof: This is trivial. - - property: locally finite proof: This is trivial. diff --git a/database/data/categories/BOn.yaml b/database/data/categories/BOn.yaml index abb23e47d..44382f926 100644 --- a/database/data/categories/BOn.yaml +++ b/database/data/categories/BOn.yaml @@ -38,52 +38,9 @@ satisfied_properties: proof: In fact, it is $\kappa$-cofiltered for every cardinal $\kappa$. By the dual of Theorem 2.2 at the nLab it suffices to prove that any set of objects has a cone (which is trivial in a one-object category) and that any set of parallel morphisms is equalized by some morphism. Here, this means that for every set of ordinals $A$ there is some ordinal $\beta$ such that $\alpha + \beta$ for $\alpha \in A$ does not depend on $\alpha$. Take $\beta$ to be any ordinal larger than $\sup(A)$ of the form $\omega^\gamma$. It is well-known that $\omega^\gamma$ has the property that $\alpha + \omega^\gamma = \omega^\gamma$ for all $\alpha < \omega^\gamma$ (Kunen's Set Theory, Exercise I.9.53), from which the claim follows. - property: ℵ₁-filtered colimits - references: - - BN_aleph1-filtered_colimits - proof: >- - The proof is similar to $B\IN$. Let $\I$ be an $\aleph_1$-filtered small category and $D : \I \to B\OnColl$ a diagram. A cocone $\lambda = (\lambda_i)_{i \in \I}$ for $D$ is a family of ordinals satisfying $\lambda_i = \lambda_j + D(f)$ for every morphism $f: i \to j$ in $\I$. - - We first observe that $D$ factors uniquely through the preorder reflection of $\I$. Indeed, any two parallel morphisms in $\I$ are coequalized by some morphism, and $B\OnColl$ is left cancellative. Thus, we may assume that $\I$ is a preordered set. Each inequality $i \leq j$ in $\I$ is mapped to an ordinal number $\alpha_{i,j} \coloneqq D(i \to j)$, and these numbers satisfy - $$\alpha_{i,k} = \alpha_{j,k} + \alpha_{i,j}$$ - for all $i \leq j \leq k$. In particular, $\alpha_{i,j} \leq \alpha_{i,k}$. - - For fixed $i \in \I$, the collection $\{\alpha_{i,j} : j \geq i\}$ is a set of ordinals because $\I$ is small, hence bounded above in $\OnColl$. We claim that it has a maximum element. Otherwise, we can find a countable chain $i = j_0 \leq j_1 \leq j_2 \leq \dotsc$ in $\I$ such that $\alpha_{i,j_n} < \alpha_{i,j_{n+1}}$ for all $n \in \IN$. Since $\I$ is $\aleph_1$-filtered, there is an upper bound $j_\infty \in \I$ of $(j_n)_{n \in \IN}$. For each $n \in \IN$, the equation - $$\alpha_{i,j_{n+1}} = \alpha_{j_n,j_{n+1}} + \alpha_{i,j_n}$$ - implies that $\alpha_{j_n,j_{n+1}} > 0$. Hence, - $$\alpha_{j_n,j_\infty} = \alpha_{j_{n+1},j_\infty} + \alpha_{j_n,j_{n+1}} > \alpha_{j_{n+1},j_\infty},$$ - so $(\alpha_{j_n,j_\infty})_{n \in \IN}$ is a strictly decreasing infinite sequence of ordinals, contradicting the well-foundedness of $\OnColl$. Thus, the maximum - $$u_i \coloneqq \max \{ \alpha_{i,j} : j \geq i \}$$ - is a well-defined ordinal number, which we regard as a morphism in $B\OnColl$. The family $(u_i)_{i \in \I}$ forms a cocone for $D$, since for all $i \leq j$ we have - $$\begin{align*} - u_i & = \max \{ \alpha_{i,k} : k \geq i \} \\ - & = \max \{ \alpha_{i,k} : k \geq j \} \\ - & = \max \{ \alpha_{j,k} + \alpha_{i,j} : k \geq j \} \\ - & = \max \{ \alpha_{j,k} : k \geq j \} + \alpha_{i,j} \\ - & = u_j + \alpha_{i,j}. - \end{align*}$$ - To establish the universal property, let $(\lambda_i)_{i \in \I}$ be any cocone for $D$, so that $\lambda_i = \lambda_j + \alpha_{i,j}$ for all $i \leq j$. The cocone relation $u_i = u_j + \alpha_{i,j}$ implies that $u_i \geq u_j$ whenever $i \leq j$. By the well-foundedness of $\OnColl$, there exists $i_0 \in \I$ such that $u_j = u_{i_0}$ for all $j \geq i_0$. For such $j$, the relation - $$u_{i_0} = u_j + \alpha_{i_0,j} = u_{i_0} + \alpha_{i_0,j}$$ - forces $\alpha_{i_0,j} = 0$. Consequently, - $$u_{i_0} = \max \{ \alpha_{i_0,j} : j \geq i_0 \} = 0.$$ - Define the mediating morphism to be the ordinal $\kappa \coloneqq \lambda_{i_0}$. We must show that $\lambda_i = \kappa + u_i$ for all $i \in \I$. Choose $j \in \I$ with $j \geq i$ and $j \geq i_0$. Since $j \geq i_0$, we have $u_j = 0$ and $\alpha_{i_0,j} = 0$. The cocone condition for $\lambda$ gives - $$\kappa = \lambda_{i_0} = \lambda_j + \alpha_{i_0,j} = \lambda_j.$$ - Applying the cocone conditions for $u$ and $\lambda$ to $i \leq j$, we obtain - $$u_i = u_j + \alpha_{i,j} = 0 + \alpha_{i,j} = \alpha_{i,j}$$ - and - $$\lambda_i = \lambda_j + \alpha_{i,j} = \kappa + \alpha_{i,j} = \kappa + u_i.$$ - This proves the existence of the mediating morphism. - - For uniqueness, suppose $\kappa'$ is any ordinal satisfying $\lambda_i = \kappa' + u_i$ for all $i \in \I$. Evaluating at $i_0$ yields - $$\lambda_{i_0} = \kappa' + u_{i_0} = \kappa' + 0 = \kappa',$$ - hence $\kappa' = \kappa$. Therefore, the cocone $(u_i)_{i \in \I}$ is the colimit of $D$ in $B\OnColl$. + proof: We apply Proposition 6 here to the large monoid $(\OnColl,+,0)$. We already know that $B\OnColl$ is left cancellative. The principal right ideals of $\OnColl$ are the collections $\alpha + \OnColl$ for $\alpha \in \OnColl$. If $\alpha + \OnColl \subseteq \alpha' + \OnColl$, then $\alpha = \alpha' + \beta$ for some ordinal $\beta$, so that $\alpha \geq \alpha'$. Hence, every ascending chain $\alpha_0 + \OnColl \subseteq \alpha_1 + \OnColl \subseteq \cdots$ satisfies $\alpha_0 \geq \alpha_1 \geq \cdots$ and therefore becomes stationary, since $\OnColl$ is well-ordered. unsatisfied_properties: - - property: one-way - proof: This is trivial. - - - property: locally essentially small - proof: This is because $\OnColl$ is a large collection. - - property: balanced proof: Every finite ordinal is both a mono- and an epimorphism (see below), but only $0$ is an isomorphism. diff --git a/database/data/categories/Set.yaml b/database/data/categories/Set.yaml index 775e66976..2cc18bd9f 100644 --- a/database/data/categories/Set.yaml +++ b/database/data/categories/Set.yaml @@ -14,6 +14,7 @@ related: - FinSet - Set_* - Set_c + - Set_c_inf - Set_bij - Set_ff - SetxSet diff --git a/database/data/categories/Set_c.yaml b/database/data/categories/Set_c.yaml index 8ad8fea07..343ec7801 100644 --- a/database/data/categories/Set_c.yaml +++ b/database/data/categories/Set_c.yaml @@ -11,6 +11,7 @@ tags: related: - FinSet + - Set_c_inf - Set - Grp_c - Vect_c diff --git a/database/data/categories/Set_c_inf.yaml b/database/data/categories/Set_c_inf.yaml new file mode 100644 index 000000000..ae6aae43c --- /dev/null +++ b/database/data/categories/Set_c_inf.yaml @@ -0,0 +1,86 @@ +id: Set_c_inf +name: category of countably infinite sets +notation: $\Set_{\c,\infty}$ +objects: countably infinite sets +morphisms: maps +description: This is the full subcategory of the category of countable sets consisting of the infinite ones. Since every object is isomorphic to $\IN$, it is equivalent to the one-object category $BM$ for the monoid $M$ of all maps $\IN \to \IN$. This entry provides an example of a non-trivial core-connected category with binary products (and binary coproducts). +nlab_link: null + +tags: + - set theory + +related: + - Set_c + - Set + - Setne + - BN + +satisfied_properties: + - property: locally small + proof: It is a full subcategory of $\Set$, which is locally small. + + - property: core-connected + proof: Every object is isomorphic to $\IN$. + + - property: binary products + proof: If $X,Y$ are countably infinite sets, then so is $X \times Y$. Hence, the product of $X,Y$ in $\Set$ is also a product in the full subcategory $\Set_{\c,\infty}$. + + - property: binary coproducts + proof: If $X,Y$ are countably infinite sets, then so is $X \sqcup Y$. Hence, the coproduct of $X,Y$ in $\Set$ is also a coproduct in the full subcategory $\Set_{\c,\infty}$. + + - property: kernel pairs + proof: 'Let $f : X \to Y$ be a map of countably infinite sets. Its kernel pair in $\Set$ is $X \times_Y X = \{(x,x'') \in X \times X : f(x) = f(x'')\}$. It is countable as a subset of $X \times X$, and infinite since it contains the diagonal. Hence, it is also a kernel pair in the full subcategory $\Set_{\c,\infty}$.' + + - property: cokernel pairs + proof: 'Let $f : X \to Y$ be a map of countably infinite sets. Its cokernel pair in $\Set$ is $Y \sqcup_X Y$. It is countable as a quotient of $Y \sqcup Y$, and infinite since the coprojections $Y \rightrightarrows Y \sqcup_X Y$ are injective, having the codiagonal $Y \sqcup_X Y \to Y$ as a left inverse. Hence, it is also a cokernel pair in the full subcategory $\Set_{\c,\infty}$.' + + - property: effective congruences + proof: 'Let $q_1,q_2 : E \rightrightarrows X$ be a congruence. Since $q_1,q_2$ are jointly monomorphic, the map $(q_1,q_2) : E \to X \times X$ is a monomorphism, hence injective (see below). Using constant maps $\IN \to X$, we see that its image $R \subseteq X \times X$ is an equivalence relation on $X$. Let $h : X \to (X/R) \sqcup \IN$ be the composite of the projection $X \to X/R$ with the coproduct inclusion (we add $\IN$ since $X/R$ may be finite). Its kernel pair in $\Set$ is $R \cong E$, hence also in the full subcategory $\Set_{\c,\infty}$.' + + - property: ℵ₁-filtered + proof: 'Every diagram $(X_i)_{i \in I}$ in $\Set_{\c,\infty}$ admits a cocone with vertex $\IN$, given by the constant maps $X_i \to \IN$ with value $0$.' + + - property: mono-regular + proof: 'Let $f : X \to Y$ be a monomorphism. It is injective (see below), hence a split monomorphism in $\Setne$. Since $\Set_{\c,\infty}$ is a full subcategory of $\Setne$, $f$ is a split monomorphism in $\Set_{\c,\infty}$ and therefore a regular monomorphism.' + + - property: epi-regular + proof: 'Let $f : X \to Y$ be an epimorphism. It is surjective (see below), hence has a right inverse $s : Y \to X$ (as in $\Set$). Thus, $f$ is a split epimorphism and therefore a regular epimorphism.' + +unsatisfied_properties: + - property: skeletal + proof: The sets $\IN$ and $\IZ$ are isomorphic, but different. + + - property: small + proof: Even the collection of all sets $\IN \times \{x\}$, where $x$ is a set, is not a set. + + - property: essentially countable + proof: The hom-set $\Hom(\IN,\IN)$ is uncountable. + + - property: Cauchy complete + proof: 'The constant map $e : \IN \to \IN$, $n \mapsto 0$ is idempotent and does not split in $\Set_{\c,\infty}$, since a splitting $\IN \to X \to \IN$ would also be a splitting in $\Set$, making $X$ isomorphic to $\im(e) = \{0\}$, which is finite.' + + - property: zero morphisms + proof: 'Assume that there is a zero morphism $0 : \IN \to \IN$. For $n \in \IN$, let $c_n : \IN \to \IN$ be the constant map with value $n$. Then $c_0 = c_0 \circ 0 = 0 = c_1 \circ 0 = c_1$, a contradiction.' + + - property: cofiltered + proof: 'The constant maps $\IN \rightrightarrows \IN$ with values $0$ and $1$ are equalized by no map $X \to \IN$ in this category.' + + - property: effective cocongruences + proof: 'The coproduct inclusions $i_1,i_2 : \IN \rightrightarrows \IN \sqcup \IN$ form a cocongruence. It is not effective, since there is no map $h : Y \to \IN$ in this category with $i_1 \circ h = i_2 \circ h$.' + +special_objects: + coproducts: + description: '[non-empty, countable case] disjoint union' + products: + description: '[non-empty, finite case] direct products' + +special_morphisms: + isomorphisms: + description: bijective maps + proof: It is a full subcategory of $\Set$, for which we know that isomorphisms are bijective maps. + monomorphisms: + description: injective maps + proof: 'For the non-trivial direction, let $f : X \to Y$ be a monomorphism and $x,x'' \in X$ with $f(x) = f(x'')$. Let $c_x, c_{x''} : \IN \rightrightarrows X$ be the constant maps with values $x, x''$. Then $f \circ c_x = f \circ c_{x''}$, hence $c_x = c_{x''}$, i.e. $x = x''$.' + epimorphisms: + description: surjective maps + proof: 'For the non-trivial direction, let $f : X \to Y$ be an epimorphism and assume that some $y \in Y$ is not in the image of $f$. Since $Y$ is infinite, there is some $y'' \in Y$ with $y'' \neq y$. Let $g : Y \to Y$ be the map with $g(y) \coloneqq y''$ and $g(z) \coloneqq z$ for $z \neq y$. Then $g \circ f = \id_Y \circ f$, hence $g = \id_Y$, which contradicts $g(y) = y'' \neq y$.' diff --git a/database/data/categories/Setne.yaml b/database/data/categories/Setne.yaml index db897c001..28ef0f5d8 100644 --- a/database/data/categories/Setne.yaml +++ b/database/data/categories/Setne.yaml @@ -13,6 +13,7 @@ tags: related: - Set + - Set_c_inf satisfied_properties: - property: locally small diff --git a/database/data/category-implications/congruences.yaml b/database/data/category-implications/congruences.yaml index 59aad90ac..c84a3a47a 100644 --- a/database/data/category-implications/congruences.yaml +++ b/database/data/category-implications/congruences.yaml @@ -50,6 +50,19 @@ # TODO: rename congruences consistently to q_1, q_2 (and cocongruences to j_1, j_2) proof: 'If $p_1, p_2 : E \rightrightarrows X$ is a congruence, the symmetry morphism $s : E \to E$ is an automorphism of $E$, hence equal to $\id_E$ by assumption. But then $p_1 = p_2 \circ s = p_2$, and simply $\id_X$ is a coequalizer. Also, for the reflexivity morphism $r : X \to E$, we have $p_1 \circ r = \id$. For the reverse composition, $p_1 \circ r \circ p_1 = p_1 \circ \id$ and $p_2 \circ r \circ p_1 = p_2 \circ \id$, so since $p_1, p_2$ are jointly monomorphic, we get $r \circ p_1 = \id$. Therefore, $p_1 = p_2$ is an isomorphism, so $E$ is the kernel pair of $\id_X$.' +- id: core-connected_locally_finite_congruences + assumptions: + - core-connected + - locally finite + conclusions: + - effective congruences + - quotients of congruences + # TODO: refactor this once we add "split-mono-trivial", cf. #383 + proof: >- + Let $q_1, q_2 : E \rightrightarrows X$ be a congruence. Since the category is core-connected, we may assume $E = X$. The reflexivity morphism $r : X \to X$ satisfies $q_1 \circ r = q_2 \circ r = \id_X$. Since the monoid $\End(X)$ is finite, every right inverse in it is a two-sided inverse (see the remark below), so that $q_1 = r^{-1} = q_2$. It follows that $\id_X$ is a coequalizer of $q_1, q_2$, and that $q_1, q_2$ is a kernel pair of $\id_X$, since $\id_X, \id_X$ is one. + + Remark: In a finite monoid $M$, every right inverse is a two-sided inverse. Indeed, let $x,y \in M$ with $xy = 1$. The map $M \to M$, $z \mapsto yz$ is injective, since $x(yz) = (xy)z = z$. Since $M$ is finite, it is also surjective, so there is some $z \in M$ with $yz = 1$. Then $x = x(yz) = (xy)z = z$, and hence $yx = yz = 1$. + - id: preadditive_kernels_normal_imply_effective_congruences assumptions: - kernels diff --git a/database/data/category-implications/connected.yaml b/database/data/category-implications/connected.yaml index dde55919a..3bac9af2d 100644 --- a/database/data/category-implications/connected.yaml +++ b/database/data/category-implications/connected.yaml @@ -81,3 +81,33 @@ conclusions: - trivial proof: Every object is isomorphic to the initial object. + +- id: core-connected_one-way_trivial + assumptions: + - core-connected + - one-way + conclusions: + - trivial + proof: Every core-connected category is equivalent to $BM$ for a (possibly large) monoid $M$. If it is one-way, $\End(*)=M$ must be trivial, in which case $BM$ is trivial. + +- id: core-connected_essentially_small + assumptions: + - core-connected + - locally essentially small + conclusions: + - essentially small + proof: There is only one isomorphism class of objects. Now use the characterization of essentially small categories as the locally essentially small categories whose collection of isomorphism classes of objects is essentially small. + +- id: core-connected_products_congruences_trivial + assumptions: + - core-connected + - binary products + - quotients of congruences + conclusions: + - trivial + proof: >- + We may assume that the category is $BM$ for a (possibly large) monoid $M$. In any category with binary products, the projections $p_1, p_2 : X \times X \rightrightarrows X$ form a congruence on $X$. Since $BM$ has only one object, we have $* \times * = *$, so that $p_1, p_2 \in M$, and for all $a,b \in M$ there is some $m \in M$ with $p_1 m = a$ and $p_2 m = b$. + + Let $c \in M$ be a quotient of $p_1, p_2$. This means that $c p_1 = c p_2$, and that every $d \in M$ with $d p_1 = d p_2$ can be written as $d = u c$ for a unique $u \in M$. For all $a,b \in M$ we choose $m \in M$ with $p_1 m = a$ and $p_2 m = b$, and get + $$c a = c p_1 m = c p_2 m = c b.$$ + For $b = 1$ this means that $c a = c$ for all $a \in M$. In particular, $c c = c = 1 c$. Since $c$ is a coequalizer, it is an epimorphism, so that $c = 1$. But then $a = c a = c = 1$ for all $a \in M$, so $M$ is trivial. diff --git a/shared/history.json b/shared/history.json index 16beb3037..140b172ab 100644 --- a/shared/history.json +++ b/shared/history.json @@ -871,6 +871,18 @@ { "type": "category", "id": "omega-Set" + }, + { + "type": "category", + "id": "BC" + }, + { + "type": "category", + "id": "BD" + }, + { + "type": "category", + "id": "Set_c_inf" } ], "properties": [