Add the category of semigroups

.vscode/settings.json

@@ -88,6 +88,7 @@
 		"coreflexivity",
 		"coregular",
 		"corelations",
+		"corestrict",
 		"corestriction",
 		"corestricts",
 		"cosifted",
@@ -201,6 +202,8 @@
 		"Rosicky",
 		"saft",
 		"Schapira",
+		"semigroup",
+		"semigroups",
 		"semisimple",
 		"setoid",
 		"Sheafifiable",
@@ -214,6 +217,7 @@
 		"subposet",
 		"subproset",
 		"subscheme",
+		"subsemigroup",
 		"subsheaf",
 		"summands",
 		"suprema",

databases/catdat/data/001_categories/002_algebra.sql

@@ -26,6 +26,15 @@ VALUES
 	'This is the prototype of a finitary algebraic category.',
 	'https://ncatlab.org/nlab/show/Grp'
 ),
+(
+	'SemiGrp',
+	'category of semigroups',
+	'$\SemiGrp$',
+	'semigroups, i.e. sets equipped with an associative binary operation',
+	'maps preserving the binary operation',
+	'In contrast to monoids, semigroups do not need to have a neutral element, and in fact, they can be empty. This small difference has a huge impact on the categorical properties. For example, we do not have a zero object anymore.',
+	'https://ncatlab.org/nlab/show/semigroup'
+),
 (
 	'Vect',
 	'category of vector spaces',

databases/catdat/data/001_categories/100_related-categories.sql

@@ -61,6 +61,7 @@ VALUES
 ('Grp', 'FinGrp'),
 ('Grp', 'Ab'),
 ('Grp', 'Mon'),
+('Grp', 'SemiGrp'),
 ('FinGrp', 'Grp'),
 ('FinGrp', 'FinAb'),
 ('Haus', 'Top'),
@@ -86,6 +87,7 @@ VALUES
 ('Mon', 'CMon'),
 ('Mon', 'Grp'),
 ('Mon', 'Cat'),
+('Mon', 'SemiGrp'),
 ('N', 'N_oo'),
 ('N', 'On'),
 ('N', 'Z_div'),
@@ -110,6 +112,8 @@ VALUES
 ('Rng', 'Ring'),
 ('Sch', 'LRS'),
 ('Sch', 'Z'),
+('SemiGrp', 'Mon'),
+('SemiGrp', 'Grp'),
 ('Set_c', 'Set'),
 ('Set_c', 'FinSet'),
 ('Set_f', 'Set'),

databases/catdat/data/001_categories/200_category-tags.sql

@@ -74,6 +74,7 @@ VALUES
 ('Ring', 'algebra'),
 ('Rng', 'algebra'),
 ('Sch', 'algebraic geometry'),
+('SemiGrp', 'algebra'),
 ('Set_c', 'set theory'),
 ('Set_f', 'set theory'),
 ('Set', 'set theory'),

databases/catdat/data/003_category-property-assignments/SemiGrp.sql

@@ -0,0 +1,134 @@
+INSERT INTO category_property_assignments (
+	category_id,
+	property_id,
+	is_satisfied,
+	reason
+)
+VALUES
+(
+	'SemiGrp',
+	'locally small',
+	TRUE,
+	'There is a forgetful functor $\SemiGrp \to \Set$ and $\Set$ is locally small.'
+),
+(
+	'SemiGrp',
+	'finitary algebraic',
+	TRUE,
+	'Take the algebraic theory of a semigroup.'
+),
+(
+    'SemiGrp',
+    'strict initial object',
+    TRUE,
+    'This is because the initial object is the empty semigroup, and a non-empty set has no map to an empty set.'
+),
+(
+    'SemiGrp',
+    'disjoint coproducts',
+    TRUE,
+    'This follows easily from the concrete description of coproducts as (a variant of) free products.'
+),
+(
+	'SemiGrp',
+	'cocartesian cofiltered limits',
+	TRUE,
+	'We need to prove that for two cofiltered diagram of semigroups $(B_i)$, $(C_i)$ the canonical map
+	$$\textstyle \alpha : \lim_i B_i + \lim_i C_i \to \lim_i (B_i + C_i)$$
+	is an isomorphism, i.e. bijective. The underlying set of the coproduct $B_i + C_i$ is identified with the disjoint union of sets of the form
+	$$B_i \times C_i \times B_i \times \cdots \times C_i$$
+	with any positive product length, starting and ending either with $B_i$ or $C_i$. Moreover, $\lim_i$ commutes with arbitrary coproducts in $\Set$, and of course also with products. This shows that $\alpha$ is bijective.'
+),
+(
+	'SemiGrp',
+	'skeletal',
+	FALSE,
+	'This is trivial.'
+),
+(
+	'SemiGrp',
+	'balanced',
+	FALSE,
+	'The inclusion of additive semigroups $\IN \hookrightarrow \IZ$ is a counterexample. Indeed, if $f,g : \IZ \to G$ are semigroup homomorphisms which agree on $\IN$, the element $e := f(0) = g(0)$ provides a neutral element for $eGe$, which therefore becomes a monoid, and $f,g$ corestrict to monoid homomorphisms $f,g : \IZ \to eGe$ which agree on $\IN$. And we already know that $f = g$ in that case (inverses are uniquely determined).<br>
+	Another example can be found in <a href="https://mathoverflow.net/a/510442/2841">MO/510431</a>.'
+),
+(
+	'SemiGrp',
+	'Malcev',
+	FALSE,
+	'Consider the subsemigroup $\{(a,b) : a \leq b \}$ of $\IN^2$ under addition.'
+),
+(
+	'SemiGrp',
+	'co-Malcev',
+	FALSE,
+	'See <a href="https://mathoverflow.net/questions/509552">MO/509552</a>: Consider the forgetful functor $U : \SemiGrp \to \Set$ and the relation $R \subseteq U^2$ defined by $R(A) := \{(a,b) \in U(A)^2 : ab = a^2\}$. Both are representable: $U$ by the free semigroup on a single generator and $R$ by the free semigroup on two generators $x,y$ subject to the relation $xy=x^2$. It is clear that $R$ is reflexive, but not symmetric.'
+),
+(
+    'SemiGrp',
+    'semi-strongly connected',
+    FALSE,
+    'Let us first remark that every non-empty finite semigroup $A$ has an idempotent element $e$, and then $B \to A$, $x \mapsto e$ does define a semigroup homomorphism for any $B$. Therefore, counterexamples need to be infinite and also without idempotent elements.<br>
+    Let $A$ be the set of positive rational numbers of the form $m/2^n$ (with $m > 0$, $n \geq 0$), and let $B$ be the set of positive rational numbers of the form $m/3^n$ (with $m > 0$, $n \geq 0$). Both are semigroups under addition. The element $1 \in A$ is $2^\infty$-divisible, meaning that for every $n \geq 0$ there is some $a \in A$ with $1 = 2^n \cdot a$. But $B$ has no $2^\infty$-divisible element. Hence, there is no semigroup homomorphism $A \to B$. Likewise, there is no semigroup homomorphism $B \to A$.'
+),
+(
+	-- TODO: find a variant of the lemma missing_cogenerating_sets (or missing_cogenerator) which handles this.
+	'SemiGrp',
+	'cogenerating set',
+	FALSE,
+	'The proof is similar to the proof for <a href="/category/Grp">$\Grp$</a>. Assume that there is a cogenerating set $S$. There is an infinite simple group $G$ larger than all the semigroups in $S$ (such as an alternating group). Since $\id_G, 1 : G \rightrightarrows G$ are different, there is a semigroup $H \in S$ and a homomorphism of semigroups $f : G \to H$ with $f \neq f \circ 1$. Then
+	$$N := \{g \in G : f(g) = f(1)\}$$
+	is a normal subgroup of $G$. It is proper, and hence trivial. But then $f$ is injective, which is a contradiction.'
+),
+(
+	'SemiGrp',
+	'cofiltered-limit-stable epimorphisms',
+	FALSE,
+	'We already know that <a href="/category/Set">$\Set$</a> does not have this property (by <a href="/category-implication/topos_no_stable_epis">this result</a>). Now apply the contrapositive of the dual of <a href="/lemma/filtered-monos">this lemma</a> to the functor $\Set \to \SemiGrp$ that equips a set with the multiplication $a \cdot b := a$.'
+),
+(
+	'SemiGrp',
+	'effective cocongruences',
+	FALSE,
+	'The proof is similar to <a href="/category/Mon">$\Mon$</a>, i.e. we adapt the counterexample from <a href="https://mathoverflow.net/a/510809" target="_blank">MO/510744</a>. Namely, consider the semigroups
+	$$\begin{align*} X & := \langle p \mid p^2 = p \rangle,\\
+	E & := \langle p, q \mid p^2 = p,\, q^2 = q,\, pq = q,\, qp = p \rangle,
+	\end{align*}$$
+	whose underlying sets are $\{p\}$ and $\{p,q\}$, respectively. Then $X$ represents the functor sending a semigroup $A$ to its idempotents, and $E$ represents the relation on idempotents $a, b$ of $A$ that $ab = b$, $ba = a$. It is easy to check that this defines an equivalence relation (see <a href="https://mathoverflow.net/a/510809" target="_blank">MO/510744</a> for details). Since $p \ne q$ in $E$, the equalizer of the two maps $X \rightrightarrows E$ is the empty semigroup. Therefore, if $E$ were effective, it would be isomorphic to the coproduct $X \sqcup X$, whose underlying set consists of non-empty words in $p,q$ with $p,q$ strictly alternating. In particular, in this coproduct, $pq \ne q$.'
+),
+(
+	'SemiGrp',
+	'natural numbers object',
+	FALSE,
+	'Assume that a natural numbers object exists. Then by <a href="/lemma/nno_distributive_criterion">this result</a>, for every semigroup $A$ the natural homomorphism
+	$$\textstyle\alpha : \coprod_{n \geq 0} A \to A \times \coprod_{n \geq 0} 1$$
+	is a split monomorphism. But this is not true: For each $n \geq 0$ let $A_n$ denote a copy of $A$. The elements of the coproduct in $\alpha$''s domain have a unique representation as
+	$$x_{n_1} * \cdots * x_{n_s}$$
+	with $x_i \in A_i$ and $n_i \neq n_{i+1}$, and we have
+	$$\alpha(x_{n_1} * \cdots * x_{n_s}) = (x_{n_1} \cdots x_{n_s}, n_1 * \cdots * n_s).$$
+	In particular, if $A$ has two non-equal commuting elements $x,y$ (for example, if $A$ is any non-trivial monoid), we have
+	$$\alpha(y_0 x_0) = \alpha(x_0 y_0),$$
+	showing that $\alpha$ is not injective.'
+),
+(
+	'SemiGrp',
+	'coregular',
+	FALSE,
+	'We will find a regular monomorphism $\iota : F \to M$ of semigroups and a homomorphism $F \to K$ such that $K \to K \sqcup_F M$ is not injective. It is similar to our example for <a href="/category/Mon">$\Mon$</a>. Consider these semigroups defined by generators and relations:
+	$$\begin{align*}
+	F & := \langle a,b,c,d \rangle \\
+	M & := \langle a,b,c,d,s : as = c, \, bs = d \rangle \cong \langle a,b,s \rangle \\
+	K & := \langle x,c,d \rangle \\
+	N & := \langle a,b,c,d,s_0,s_1 : as_i = c, \, bs_i = d \rangle \\
+	& \cong \langle a,b,s_0,s_1 : a s_0 = a s_1, \, b s_0 = b s_1 \rangle \\
+	\end{align*}$$
+	There is a canonical homomorphism $\iota : F \to M$, which is the equalizer of the two canonical homomorphisms $M \rightrightarrows N$ defined by $s \mapsto s_i$. We define $F \to K$ by $a \mapsto x$, $b \mapsto x$, $c \mapsto c$, $d \mapsto d$. Then
+	$$K \sqcup_F M \cong \langle x,c,d,s : x s = c,\, x s = d \rangle$$
+	shows that $c,d \in K$ have the same image in the pushout.'
+),
+(
+	'SemiGrp',
+	'regular subobject classifier',
+	FALSE,
+	'Assume that a regular subobject classifier $\Omega$ exists in $\SemiGrp$. The universal regular monomorphism $\top : 1 \to \Omega$ corresponds to an idempotent element $e \in \Omega$. It follows that $e \Omega e$ is a monoid with neutral element $e$. We claim that it is a regular subobject classifier in <a href="/category/Mon">$\Mon$</a>, which we know does not exist. Indeed, let $\iota : A \to B$ be a regular monomorphism of monoids. Since the forgetful functor $\Mon \to \SemiGrp$ preserves limits, we can also see $\iota$ as a regular monomorphism of semigroups. Hence, there is a unique homomorphism of semigroups $f : B \to \Omega$ with $\iota(A) = \{b \in B : f(b) = e\}$. Since $1 \in \iota(A)$, we have $f(1) = e$. Then $f$ corresponds to a homomorphism of monoids $f : B \to e \Omega e$ with kernel $\iota$, which proves our claim.'
+);

databases/catdat/data/005_special-objects/002_initial_objects.sql

@@ -44,6 +44,7 @@ VALUES
 ('Ring', 'ring of integers'),
 ('Rng', 'trivial ring'),
 ('Sch', 'empty scheme'),
+('SemiGrp', 'empty semigroup'),
 ('Set_c', 'empty set'),
 ('Set_f', 'empty set'),
 ('Set', 'empty set'),

databases/catdat/data/005_special-objects/003_terminal_objects.sql

@@ -41,6 +41,7 @@ VALUES
 ('Ring', 'zero ring'),
 ('Rng', 'zero ring'),
 ('Sch', '$\Spec(\IZ)$'),
+('SemiGrp', 'trivial semigroup'),
 ('Set_c', 'singleton set'),
 ('Set', 'singleton set'),
 ('Set*', 'singleton pointed set'),

databases/catdat/data/005_special-objects/004_coproducts.sql

@@ -37,6 +37,7 @@ VALUES
 ('Ring', 'see <a href="https://math.stackexchange.com/questions/625874" target="_blank">MSE/625874</a>'),
 ('Rng', 'see <a href="https://math.stackexchange.com/questions/4975797" target="_blank">MSE/4975797</a>'),
 ('Sch', 'disjoint union with the product sheaf'),
+('SemiGrp', 'similar to free products of monoids, but where the empty word is not allowed, and of course we do (and can) not demand that the factors are $\neq 1$'),
 ('Set', 'disjoint union'),
 ('Set*', 'wedge sum, aka one-point union'),
 ('SetxSet', 'component-wise disjoint union'),

databases/catdat/data/005_special-objects/005_products.sql

@@ -33,6 +33,7 @@ VALUES
 ('Rel', 'disjoint unions (!)'),
 ('Ring', 'direct products with pointwise operations'),
 ('Rng', 'direct products with pointwise operations'),
+('SemiGrp', 'direct products with pointwise operations'),
 ('Set', 'direct products with pointwise operations'),
 ('Set*', 'direct products with pointwise operations'),
 ('Setne', 'direct products'),

databases/catdat/data/006_special-morphisms/002_isomorphisms.sql

@@ -255,6 +255,11 @@ VALUES
 	'pairs $(f,f^{\sharp})$ consisting of a homeomorphism $f$ and an isomorphism of sheaves $f^{\sharp}$',
 	'This is easy.'
 ),
+(
+	'SemiGrp',
+	'bijective ring homomorphisms',
+	'This characterization holds in every algebraic category.'
+),
 (
 	'Set_c',
 	'bijective maps',

databases/catdat/data/006_special-morphisms/003_monomorphisms.sql

@@ -245,6 +245,11 @@ VALUES
 	'injective rng homomorphisms',
 	'This holds in every finitary algebraic category: the forgetful functor to $\Set$ is faithful, hence reflects monomorphisms, and it is continuous (even representable), hence preserves monomorphisms.'
 ),
+(
+	'SemiGrp',
+	'injective semigroup homomorphisms',
+	'This holds in every finitary algebraic category: the forgetful functor to $\Set$ is faithful, hence reflects monomorphisms, and it is continuous (even representable), hence preserves monomorphisms.'
+),
 (
 	'Set_c',
 	'injective maps',

databases/catdat/data/006_special-morphisms/004_epimorphisms.sql

@@ -231,6 +231,11 @@ VALUES
 	'A relation $R : A \to B$ is an epimorphism iff the map $R^* : P(B) \to P(A)$ defined by $S \mapsto \{a \in A : \exists \, b \in S: (a,b) \in R \}$ is injective.',
 	'See <a href="https://math.stackexchange.com/questions/350716/" target="_blank">MSE/350716</a>.'
 ),
+(
+	'SemiGrp',
+	'A semigroup homomorphism $f : T \to S$ is an epimorphism iff $S$ equals the <i>dominion</i> of $U := f(T) \subseteq S$, meaning that for every $s \in S$ there are $u_1,\dotsc,u_{m+1} \in U$, $v_1,\dotsc,v_m \in U$, $x_1,\dotsc,x_m \in S$ and $y_1,\dotsc,y_m \in S$ such that $s = x_1 u_1$, $u_1 = v_1 y_1$, $x_{i-1} v_{i-1} = x_i u_i$, $u_i y_{i-1} = v_i y_i$, $x_m v_m = u_{m+1}$ and $u_{m+1} y_m = s$.',
+	'This is <a href="https://en.wikipedia.org/wiki/Isbell''s_zigzag_theorem" target="_blank">Isbell''s zigzag Theorem</a>, see references there.'
+),
 (
 	'Set_c',
 	'surjective maps',

databases/catdat/data/006_special-morphisms/006_regular-epimorphisms.sql

@@ -180,6 +180,11 @@ VALUES
     'surjective homomorphisms',
     'This holds in every finitary algebraic category.'
 ), 
+(
+    'SemiGrp',
+    'surjective homomorphisms',
+    'This holds in every finitary algebraic category.'
+),  
 (
     'Set_c',
     'same as epimorphisms',

databases/catdat/data/009_lemmas/001_lemmas.sql

@@ -160,5 +160,28 @@ INSERT INTO lemmas (
     Now if this congruence is the kernel pair of $h : A+X \to Z$ in $A \backslash \C$, then $E$ is the kernel pair of $h \circ i_2 : X \to Z$ in $\C$. Namely, if we have two generalized elements $x_1, x_2 : T \rightrightarrows X$ such that $h \circ i_2 \circ x_1 = h \circ i_2 \circ x_2$, then we can construct a map pair
     $$\id_A + x_1,\, \id_A + x_2 : A+T \rightrightarrows A+X$$
     in $A \backslash \C$ with $h \circ (\id_A + x_1) = h \circ (\id_A + x_2)$. Therefore, $\id_A + x_1, \id_A + x_2$ factors through $A+E$ in $A \backslash \C$, so $x_1, x_2$ factors through $A+E$ in $\C$; and using disjoint coproducts, we may conclude $x_1, x_2$ factors through $E$.'
+),
+(
+    'nno_distributive_criterion',
+    'Natural number objects indicate distributivity',
+    'Let $\C$ be a category with finite products, arbitrary copowers (denoted $\otimes$), and a natural numbers object $1 \xrightarrow{z} N \xrightarrow{s} N$. Then there is an isomorphism $N \cong \IN \otimes 1$, and for every object $A$ the natural morphism
+    $$\alpha : \IN \otimes A \to A \times (\IN \otimes 1)$$
+    is a split monomorphism.',
+    'We will use generalized elements extensively. In particular, for every element $a \in A$ and $n \in \IN$ there is an element $n \otimes a \in \IN \otimes A$, formally defined by the $n$th coproduct inclusion. The morphism $\alpha$ is defined by
+    $$\alpha(n \otimes a) = (a , n \otimes 1).$$
+    In any category with finite products and arbitrary copowers, we can construct the non-parameterized NNO $\IN \otimes 1$ with the element $0 \otimes 1 \in \IN \otimes 1$ and the map
+    $$s : \IN \otimes 1 \to \IN \otimes 1, \quad s(n \otimes 1) := (n+1) \otimes 1.$$
+    Its universal property states that it is initial in the category of pairs $1 \xrightarrow{x_0} X \xrightarrow{r} X$. Hence, it is unique up to isomorphism. Since by assumption $1 \xrightarrow{z} N \xrightarrow{s} N$ is a full, parameterized NNO, it is also a non-parameterized NNO and therefore isomorphic to the one described. We will assume w.l.o.g. that it is equal to it and continue to work with $N = \IN \otimes 1$.<br><br>
+    Next, we apply the (parameterized) universal property of the NNO to the diagram
+    $$A \xrightarrow{f} \IN \otimes A \xrightarrow{g} \IN \otimes A$$
+    defined by $f(a) := 0 \otimes a$ and $g(n \otimes a) := (n+1) \otimes a$. It tells us that there is a map
+    $$\Phi : A \times N \to \IN \otimes A$$
+    with
+    $$\Phi(a,0 \otimes 1) = 0 \otimes a, \quad \Phi(a, s(m)) = g(\Phi(a,m)).$$
+    For $m := n \otimes 1 \in N$ (where $n \in \IN$) the second equation reads
+    $$\Phi(a, (n+1) \otimes 1) = g(\Phi(a, n \otimes 1)).$$
+    By classical induction on $n \in \IN$ it follows that
+    $$\Phi(a, n \otimes 1) = n \otimes a,$$
+    which exactly means $\Phi \circ \alpha = \id_{\IN \otimes A}$.'
 );
 

src/lib/server/macros.ts

@@ -97,4 +97,5 @@ export const MACROS = {
 	'\\RS': '\\mathbf{RS}',
 	'\\CoAlg': '\\mathbf{CoAlg}',
 	'\\Cone': '\\mathbf{Cone}',
+	'\\SemiGrp': '\\mathbf{SemiGrp}',
 }