The monads of classical algebra are seldom weakly cartesian
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Abstract
This paper begins a systematic study of weakly cartesian properties of monads that determine familiar varieties of universal algebras. While these properties clearly fail to hold for groups, rings, and many other related classical algebraic structures, their analysis becomes nontrivial in the case of semimodules over semirings, to which our main results are devoted. In particular necessary and sufficient conditions on a semiring S, under which the free semimodule monad has: (a) its underlying functor weakly cartesian, (b) its unit a weakly cartesian natural transformation, (c) its multiplication a weakly cartesian natural transformation, are obtained.
Keywords
Monad Cartesian Weakly cartesian Weak pullback Variety of algebras Semiring Semimodule SubtractiveMathematics Subject Classification (1991)
18C15 18C20 08A62 16Y601 Introduction
There are several areas of applied category theory, from Grothendieck descent theory to the study of higherdimensional categorical structures, which use socalled cartesian monads and monads satisfying weaker conditions, where the relevant pullbacks are replaced with weak pullbacks. In particular, previous work of the first two authors [6, 7] extensively used the fact that the ultrafilter monad on the category Set of sets preserves weak pullbacks and its multiplication is weakly cartesian (that is, satisfying (BC) in the terminology of [4, 5]).

the functor \(T\) preserves weak pullbacks;

the natural transformations \(\eta \) and \(\mu \) are weakly cartesian.

Observation 3.1, listing some known varieties determined by cartesian monads.

Theorem 3.3, which in fact gives wide classes of monads \(T = (T,\eta ,\mu )\), where \(T\) does not preserve weak pullbacks, and where \(\mu \) is not weakly cartesian—although we then consider only one specific example.

Proposition 4.4, which gives another wide class of monads \(T\), where the functor \(T\) does not preserve weak pullbacks; that class contain monads that determine varieties of groups, rings, modules and all kinds of algebras over rings, and, much more generally, all universal algebras that admit a very weak form of subtraction operation. Some intermediate levels of generality are also mentioned in the same section.

Theorems 6.4, 7.1, and 8.10, which give necessary and sufficient conditions on a semiring \(S\), under which the free \(S\)semimodule monad satisfies any of the abovementioned properties. In some sense the case of semimodules is the only case, found so far, where those properties are characterized in terms of nontrivial algebraic conditions rather than just either obviously hold or obviously fail. We therefore devoted the last section to consider semimodules over some particular semirings. Note also that considering varieties of semimodules is the same as considering arbitrary varieties, in which finite coproducts (=free products) canonically coincide with finite products (see e.g. [9]), or, equivalently, pointed varieties with binary biproducts.
 2.
Weakly cartesian monads
 3.
Some cartesian cases and a related negative result
 4.
Normal, 0regular, subtractive, and weakly subtractive varieties
 5.
A condition that excludes cartesianness
 6.
Semimodules I: Preliminary remarks and weak cartesianness of the functor \(T\)
 7.
Semimodules II: Weak cartesianness for \(\eta \)
 8.
Semimodules III: Weak cartesianness of \(\mu \) and the main results
 9.
Semimodules IV: Examples.
2 Weakly cartesian monads
Definition 2.1
 (a)
the functor \(T\) is weakly cartesian if it preserves weak pullbacks, or, equivalently, transforms pullbacks into weak pullbacks.
 (b)the natural transformation \(\eta \) is weakly cartesian if, for every map \(u : A\rightarrow B\), the naturality square(2.3)
 (c)analogously, the natural transformation \(\mu \) is weakly cartesian if, for every map \(u:A\rightarrow B\), the naturality square(2.4)
Remarks 2.2
 (a)
Replacing weak pullbacks with (ordinary) pullbacks in Definitions 2.1, we obtain the conditions that define cartesian monads. Moreover, every cartesian monad obviously satisfies the conditions of Definition 2.1. Monads satisfying these conditions will be called weakly cartesian, as it is done in [18, 24, 27].
 (b)
The monads with weakly cartesian functor \(T\) (called also condition (BC) on the functor \(T\)) and weakly cartesian \(\mu \) (called condition (BC) on the natural transformation \(\mu \)) were considered many times, starting from [4], and including [8]. Weak cartesianness of \(\eta \) was not used in those papers since it does not hold for the ultrafilter monad, which was the main example there—except [7], where a special case was used. We will say that the monad \((T,\eta ,\mu )\) satisfies (BC) if \(T\) and \(\mu \) satisfy (BC).
 (c)
As, e.g., Manes does in his papers [20, 21], we call a functor taut if it preserves pullbacks along monomorphisms, while a natural transformation is taut if the naturality squares determined by monomorphisms are pullbacks. Clearly every weakly cartesian functor is taut, and every weakly cartesian natural transformation is taut. A monad \(T=(T,\eta ,\mu )\) is taut if \(T\), \(\eta \) and \(\mu \) are taut. We also remark that Möbus introduced this type of monads under the name Alexandrov monad in [22].
In particular, there are two trivial cases:
Example 2.3
 (a)
the monad \(T\) with \(T(X) = 1\) (oneelement set) for every set \(X\)—this is the case if and only if the set \(\Omega _0\) of \(0\)ary operators in any (equivalently, in every) presentation (2.7) is nonempty; and
 (b)the monad \(T\) withwhich is the case if and only if the set \(\Omega _0\) is empty.$$\begin{aligned} T(X)={\left\{ \begin{array}{ll}1,&{}\hbox { if }X\ne \emptyset ,\\ \emptyset ,&{}\hbox { if }X=\emptyset \end{array}\right. } \end{aligned}$$
In both cases it is easy to see that \(T\) satisfies (BC), but \(\eta \) is not weakly cartesian.
From now on we shall always assume that the monad \(T\) is nontrivial, or, equivalently, that \(\mathrm{\mathbf C}\) is nontrivial, that is, that \(\mathrm{\mathbf C}\) contains algebras of arbitrarily large cardinality. In particular this implies that every component of \(\eta \) is injective.
Remarks 2.4
 (a)
Since every component of \(\eta \) is injective, weak cartesianness of \(\eta \) is equivalent to the requirement that every diagram of the form (2.3) is a pullback, that is, it is equivalent to cartesianness of \(\eta \). Hence, when \(T1\cong 1\), weak cartesianness of \(\eta \) gives that diagram (2.3) for \(u:X\rightarrow 1\) is a pullback, that is, \(\eta _X\) is an isomorphism for every \(X\), and so \(T\) is the identity monad.
 (b)In contrast to this and Proposition 2.3 of [21] we point out that, if \(T\) is nontrivial, \(\eta \) is taut. Indeed, consider our diagram (2.3) assuming that \(u\) is injective, and suppose \(\eta _B(b) = T(u)(t)\) for some \(b\in B\) and \(t\in T(A)\). We shall separately consider the case \(b\in u(A)\), and the case \(b\not \in u(A)\).
 Case 1: \(b\in u(A)\). We have \(b = u(a)\) for some \(a\in A\). This gives:and so \(\eta _A(a) = t\), since \(T(u)\) is injective.$$\begin{aligned} T(u)\eta _A(a) =\eta _Bu(a) =\eta _B(b) = T(u)(t) \end{aligned}$$
 Case 2: \(b\not \in u(A)\). Let \(C\) be an arbitrary nonempty \(T\)algebra, and \(c\) and \(d\) elements in \(C\). Let \(f\) and \(g\) be any two maps from \(B\) to \(C\) with \(f(b) = c\), \(g(b) = d\), and \(f(x) = g(x)\) for every \(x\in u(A)\), and let \(\overline{f}\) and \(\overline{g}\) be \(T\)algebra homomorphisms \(T(B)\rightarrow C\) induced by \(f\) and \(g\) respectively. We have:That is, every \(T\)algebra has at most one element, making \(T\) a trivial monad. This excludes Case 2.$$\begin{aligned} \begin{array}{rcll} c &{}=&{} f(b)&{}\hbox {(by the assumption on f)}\\ &{}=&{} \overline{f}\eta _B(b)&{}\hbox {(by definition of}\; \overline{f})\\ &{}=&{}\overline{f}T(u)(t)&{}\hbox {(by the assumption on}\; b\; \hbox {and}\; t)\\ &{}=&{}\overline{g}T(u)(t)&{}\hbox {(since} \overline{f}T(u),\overline{g}T(u):T(A)\rightarrow C\; \hbox {are}\; T\hbox {algebra} \\ &{}&{}&{}\hbox {homomorphisms induced by the maps} fu,gu : A\rightarrow C \\ &{}&{}&{}\hbox {respectively, and since these two maps coincide by the}\\ &{}&{}&{}\hbox {assumptions on} f,g)\\ &{}=&{}d&{}\hbox {(by similar arguments).} \end{array} \end{aligned}$$

3 Some cartesian cases and a related negative result
Following Carboni and Johnstone [2], Leinster [19] defines a finitary algebraic theory to be strongly regular if it can be presented with identities in which “the same variables appear in the same order, without repetition, on each side”. In our notation (2.7) this means that \(\Phi \) consists of such identities, and if so, then, motivated by Theorem C.1.1 in [19], we could call \(\mathrm{\mathbf C}\) operadic. As we know from [2] (see also [19] and [3]), the corresponding monad is then cartesian, which, among other things, obviously implies the following wellknown observation:
Observation 3.1
 (a)
All varieties of the form \(\mathrm{Alg}(\Omega ,\Phi )\) with \(\Phi =\emptyset \). In particular, all varieties in which \(\Omega =\Omega _0\), that is, each basic operation is \(0\)ary, can be presented this way. This includes sets and pointed sets.
 (b)
All varieties \(\mathrm{\mathbf C}\) of the form \(\mathrm{\mathbf C}=\mathrm{\mathbf Set}^M\), where \(M\) is a monoid. In fact, more generally, the same is obviously true for all varieties of the form \((C\downarrow \mathrm{\mathbf Set}^M)\), where \(C\) is any \(M\)set.
 (c)
The variety of semigroups.
 (f)
The variety of monoids.
We should also mention an analogy with Manes’ Theorem 4.14 of [20] and Proposition 3.11 of [21], which say that a finitary monad is taut if and only if the corresponding algebraic theory is balanced. Here “balanced”, also called “regular” by some other authors, is a notion weaker than what we referred to as “strongly regular”: it only requires having the same set of variables on each side. This fact is very important for us since all easy consequences of our results involving taut monads can also be obtained from it.

The universalalgebraic expression \(t = t(x)\), for a unary term \(t\), simply means that \(t\) belongs to \(T(\{x\})\). And if \(t\) belongs to \(T(\{x\})\), \((A,h_A)\) is a \(T\)algebra, and \(a\) is an element in \(A\), then \(t(a) = h_AT(\underline{a})(t)\) is the image of \(t\) under the \(T\)algebra homomorphism \((T(\{x\}),\mu _{\{x\}})\rightarrow (A,h_A)\), induced by the map \(\underline{a} :\{x\}\rightarrow A\) sending \(x\) to \(a\). In particular, if \(t'\in T(\{x\})\) is another such unary term, then \(t(t') =\mu _{\{x\}}T(\underline{t'})(t)\), that is, \(t(t')\) is the image of \(t\) under the \(T\)algebra homomorphism \((T(\{x\}),\mu _{\{x\}})\rightarrow (T(\{x\}),\mu _{\{x\}})\), induced by the map \(\underline{t'}: \{x\}\rightarrow T(\{x\})\) sending \(x\) to \(t'\). Furthermore, \(t' =\eta _{\{x\}}(x)\) makes \(\mu _{\{x\}}T(\underline{t'})(t) =\mu _{\{x\}}T(\eta _{\{x\}})(t) = t\), and so \(t=t(\eta _{\{x\}}(x))\); this nicely agrees with the universalalgebraic expression \(t = t(x)\) up to identifying \(x\) with its image in \(T(\{x\})\).
 In the notation above, if \(u : A\rightarrow B\) is a \(T\)algebra homomorphism, thenIndeed, \(u(t(a)) = uh_AT(\underline{a})(t) = h_BT(u)T(\underline{a})(t) = h_BT(u(\underline{a}))(t) = h_BT(\underline{u(a)})(t) = t(u(a))\).$$\begin{aligned} u(t(a)) = t(u(a)). \end{aligned}$$(3.1)
 When \(A\) is equipped with a \(T\)algebra structure \(h_A\), the (welldefined) element \(t(a)\) of \(A\) should not be confused with the element \(t(\eta _A(a))\) of \(T(A)\). These elements are only related bysince$$\begin{aligned} t(a) = h_A(t(\eta _A(a))). \end{aligned}$$(3.2)which is easy to check:$$\begin{aligned} t(\eta _A(a)) = T(\underline{a})(t), \end{aligned}$$(3.3)$$\begin{aligned} t(\eta _A(a)) =\mu _AT(\underline{\eta _A(a)})(t) =\mu _AT(\eta _A\underline{a})(t) =\mu _AT(\eta _A)T(\underline{a})(t) = T(\underline{a})(t). \end{aligned}$$

By a pseudoconstant unary term we mean a term \(t\in T(\{x\})\) such that, for every \(T\)algebra \(A\) and every two elements \(a\) and \(a'\) in \(A\), we have \(t(a) = t(a')\) in \(A\). In the universalalgebraic terminology, a unary term \(t\) is pseudoconstant if and only if, for some distinct variables \(x\) and \(y\), the identity \(t(x) = t(y)\) holds in \(\mathrm{\mathbf C}\). Note that whenever \(T(\emptyset )\) is nonempty, or, equivalently, \(\Omega _0\) is nonempty, every pseudoconstant term \(t\in T(\{x\})\) is constant, that is, it belongs to the image of \(T(\emptyset )\) in \(T(\{x\})\). The notion of pseudoconstant term has an obvious nonunary version of course.
Lemma 3.2
Let \(S\) be a set and \(s\) and \(s'\) elements in \(S\). If \(s\ne s'\) and \(t\) is not a pseudoconstant term, then \(t(\eta _A(s))\ne t(\eta _A(s'))\).
Proof
Theorem 3.3
 (a)
If \(\mathrm{\mathbf C}\) admits a pseudoconstant term that is not constant, then the functor \(T\) is not weakly cartesian (and not even taut).
 (b)
Let \(\mathrm{\mathbf C}= \mathrm{\mathbf Set}^T = \mathrm{Alg}(\Omega ,\Phi )\) be a variety of universal algebras that admits nonpseudoconstant unary terms \(t, t'\in T(\{x\})\), for which \(t(t')\) is a pseudoconstant term. Then the natural transformation \(\mu \) is not weakly cartesian (and not even taut).
Proof

\(T(W) = T(\emptyset )\) is empty – otherwise there would be no pseudoconstant nonconstant terms;

\(T(X)\times _{T(Z)}T(Y)\) is nonempty since it contains the element determined by the pair \((t(\eta _X(x)), t(\eta _Y(y)))\).
The simplest example seems to be:
Example 3.4

\(T(\emptyset ) =\emptyset \);

If \(X\) is a nonempty set, then \(T(X)\) is the disjoint union of \(\{0,1\}\times X\) with a oneelement set, say \(\{\infty \}\);

\(\eta _X : X\rightarrow T(X)\) is defined by \(\eta _X(x) = (0,x)\);
 \(\mu _X : T^2(X)\rightarrow T(X)\) is defined by$$\begin{aligned} \begin{array}{ll}\mu _X(k,(l,x))&{}={\left\{ \begin{array}{ll} (k+l,x),&{}\hbox { if }k+l\le 1,\\ \infty ,&{}\hbox { if }k+l=2,\end{array}\right. }\\ \mu _X(k,\infty )&{}=\mu _X(\infty )=\infty .\end{array} \end{aligned}$$
Remark 3.5
It is easy to check that modifying Example 3.4 by only adding an arbitrary (nonzero) number of constants and still requiring only the identity \(2x = 2y\), and possibly identities involving only constants, we obtain a monad with \(T\) and \(\eta \) weakly cartesian, but, still by Theorem 3.3(b), \(\mu \) is not weakly cartesian.
4 Normal, \(0\)regular, subtractive, and weakly subtractive varieties

\(\mathrm{\mathbf C}\) is pointed as a category, that is, each homset \(\hom _C(A,B)\) has a distinguished element \(0_{A,B}\), such that, for every morphism \(\alpha :A'\rightarrow A\) and every morphism \(\beta :B\rightarrow B'\), we have \(\beta 0_{A,B}\alpha = 0_{A',B'}\);

the free algebra in \(\mathrm{\mathbf C}\) on the empty set (=the initial object in \(\mathrm{\mathbf C}\)) has exactly one element (=is isomorphic to the terminal object in \(\mathrm{\mathbf C}\));

the set \(\Omega _0\) of \(0\)ary operators in \(\mathrm{\mathbf C}\) is nonempty, and every two constant terms are equal to each other in each object in \(\mathrm{\mathbf C}\);

every algebra in \(\mathrm{\mathbf C}\) has a oneelement subalgebra;

every algebra in \(\mathrm{\mathbf C}\) has a unique oneelement subalgebra, and, for every morphism \(\alpha :A'\rightarrow A\), that subalgebra is the image of \(0_{A',A}\).

each of the morphisms \(0_{A,B}\);

the initial object in \(\mathrm{\mathbf C}\);

the unique subalgebra in any algebra in \(\mathrm{\mathbf C}\);

the unique element in that subalgebra;
Slightly more generally, instead of normal varieties we could consider what universal algebraists call \(0\)regular varieties (see [14, 15, 16] for explanations and references). Those are not necessarily pointed, but have a distinguished constant \(0\) such that every congruence on any given algebra is completely determined by its \(0\)class. An obvious reason to consider the nonpointed case is to include important examples, such as unital rings and unital algebras over rings, Boolean and Heyting algebras, etc.
Definition 4.1
A variety \(\mathrm{\mathbf C}\) is said to be weakly subtractive if it admits a constant \(0\) and a nonpseudoconstant \(n\)ary term \(s\) satisfying the identity \(s(x,\ldots ,x) = 0\) and not satisfying the identity \(s(x_1,\ldots ,x_n) = 0\).
Observation 4.2
 (a)
Every subtractive variety is weakly subtractive indeed, as trivially follows from the definitions.
 (b)
Every \(0\)regular variety is also weakly subtractive as follows, e.g., from Corollary 1.7 in [11].
 (c)
For the subtractive and \(0\)regular varieties the desired term \(s\) there is actually binary. On the other hand, in Definition 4.1, the cases \(n= 0\) and \(n = 1\) are automatically excluded, since in those cases our requirements on \(s\) become contradictory.
For a morphism \(u : A\rightarrow B\) in a variety \(\mathrm{\mathbf C}\) having a distinguished constant \(0\), we shall still write \(\mathrm{Ker}(u) = \{a\in A\;\;u(a) = 0\}\), even though this \(\mathrm{Ker}(u)\) does not have to be a subalgebra in \(A\) in general. The following simple observation characterizes weak subtractivity of \(\mathrm{\mathbf C}\) in terms of the corresponding monad on \(\mathrm{\mathbf Set}\):
Observation 4.3
 (i)
\(\mathrm{\mathbf C}\) is weakly subtractive;
 (ii)
there exists a map \(f : X\rightarrow Y\) with \(\mathrm{Ker}(T(f))\ne 0\);
 (iii)
for each cardinal number \({\mathfrak {m}}\) there exists a map \(f : X\rightarrow Y\) with \(\mathrm{Ker}(T(f))\) having cardinality \(\ge \mathfrak {m}\).

a set \(S\) of cardinality \(\ge \mathfrak {m}\);

the map \(S\times f : S\times X \rightarrow S\times Y\) defined by \((S\times f)(s,x) = (s,f(x))\).
 (a)\(t(0,\ldots ,0) = 0\) in any \(T\)algebra \(A\). Indeed, since \(T(Y)\) is the free \(T\)algebra on \(Y\), the map \(u :Y \rightarrow T(A)\), sending all elements of \(Y\) to \(0\), extends to a homomorphism \(v : T(Y) \rightarrow T(A)\), and we have$$\begin{aligned} t(0,\ldots ,0)&= t(uf(x_1),\ldots ,uf(x_n)) = t(vf(x_1),\ldots ,vf(x_n)) \\&= v(t(f(x_1),\ldots , f(x_n))) = v(0) = 0. \end{aligned}$$
 (b)\(t((S\times f)(s,x_1),\ldots , (S\times f)(s,x_n)) = 0\) in \(T(S\times X)\) for every \(s\) in \(S\). Indeed, let \(w : Y \rightarrow S\times Y\) be defined by \(w(y) = (s,y)\)—thensince \(T(w)\) is a homomorphism of algebras.$$\begin{aligned} t((S\times f)(s,x_1),\ldots , (S\times f)(s,x_n))&= T(w)(t(f(x_1),\ldots , f(x_n))) \\&= T(w)(0) = 0, \end{aligned}$$
 (c)As follows from (b), \(t((s,x_1),\ldots ,(s,x_n))\) is in \(\mathrm{Ker}(T(S\times f))\) for each \(s\) in \(S\), after which it suffices to show thatFor, given \(s \ne s'\) in \(S\), consider any map \(g : S\times X \rightarrow X\) with \(g(s,x) = x\) and \(g(s',x) = 0\), for each \(x\) in \(X\). We have$$\begin{aligned} s \ne s' \implies t((s,x_1),\ldots ,(s,x_n)) \ne t((s',x_1),\ldots ,(s',x_n)). \end{aligned}$$$$\begin{aligned} \begin{array}{{l}} T(g)(t((s,x_1),\ldots ,(s,x_n))) = t(x_1,\ldots ,x_n) \ne 0,\hbox { and}\\ T(g)(t((s',x_1),\ldots ,(s',x_n))) = t(0,\ldots ,0) = 0 \hbox { by (a)}.\end{array} \end{aligned}$$
And this observation helps to prove:
Proposition 4.4
If \(\mathrm{\mathbf C}\) is weakly subtractive, then \(T\) is not a weakly cartesian functor (not even taut).
Proof
5 A condition that excludes cartesianness
Unlike Sect. 2, due to the simplicity of the data considered in this section, we shall freely use the universalalgebraic expressions \(t = t(x), t(y)\) instead of \(t(\eta _Y(y))\), etc. The section is devoted to a simple result showing that many familiar monads are not cartesian:
Proposition 5.1
Suppose \(\mathrm{\mathbf C}\) admits a commutative binary operation which is not pseudoconstant, that is, admits a binary term \(t\) satisfying \(t(x,y) = t(y,x)\) and not \(t(x,x') = t(y,y')\), where \(x, x', y, y'\) are pairwise distinct variables. Then the functor \(T\) does not preserve pullbacks.
Proof

\(X\) is any set containing \(x\) and \(x'\);

\(Y\) is any set containing \(y\) and \(y'\);

\(Z\) is a oneelement set (which uniquely determines \(f\) and \(g\)).
Marek Zawadowski told us that this proposition also follows from results of [23].
6 Semimodules I: preliminary remarks and weak cartesianness of the functor \(T\)
From now on, \(\mathrm{\mathbf C}\) will denote the variety \(S\)\(\mathrm{\mathbf SMod}\) of semimodules over a semiring \(S\). This and the next two sections will be devoted to the analysis of weak cartesianness of the functor \(T\) and of the natural transformations \(\eta \), \(\mu \), respectively, for the corresponding (free \(S\)semimodule) monad \(T = (T,\eta ,\mu )\).

for any set \(X\), \(T(X)\) is the set of all maps \(t : X\rightarrow S\) for which the set \(\{x\in X\;\;t(x)\ne 0\}\) is finite;

for a map \(f : X\rightarrow Y\), we have \(T(f)(t)(y) = \sum _{f(x)=y}t(x)\);

\(\eta _X : X\rightarrow T(X)\) is the map that has \(\eta _X(x)(x) = 1\) and \(\eta _X(x)(y) = 0\) for \(x\ne y\);

\(\mu _X : T^2(X)\rightarrow T(X)\) is the map defined by \(\mu _X(u)(x) = \sum _{t\in T(X)}u(t)t(x)\), where all possibly infinite sums make sense since they involve only finitely many nonzero summands.
Lemma 6.1
Proof
If \(Y =\emptyset \), then \(W =\emptyset \) and \(T(W) = T(Y) = \{0\}\), and so the \(T\)image of (2.1) is a weak pullback if and only if the kernel of \(T(f) : T(X)\rightarrow T(Z)\) is \(\{0\}\). And when \(X\) and \(Z\) are as in (ii), the map \(T(f) : T(X)\rightarrow T(Z)\) can be identified with the addition map \(+: S\times S\rightarrow S\), which makes the equivalence (ii) \(\Leftrightarrow \) (iii) obvious. Since the implication (i) \(\Rightarrow \) (ii) is trivial, it only remains to prove the implication (iii) \(\Rightarrow \) (i).
Suppose \(\alpha :X\rightarrow S\) belongs to the kernel of \(T(f)\). Then, for every \(z\) in \(Z\), we have \(\sum _{f(x)=z}\alpha (x) = 0\), and, as follows from (iii), this means that \(\alpha (x) = 0\) for every \(x\) in \(X\). That is, \(\alpha =0\) in \(T(X)\). \(\square \)
Lemma 6.2
Proof
Just observe that the free \(S\)semimodule functor preserves coproducts and that finite coproducts of \(S\)semimodules are canonically isomorphic to products. \(\square \)
Lemma 6.3
 (i)
the \(T\)image of (2.1) is a weak pullback whenever \(f\) and \(g\) are surjective;
 (ii)
the \(T\)image of (2.1) is a weak pullback whenever the numbers of elements in \(X\), \(Y\) and \(Z\) are \(2\), \(2\) and \(1\) respectively.
 (iii)for every \(a, b, c, d\) in \(S\) with \(a+b = c+d\), there exist \(x, y, z, t\) in \(S\) with$$\begin{aligned} x+y = a,\; z+t = b,\; x+z = c,\; y+t = d. \end{aligned}$$(6.4)
 (iv)for every natural \(m\) and \(n\), and \(s_1,\ldots , s_m\) and \(t_1,\ldots , t_n\) in \(S\) with \(s_1+\cdots +s_m = t_1+\cdots +t_n\), there exists a \(\{1,\ldots ,m\}\,\times \, \{1,\ldots ,n\}\)indexed family \((x_{ij})\) of elements in \(S\) with$$\begin{aligned}&x_{i1}+\cdots +x_{in} = s_i\quad \hbox { for all } i = 1,\ldots ,m\quad \hbox { and }\nonumber \\&\quad x_{1j}+\cdots +x_{mj}= t_j \quad \hbox { for all } j = 1,\ldots ,n. \end{aligned}$$(6.5)
Proof
(iv) \(\Rightarrow \) (i): Since the functor \(T\) preserves filtered colimits, it suffices to consider the case of finite \(X\), \(Y\), and \(Z\). Moreover, thanks to Lemma 6.2, we can assume that \(Z\) is a oneelement set. But assuming \(Z\) in (i) to be a oneelement set (and \(X\ne \emptyset \ne Y\)) makes (i) obviously equivalent to (iv), just like (ii) is equivalent to (iii). \(\square \)
Theorem 6.4
 (i)
\(T\), as a functor, is weakly cartesian;
 (ii)
\(S\) satisfies the conditions of Lemmas 6.1 and 6.3.
Proof
(i) \(\Rightarrow \) (ii) follows from Lemmas 6.1 and 6.3.
(ii) \(\Rightarrow \) (i): Assuming that conditions 6.1(iii) and 6.3(iii) hold, we have to show that \(T\)images of arbitrary pullbacks (2.1) are weak pullbacks. Using filtered colimits again this reduces to the case of pullbacks of finite sets, and then, using Lemma 6.2, we can assume that \(Z\) in (2.1) is a oneelement set. After this we just apply Lemmas 6.1 and 6.3. \(\square \)
Remark 6.5
Using similar arguments it is easy to show that the functor \(T\) is taut if and only if no nonzero element in \(S\) has an additive inverse.
7 Semimodules II: weak cartesianness for \(\eta \)
Theorem 7.1
 (i)
\(\eta \) is weakly cartesian;
 (ii)
If \(s+s' = 1\) in \(S\), then either \(s = 0\), or \(s' = 0\). In particular no nonzero element of \(S\) has an additive inverse.
Proof
8 Semimodules III: weak cartesianness of \(\mu \) and the main results
For \(S\) and \(T = (T,\eta ,\mu )\) as in the two previous sections, the analysis of weak cartesianness for \(\mu \), which this section is devoted to, is rather complicated and we will use weak cartesianness for the functor \(T\) in it. The first of our auxiliary results is obvious and surely known to many readers:
Lemma 8.1
 (a)
if \(\fbox {1}\) and \(\fbox {2}\) are weak pullbacks, then so is the rectangle \(\fbox {1}\fbox {2}\);
 (b)
if \(\fbox {1}\fbox {2}\) is a weak pullback and Open image in new window are jointly monic, then \(\fbox {1}\) is a weak pullback;
 (c)
if \(\fbox {1}\fbox {2}\) is a weak pullback and the map \(X\rightarrow Y\) is surjective, then \(\fbox {2}\) is a weak pullback.
 (d)
the class of weak pullback diagrams of sets is closed under products in the category of all square diagrams in the category of sets;
 (e)for any weak pullback diagram (2.1) and any map \(h : X\rightarrow E\), the diagram(8.2)
Lemma 8.2
 (a)
it contains all bijections and is closed under composition and filtered colimits;
 (b)
if \(T\) is a weakly cartesian functor, then \({\mathcal {U}}\) is closed under coproducts with bijections.
Proof
(a): The fact that \({\mathcal {U}}\) contains all bijections is obvious. Closedness under composition easily follows from Lemma 8.1(a). Closedness under filtered colimits follows from the fact that the functor \(T\) preserves filtered colimits.

the square \(\fbox {1}\) is a weak pullback, which follows from weak cartesianness of the functor \(T\) and Lemma 8.1(e);

the square \(\fbox {2}\) is a weak pullback, which follows from the fact that so is diagram (2.4) and Lemma 8.1(d);

therefore \(\fbox {1}\fbox {2}\) is a weak pullback by Lemma 8.1(a);

Since the two top and two bottom vertical arrows in (8.4) are bijections, and \(\fbox {1}\fbox {2}\) is a weak pullback, diagram (8.3) also is a weak pullback. \(\square \)
Corollary 8.3
If \(T\) is a weakly cartesian functor, then the natural transformation \(\mu \) is weakly cartesian if diagram (2.4) is a weak pullback whenever \(A\) is either empty or is a twoelement set, and \(B\) is a oneelement set.
Corollary 8.4
If \(T\) is a taut functor, then the natural transformation \(\mu \) is taut if diagram (2.4) is a (weak) pullback whenever \(A\) is empty and \(B\) is a oneelement set.
Observation 8.5
 (a)Let \(A\) and \(B\) be the empty and a oneelement set, respectively. Then diagram (2.4) can be described as(8.5)
 (b)Let \(A\) and \(B\) be a twoelement set and a oneelement set, respectively (as in the proof of Theorem 7.1). Then diagram (2.4) can be described as(8.6)respectively.$$\begin{aligned} \mu _2(t) = \left( \sum _{(x,y)\in S\times S} t(x,y)x,\sum _{(x,y)\in S\times S} t(x,y)y\right) , \end{aligned}$$
Lemma 8.6
 (a)
\(s+ s' = 0 \;\Rightarrow \; s = 0 = s'\) (condition 6.1(iii));
 (b)
\(ss' = 0 \;\Rightarrow \; (s = 0\,\vee \,s' = 0)\).
Proof
Combining this lemma with Corollaries 6.5 and 8.4 we obtain the following characterization of taut free \(S\)semimodule monads:
Theorem 8.7
The free \(S\)semimodule monad \(T\) is taut if and only if no nonzero element in \(S\) has an additive inverse and no nonzero element of \(S\) is a zero divisor.
Lemma 8.8
Proof
 Since \(\mu _1(m_1)+\mu _1(m_2) = a+b\), by Lemma 6.3 there exist \(a_1, a_2, b_1, b_2\in S\) with$$\begin{aligned} \mu _1(m_1) = a_1+b_1,\;\mu _1(m_2) = a_2+b_2,\; a = a_1+a_2,\; b = b_1+b_2. \end{aligned}$$
 Then, since \(m_1\) and \(m_2\) are in \(M(S)\), there exist \(t_1\) and \(t_2\) in \(T^2(S)\) with$$\begin{aligned} T(+)(t_1) = m_1,\;\mu _2(t_1) = (a_1,b_1),\; T(+)(t_2) = m_2,\;\mu _2(t_2) = (a_2,b_2). \end{aligned}$$

Therefore \(T(+)(t_1+t_2) = m_1+m_2\) and \(\mu _2(t_1+t_2) = (a_1,b_1)+(a_2,b_2) = (a,b)\), which proves that \(m_1+m_2\) is in \(M(S)\).\(\square \)
Lemma 8.9
 (i)
diagram (2.4) is a weak pullback whenever \(A\) is a twoelement set, and \(B\) is a oneelement set;
 (ii)for every \(a, b, c, d\in S\) with \(c\) in \(\mathrm{\mathbf C}\) and \(cd = a+b\), there exists a map with$$\begin{aligned} \sum _{x+y=d} t(x,y)x = a,\;\sum _{x+y=d} t(x,y)y = b,\;\sum _{x+y=d} t(x,y) = c. \end{aligned}$$(8.11)
Proof
Now we are ready to present our main results on the free \(S\)semimodule monad \(T\):
Theorem 8.10
 (a)
no nonzero element in \(S\) has an additive inverse;
 (b)for every \(a, b, c, d\in S\) with \(a+b = c+d\), there exist \(x, y, z, t\) in \(S\) with$$\begin{aligned} x+y = a,\, z+t = b,\; x+z = c,\; y+t = d; \end{aligned}$$
 (c)
\(S\) has no zero divisors, that is, \(ss'=0\;\Rightarrow \;(s=0\,\vee \,s'=0)\) for \(s, s'\in S\);
 (d)
for every \(a, b, c, d\in S\) with \(cd = a+b\), there exists a map \(t : \{(x,y)\in S\times S\;\; x + y = d\}\rightarrow S\) satisfying (8.11).
 (e)
there exists a subset \(C\) of \(S\) that generates \(S\) as an additive monoid, and such that, for every \(a, b, c, d\in S\) with \(c\) in \(C\) and \(cd = a+b\), there exists a map \(t : \{(x,y)\in S\times S\;\; x+ y = d\}\rightarrow S\) satisfying (8.11).
Proof
Note that conditions (a)–(e) coincide with conditions 6.1(iii), 6.3(iii), 8.6(b), 8.9(ii) with \(C = S\), and 8.9(ii) respectively; for (e) \(\Leftrightarrow \) 8.9(ii) note, however, that “any” can be understood as either existential or universal quantifier—which follows from Lemma 8.9. After that all we need is to recall that the functor \(T\) satisfies (BC) if and only if \(S\) satisfies 6.1(iii) and 6.3(iii) (Theorem 6.4), and to put together Corollary 8.3 and Lemmas 8.6 and 8.9. \(\square \)
From here and Theorem 7.1 we obtain:
Theorem 8.11
The free \(S\)semimodule monad \(T\) is weakly cartesian if and only if it satisfies (BC) and the implication \(s+s' = 1\;\Rightarrow \;(s = 0 \,\vee \, s' = 0)\) for \(s, s'\in S\).
9 Semimodules IV: examples
Conditions (b), (d), and (e) of Theorem 8.10 can be complicated to check, and we will use simple stronger conditions provided by
Proposition 9.1
 (a)
If for every \(a, b\in S\) there exists \(x\in S\) such that either \(a+x = b\) or \(b+x = a\), and the additive monoid of \(S\) is a monoid with cancelation, then condition 8.10 (b) holds.
 (b)
If every nonzero element in \(S\) has a multiplicative inverse and no nonzero element in \(S\) has an additive inverse, then condition 8.10(d) holds.
 (c)
If \(S\) is generated by \(\{1\}\), then condition 8.10(e) holds.
Proof
(a): Let \(a, b, c, d\in S\) be as in Theorem 8.10(b). Without loss of generality we can assume that there exist \(p, q, r\in S\) with \(a+p = b\), \(a+q = c\), \(a+r = d\), after which we can take \(x = 0\), \(y = a\), \(z = c\), and \(t = r\). Indeed, we will obtain \(x+y = 0+a = a\), \(z+t = c+r = b\) since \(a+c+r = c+d = a+b\), \(x+z = 0+c = c\), and \(y+t = a+r = d\).
Example 9.2
If \(S = {\mathbb {N}}\) is the semiring of natural numbers, then all our conditions are satisfied. In particular for conditions 8.10(b) and 8.10(e) this follows from Propositions 9.1(a) and 9.1(c) respectively. In this case \(S\)semimodules are nothing but commutative monoids, and we conclude that the free commutative monoid monad is weakly cartesian. Note that it is not cartesian by Proposition 5.1.
Example 9.3
Let \(S = \{0,1\}\) be the quotient semiring of \({\mathbb {N}}\) determined by \(1+1 = 1\). Then the \(S\)semimodules are nothing but commutative idempotent monoids, and so, as trivially follows from Theorem 8.11, the free commutative idempotent monoid monad is not weakly cartesian. However, it satisfies (BC), since \(S = \{0,1\}\) satisfies all conditions of Theorem 8.10. For condition 8.10(e) it follows again from Proposition 9.1(c), while condition 8.10(b) can be easily checked directly. Note that Proposition 9.1(a) cannot be used here since the additive monoid of \(S\) does not admit cancelation. The fact that the free commutative idempotent monoid monad satisfies (BC) is in fact known and was already mentioned e.g. in [8].
Example 8.4
Example 8.5
If \(S = {\mathbb {Q}}_+\) is the semiring of nonnegative rational numbers, then, as in Example 9.3, the free \(S\)semimodule monad is not weakly cartesian (again, trivially from Theorem 8.11) but satisfies (BC). However, Proposition 8.10(c) cannot be used here, and Proposition 9.1(b) should be used instead, which makes condition 8.10(d) trivial. The same is true if \(S = {\mathbb {R}}_+\) is the semiring of nonnegative real numbers.
Example 8.6
If the semirings \(S\) and \(S'\) satisfy conditions 8.10(a) and 8.10(b), then the product \(S\times S'\) also does. But it never satisfies condition 8.10(c) unless one of the two rings is trivial. This gives free semimodule monads which do not satisfy (BC) although the functor \(T\) does.
Notes
Acknowledgments
We are grateful to Lurdes Sousa, who showed us that Theorem 6.4 is essentially the same as Theorem 5.13 in the paper [10] of Gumm and Schröder. We are also grateful to the referee for mentioning to us the relationship between regularity and tautness.
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