The Zappa-Szép product, strict factorization systems and distributive laws

$\newcommand{\con}{\mathbf{Set}} \newcommand{\uno}{\mathbf{1}} \newcommand{\Cat}{\mathbf{Cat}}\newcommand{\mon}{\mathbf{Mon}} \newcommand{\ab}{\mathbf{Ab}} \newcommand{\an}{\mathbf{An}}\newcommand {\matcon}{\mathbf{Set\text{-}Mat}} \newcommand{\grp}{\mathbf{Grp}} \newcommand{\ob}{\mathrm{Ob}}$

The Zappa-Szép product and distributive laws

The Zappa-Szép product, strict factorization systems and distributive laws are related in a certain way. Let us talk first about the Zappa-Szép product (also known as knit product or matched pair of groups).
     The Zappa-Szép product is a generalization of the semidirect product for groups in the same way as this product is a generalization of the direct product for groups, and the zappa-szép product is the most general way in which a group is presented as a product of two groups.
     The internal Zappa-Szép product is defined as follows. Given a group $G$ and two subgroups $H$ and $K$ of $G$, the following statements are equivalent:

• $G=KH$ y $K\cap H=\{e\}$,
• for every $g\in G$ there is a unique $k\in K$ and a unique $h\in H$ such that $g=kh$.

If either of these statements is satisfied, then $G$ is said to be an internal Zappa-Szép product of $K$ and $H$.
     There is an external version of the Zappa-Szép product, which is the one we are interested in, because it's by means of this product that a link between distributive laws and strict factorization systems is established.
     Given two groups $K$ and $H$, suppose there are funtions $\alpha:H\times K\rightarrow K$ and $\beta:H\times K\rightarrow H$ such that

1. $\alpha(h_1h_2,k)=\alpha(h_1,\alpha(h_2,k))$,
2. $\beta(h_1h_2,k)=\beta(h_1,\alpha(h_2,k))\beta(h_2,k)$,
3. $\beta(h,k_1k_2)=\beta(\beta(h,k_1),k_2)$,
4. $\alpha(h,k_1k_2)=\alpha(h,k_1)\alpha(\beta(h,k_1),k_2)$,
5. $\alpha(e,k)=k$,
6. $\beta(h,e)=h$

for every $h,h_1,h_2\in H$ y $k,k_1,k_2\in K$ (cf. [4]). Note that from (ii) and (v) and from (iv) and (vi) follows respectively:

7. $\beta(e,k)=e$ and
8. $\alpha(h,e)=e$.

We can define from (i)-(vi) a multiplication and an inverse on $K\times H$ as $$(k_1,h_1)\gamma(k_2,h_2):=(k_1\alpha(h_1,k_2),\beta(h_1,k_2)h_2))$$ and $$(k,h)^{-1}:=(\alpha(h^{-1},k^{-1}),\beta(h^{-1},k^{-1})).$$ $(\gamma,K\times H)$ is called an external Zappa-Szép product of $K$ and $H$.
     Note that if $G$ is an internal Zappa-Szép product of its subgroups $K$ y $H$, then there are funtions $\alpha:H\times K\rightarrow K$ and $\beta:H\times K\rightarrow H$ such that (i), (ii), (iii), (iv), (v) and (vi) in the previous definition are satisfied: their existence follows from the fact that every element $g\in G$ can be written uniquely as a product $kh$, and the fact that if $G$ is an external Zappa-Szép product of the groups $K$ and $H$, then $G$ is an internal Zappa-Szép product of its subgroups $K\times e_H$ and $e_K\times H$.
     Let $H,K\in\grp$. Let $S$ and $T$ be the endofunctors $H\times(-)$ and $K\times(-)$ on $\con$, respectively. Then we have the monads $$H\times X,\quad\xymatrix{X\ar[r]^(.4){e_H\times 1_X} & H\times X},\quad\xymatrix{H\times H\times X\ar[r]^(.6){m_H\times 1_X} & H\times X}$$ and $$K\times X,\quad\xymatrix{X\ar[r]^(.4){e_K\times 1_X} & K\times X},\quad\xymatrix{K\times K\times X\ar[r]^(.6){m_K\times 1_X} & K\times X}.$$ Denote their units and multiplicationss as $\eta',\mu'$ y $\eta,\mu$, resp. Let $\gamma$ be a group structure on $K\times H$ such that $K\times e_H,e_K\times H$ are subgroups of $(\gamma,K\times H)$ and $(K\times e_H)\gamma(e_K\times H)=(\gamma,K\times H)$. In other words, such that $(\gamma,K\times H)$ is an internal Zappa-Szép product of $K\times e_H$ and $e_k\times H$, or that $(\gamma, K\times H)$ is an external Zappa-Szép product of $K$ and $H$.
     Without loss of generality, we can assume $(k,h)=(k,e_H)\gamma(e_K,h)$ for any $k\in K$ and $h\in H$ (consider the definition of external Zappa-Szép product and the properties of $\alpha$ and $\beta$). Then we have the monad $$(K\times H\times(-),(e_K,e_H)\times 1_{(-)}, \gamma\times 1_{(-)})$$ on $\con$. Since $K\times e_H$ and $e_K\times H$ are subgroups of $(\gamma,K\times H)$, then $\eta S$ and $T\eta'$ are morphisms of monads, and since $(k,h)=(k,e_H)\gamma(e_K,h)$, the previous monad satisfies the middle unitary law, so the monad above induces a distributive law $ST\Rightarrow TS$; namely, $$(\gamma\times 1_{(-)})\cdot\eta ST\eta':H\times K\times(-)\Rightarrow K\times H\times(-);$$ explicitly, $$\xymatrix{(h,k,-)\ar@{|->}[r] & (e_K,h,k,e_H,-)\ar@{|->}[r] & ((e_K,h)\gamma(k,e_H),-)}.$$
     Conversely, let $\lambda:ST\Rightarrow TS$ be a distributive law of $S$ over $T$, and consider $\lambda 1:H\times K\times 1\rightarrow K\times H\times 1$. Neglect the singleton, and put $$\lambda 1=(\alpha,\beta),$$ where $\alpha$ and $\beta$ are determined by the following commutative diagram: $$\xymatrix{ & H\times K\ar[d]^{\lambda 1}\ar[dr]^\alpha\ar[dl]_\beta &\\ K & K\times H\ar[l]^{p_K}\ar[r]_{p_H} & H. }$$ Then, by the compatibility of $\lambda$ with the unit of $S$, $$\xymatrix{ & H\times K\times 1\ar[dd]^{\lambda 1}\\ K\times 1\ar[ur]^{e_H\times K\times 1}\ar[dr]_{K\times e_H\times 1} &\\ & K\times H\times 1 }$$ commutes; hence, $\alpha(e_H,k)=k$ y $\beta(e_H,k)=e_H$.
     Now, by the compatibility of $\lambda$ with the multiplication of $S$, $$\xymatrix{ H\times H\times K\times 1\ar[rr]^{H\times\lambda 1}\ar[d]_{m_H\times K\times 1} & & H\times K\times H\times 1\ar[rr]^{\lambda_{H\times 1}} & & K\times H\times H\times 1\ar[d]^{K\times m_H\times 1}\\ H\times K\times 1\ar[rrrr]_{\lambda 1} & & & & K\times H\times 1 }$$ commutes; whence, $$\alpha(h_1h_2,k)=\alpha(h_1,\alpha(h_2,k))\quad\text{y}\quad \beta(h_1h_2,k)=\beta(h_1,\alpha(h_2,k))\beta(h_2,k).$$ Similarly, by the compatibility of $\lambda$ with the unit and the multiplication of $T$, $\alpha(h,e_K)=e_K$, $\beta(h,e_K)=h$ and $$\alpha(h,k_1k_2)=\alpha(h,k_1)\alpha(\beta(h,k_1),k_2)\quad\text{y}\quad\beta(h,k_1k_2)=\beta(\beta(h,k_1),k_2).$$
     Therefore, we have a Zappa-Szép product of $K$ and $H$.
     Thus we have a correspondence between the Zappa-Szép products of $K$ and $H$, and the distributive laws of $H\times(-)$ over $K\times(-)$. Indeed, define $L:\mathbf{Zappa\text{-}Sz\acute{e}p}(K,H)\rightarrow\mathbf{DistLaw}(H,K)$, a function which goes from the Zappa-Szép products of $K$ and $H$ to the distributive laws of $H\times(-)$ ovoer $K\times(-)$ (let $\alpha:H\times K\rightarrow K$ and $\beta:H\times K\rightarrow H$ be the functions which define a Zappa-Szép product of $K$ and $H$): $$\xymatrix{ \mathbf{Zappa\text{-}Sz\acute{e}p}(K,H)\ar[r]^(.55)L & \mathbf{DistLaw}(H,K) }\qquad\qquad\quad$$ $$\xymatrix{ (\alpha,\beta)\ar@{|->}[r] & (\alpha,\beta)\times 1_{(-)}. }$$ Define $N:\mathbf{DistLaw}(H,K)\rightarrow\mathbf{Zappa\text{-}Sz\acute{e}p}(K,H)$ as $$\xymatrix{ \mathbf{DistLaw}(H,K)\ar[r]^(.45)N & \mathbf{Zappa\text{-}Sz\acute{e}p}(K,H) }\qquad\qquad\quad$$ $$\xymatrix{ \lambda\ar@{|->}[r] & (p_K\circ\lambda 1,p_H\circ\lambda 1). }$$ Obviously $NL=1$. Less obvious is that $LN=1$. Since $\con$ is a distributive category, \begin{equation}\label{D:distcatset} Z\times X=\sum\nolimits_{x\in X}Z\times H\times\{x\}. \end{equation}
     Consider now the injection $i_x:\{x\}\rightarrow X$; then by naturality of $\lambda$, the following diagram commutes: $$\xymatrix{ H\times K\times\{x\}\ar[rr]^{H\times K\times i_x}\ar[d]_{\lambda_{\{x\}}} & & H\times K\times X\ar[d]^{\lambda_X}\\ K\times H\times\{x\}\ar[rr]_{K\times H\times i_x} & & K\times H\times X. }$$ On the other hand, since $H\times K\times i_x$ is the injection $$\xymatrix{H\times K\times\{x\}\ar[r] & \sum_{x\in X}H\times K\times\{x\}},$$ $K\times H\times i_x$ is the injection $$\xymatrix{K\times H\times\{x\}\ar[r] & \sum_{x\in X}K\times H\times\{x\}}$$ and the equality \eqref{D:distcatset} holds, $\lambda_X=\sum_{x\in X}\lambda_{\{x\}}$. However $\lambda_{\{x\}}=\lambda 1\times 1_{\{x\}}$ for every $x\in X$, so $\lambda_X=\lambda 1\times 1_X$. Hence, $\lambda_X=(p_K\circ\lambda 1,p_H\circ\lambda 1)\times 1_X$.

The bicategory of set-valued matrices

The Zappa-Szép product is generalized in [3] by showing the equivalence of the concept of distributive law in the bicategory of set-valued matrices and the concept of strict factorization system. Let's see how that is done. First describe the bicategory of set-valued matrices $\matcon$ as follows: the objects (the 0-cells) of $\matcon$ are sets, a 1-cell $M:A\rightarrow B$ is a set-valued matrix, i. e., $M(b,a)\in\con$ for every $a\in A$ and $b\in B$, a 2-cell $\tau:M\Rightarrow N:A\rightarrow B$ is a matrix of functions $\tau(b,a):M(b,a)\rightarrow N(b,a)$. The composite of 1-cells $$\xymatrix{ A\ar[r]^M & B\ar[r]^E & C\ar@{}|{=}[r] & A\ar[r]^{EM} & C }$$ is defined as $$EM(c,a):=\sum_{a\in A}E(c,b)\times M(b,a).$$ Given $A\in\matcon$, we define $$1_A(b,a):= \begin{cases} 1,\text{ the singleton, if $b=a$};\\ \emptyset,\text{ if $b\neq a$}. \end{cases}$$ It is clear that $M 1_A\overset{r}{\cong} M$ and $1_B M\overset{l}{\cong} M$.
     A monad $T$ on an object $A$ in this bicategory is precisely a category with set of objects $A$. Let's shed some light on what is happening. Diagrammatically $T$ is determined by the following commutative diagrams: $$\xymatrix{ (TT)T\ar@{}|{\cong}[r]\ar[d]_{\mu T} & T(TT)\ar[r]^(.55){T\mu} & TT\ar[d]^\mu\\ TT\ar[rr]_\mu & & T }$$ and $$\xymatrix{ 1_AT\ar[r]^{\eta T}\ar[dr]_l & TT\ar[d]^\mu & T1_A\ar[l]_{T\eta}\ar[ld]^r\\ & T &, }$$ where $l$ and $r$ are the isomorphisms above induced by the product and the terminal object of $\con$. Now, what $T$ does is to assign to each pair of elements $b,a\in A$ a set $T(b,a)$ of arrows, to give a composite to each composable pair by means of $\mu$, to choose an identity for each element $a\in A$ via $\eta$ and finally, with the previous diagrams, to make composition associative and make composition with an identity a unit law for the arrows of the small category $A$ with objects its elements and its arrows in $T(b,a)$; put differently, a monad in $\matcon$ is a category.
     Now let $M$ and $E$ be two categories with set of objects $A$ and let $\lambda:ME\Rightarrow EM$ be a distributive law of $M$ over $E$; $\lambda$ yields a function $$\xymatrix{ ME(a,c)\ar[r]^{\lambda(a,c)} & EM(a,c) }$$ for every pair $(a,c)\in A\times A$. Thus we have a family of functions $$(\xymatrix{ M(a,b)\times E(b,c)\ar[r] & \sum_{i\in A}E(a,i)\times M(i,c) })_{b\in A}.$$ If we write $m:\xymatrix{a\ \ar@{>->}[r] & b}$ for an arrow in $M$ and $e:\xymatrix{b\ar@{->>}[r] & c}$ for an arrow in $E$, then $\lambda$ yields an object $e_\lambda m$ and a composable pair $(e_\alpha m,e_\beta m)$ as shown by: $$\xymatrix{ a\; \ar@{>->}[r]^m\ar@{->>}[d]_{e_\alpha m}\ar@{}|{\triangleleft\scriptscriptstyle{\dashv}}[dr] & b\ar@{->>}[d]^e\\ e_\lambda m\;\ar@{>->}[r]_{e_\beta m} & c. }$$ We call a diagram such as this a $\lambda$-square. Consider the compatibility diagrams for the distributive law $\lambda:ME\Rightarrow EM$: \begin{equation} \vcenter{\xymatrix{ & ME\ar[dd]^\lambda\\ E\ar[ru]^{1E}\ar[dr]_{E1} & \\ & EM, }}\quad\text{compatibility of $\lambda$ with $1$ (CuM)}\notag \end{equation} \begin{equation} \vcenter{\xymatrix{ & ME\ar[dd]^\lambda\\ M\ar[ur]^{M1}\ar[dr]_{1 M} &\\ & EM, }}\quad\text{compatibility of $\lambda$ with $1$ (CuE)}\notag \end{equation} \begin{equation} \vcenter{\xymatrix{ MME\ar[r]^{M\lambda}\ar[d]_{\bullet E} & MEM\ar[r]^{\lambda M} & EMM\ar[d]^{E\bullet}\\ ME\ar[rr]_\lambda & & EM, }}\quad\text{compatibility of $\lambda$ with $\bullet$ (CmM)}\notag \end{equation} \begin{equation} \vcenter{\xymatrix{ MEE\ar[r]^{\lambda E}\ar[d]_{M\bullet} & EME\ar[r]^{E\lambda} & EEM\ar[d]^{\bullet M}\\ ME\ar[rr]_\lambda & & EM, }}\quad\text{compatibility of $\lambda$ with $\bullet$ (CmE)}\notag \end{equation} where we denote by 1 the transformations that provide identities and by $\bullet$ the transformations that provide the composites (the units and the multiplications of the monads $M$ and $E$). Now, in terms of $\lambda$-squares, the compatibility of $\lambda$ with the units is expressed by $$\xymatrix{ b\;\ar@{>->}[r]^{1_b}\ar@{->>}[d]_e\ar@{}|{\triangleleft\scriptscriptstyle{\dashv}}[dr] & b\ar@{->>}[d]^e & & a\;\ar@{>->}[r]^m\ar@{->>}[d]_{1_a}\ar@{}|{\triangleleft\scriptscriptstyle{\dashv}}[dr] & b\ar@{->>}[d]^{1_b}\\ c\;\ar@{>->}[r]_{1_c} & c & & a\;\ar@{>->}[r]_m & b; }$$ i. e., CuM and CuE state that ${1_b}_\lambda e=c,m_\lambda 1_b=a$ and that

1. ${1_b}_\alpha e=e$,
2. ${1_b}_\beta e=1_c$,
3. $m_\alpha 1_b=1_a$,
4. $m_\beta 1_b=m$.

Chasing the upper right path of the diagram of CmM yields $$\xymatrix{ a\;\ar@{>->}[rr]^m\ar@{->>}[d]_{m_\alpha(n_\alpha e)}\ar@{}|{\triangleleft\scriptscriptstyle{\dashv}}[drrrr] & & b\;\ar@{>->}[rr]^n & & b''\ar@{->>}[d]^e\\ m_\lambda(n_\alpha e)\;\ar@{>->}[rr]_{m_\beta(n_\alpha e)} & & n_\lambda e\;\ar@{>->}[rr]_{n_\beta e} & & c; }$$ chasing the left lower path, $(mn)_\lambda e=m_\lambda(n_\alpha e)$ and

5. $(mn)_\alpha e=m_\alpha(n_\alpha e)$,
6. $(mn)_\beta e=m_\beta(n_\alpha e)\bullet n_\beta e$.

Similarly, for CmE, if we chase the upper right paht of the diagram, then $$\xymatrix{ a\;\ar@{>->}[rr]^m\ar@{->>}[d]_{m_\alpha e}\ar@{}|{\triangleleft\scriptscriptstyle{\dashv}}[ddrr] & & b\;\ar@{->>}[d]^e\\ m_\lambda e\ar@{->>}[d]_{(m_\beta e)_\alpha f} & & b'\ar@{->>}[d]^f\\ (m_\beta e)_\lambda f\;\ar@{>->}[rr]_(.55){(m_\beta e)_\beta f} & & c; }$$ chasing the left lower path, $m_\lambda(ef)=(m_\beta e)_\lambda f$ and

7. $m_\alpha(ef)=m_\alpha e\bullet(m_\beta e)_\alpha f$,
8. $m_\beta(ef)=(m_\beta e)_\beta f$.

If $A$ has a single element, then $M$ and $E$ are monoids, the equations (I)-(VIII) are the equations (i)-(viii) and the equalities for objects are trivial.
     From the general theory of distsributive law [1], $\lambda$ induces a composite monad $E_\lambda M$, a category with set of objects $A$ in which an arrow from $a$ to $c$ is given by specifying a third object $b$ and a pair $$\xymatrix{ a\ar@{->>}[r]^e & b\;\ar@{>->}[r]^m & c }$$ with $e$ in $E$ and $m$ in $M$; i. e., the arrows in $E_\lambda M$ are described as a formal composition $e\circ m$. The composition in $E_\lambda M$ is given by the multiplication for the monad $E_\lambda M$; namely, by $$\xymatrix{ & & EEM\ar[dr]^{\bullet M} & \\ EMEM \ar[r]^{E\lambda M} & EEMM\ar[ur]^{EE\bullet}\ar[dr]_{\bullet MM}\ar[rr]^(.55){\bullet\;\bullet} & & EM\\ & & EMM\ar[ur]_{E\bullet} &, }$$ thus the composite of $a\overset{e}{\twoheadrightarrow} b\overset{m}{\rightarrowtail} c$ and $c\overset{f}{\twoheadrightarrow} d\overset{n}{\rightarrowtail} x$ is given by $$\xymatrix{ a\ar@{->>}[r]^e\ar@{->>}[dr]_{e\bullet m_\alpha f} & b\;\ar@{>->}[r]^m\ar@{->>}[d]_(.35){m_\alpha f}\ar@{}|{\triangleleft\scriptscriptstyle{\dashv}}[dr] & c\ar@{->>}[d]^f\\ & m_\lambda f\;\ar@{>->}[r]_(.6){m_\beta f}\ar@{>->}[dr]_{m_\beta f\bullet n} & d\ar@{>->}[d]^n\\ & & x; }$$ i. e., $(e\circ m)\bullet(f\circ n)=(e\bullet m_\alpha f)\circ(m_\beta f\bullet n)$.
     It is easy to check that the unit of the composite monad $E_\lambda M$ characterizes the identities in $E_\lambda M$ by $$a\overset{1_a}{\twoheadrightarrow}a\overset{1_a}{\rightarrowtail}a.$$
     The morphisms of monads $1M:M\rightarrow E_\lambda M$ and $E1:E\rightarrow E_\lambda M$ are given by $m\mapsto 1\circ m$ and $e\mapsto e\circ 1$, resp. The middle unitary law yields that for every $e\circ m$ in $E_\lambda M$, $$(e\circ 1)\bullet(1\circ m)=e\circ m$$ holds.
     Note that if $M$ and $E$ are monoids, the composition in $E_\lambda M$ is the multiplication defined in the case of the Zappa-Szép product for groups.
     There is a greater generalization of the Zappa-Szép product in [2]; however, things there are done more à la Ehresmann.

Strict factorization systems

Given a category $C$ with $\ob(C)=:A$, a strict factorization system on $C$ is a pair of subcategories $S:=(E,M)$ of $C$ such that $\ob(M)=\ob(E)=\ob(C)$ and such that for every $f$ in $C$, there is a unique factorization $f=e_fm_f$ with $e_f$ in $E$ and $m_f$ in $M$. Consider $M$ and $E$ as monads on $A$ in $\matcon$. The pair $(E,M)$ induces a distributive law $\lambda_S:ME\Rightarrow EM$; indeed, define $\lambda_S$ by $$\xymatrix{ ME\ar[r]^{\lambda_S} & EM }$$ $$\xymatrix{ a\ar[r]^n & b\ar[r]^f & c\ar@{|->}[r] & a\ar[r]^{e_{n\cdot f}} & i\ar[r]^{m_{n\cdot f}} & c. }\ $$ Its compatibility with the unit of $M$ is obvious, for $m\cdot 1_b=m$. Now, the upper right path of the compatibility diagram of $\lambda_S$ with respect to the multiplication of $M$ (check the diagram above) produces the following diagram: $$\xymatrix{ a\ar[rr]^f\ar[dr]_{e_{f\cdot e_{g\cdot h}}} & & b\ar[r]^g\ar[dr]_(.45){e_{g\cdot h}} & c\ar[r]^h & d\\ & k\ar[rr]_{m_{f\cdot e_{g\cdot h}}} & & j\ar[ur]_{m_{g\cdot h}} &\ ; }$$ so $f\cdot g\cdot h$ has as factorization $e_{f\cdot e_{g\cdot h}}\cdot m_{f\cdot e_{g\cdot h}}\cdot m_{g\cdot h}$, which is unique; whence, the left lower path in the compatibility diagram of $\lambda_S$ with respect to the multiplication of $M$ produces the same result, and in consequence the compatibility of $\lambda_S$ with $M$ is satisfied. Similarly, the compatibility of $\lambda_S$ with $E$ is satisfied.
     Conversely, let $\lambda:ME\Rightarrow EM$ be a distributive law in $\matcon$, and consider the subcategories of $E_\lambda M$ defined by $$\lambda E:=\{e\circ 1\mid e\in E\}\quad\text{y}\quad M\lambda:=\{1\circ m\mid m\in M\}.$$ Each of these subcategories contains all the identities of $E_\lambda M$ and thus each contains all the objects of $E_\lambda M$. By the middle unitary law, $e\circ m$ is factorized as $(e\circ 1)\bullet(1\circ m)$. It is clear that this factorization is unique, and therefore $(\lambda E, M\lambda)$ is a strict factorization system $S_\lambda$ for $E_\lambda M$.
     We leave this correspondence just right there; i. e., we won't show that $S_{(-)}$ and $\lambda_{(-)}$ are biequivalences inverse of each other, because this is not our objetive right now.
     The way a distributive law is induced by the Zappa-Szép product in the group case makes us wonder whether a distributive law in $\matcon$ participates in a correspondence of that kind...

References

[1] Beck, J. [1969]: Distributive laws, Seminar on Triples and Categorical Homological Theory, ETH 1966/67, 80, 119-140 (1969). ↩
[2] Brin, M. G. [2005]: On the Zappa-Szép Product, Communications in Algebra, 33, 393-424 (2005). ↩
[3] Rosebrugh, R., Wood, R. J. [2002]: Distributive laws and factorization, Journal of Pure and Applied Algebra, 175(1-3), 327-353 (2002). ↩
[4] Takeuchi, M. [1981]: Matched pairs of groups and bismash products of Hopf algebras, Communications in Algebra, 9(8), 841-882 (1981). ↩

No hay álgebras booleanas completas libres sobre conjuntos infinitos

Mac Lane en su Categories for the working mathematician da dos ejemplos de funtores continuos con dominio pequeño-completo que no satisfacen la condición conjunto solución del Teorema de Freyd del Funtor Adjunto. Tal teorema dice lo siquiente.

Teorema de Freyd del Funtor Adjunto. Si $A$ es una categoría pequeño-completa con homoconjuntos pequeños, entonces un funtor $G:A\rightarrow X$ tiene adjunto izquierdo si y sólo si preserva todo límite pequeño y satisface lo siguiente.
     Condición conjunto solución. Para todo objeto $x\in X$, existe un conjunto pequeño $I$ y una familia de flechas $f_i:x\rightarrow Ga_i$ indexada por $I$ tal que toda flecha $h:x\rightarrow Ga$ se puede escribir como la composición $h=Gt\circ f_i$ para algún índice $i$ y alguna $t:a_i\rightarrow a$.

El ejemplo que me sorprendió e intrigó fue el segundo. Este empieza diciendo: “Dado un conjunto numerable $D$, uno puede construir un álgebra booleana completa arbitrariamente grande generada por $D$”. Tal afirmación la demostraron primero Gaifman y Hales de manera independiente utilizando argumentos de la lógica infinitaria y luego la demostró Solovay haciendo uso del álgebra abierta regular. La demostración de Solovay es mucho más simple. Hago un recuento detallado de esta demostración.

Sea $\kappa$ un cardinal infinito y dótese a $\kappa$ de la topología discreta. Considérese ahora a $\kappa^\omega$ con la topología producto. Sea $X:=\mathbf{AR}(\kappa^\omega)$, el álgebra abierta regular del espacio $\kappa^\omega$. Notemos que todo cerrabierto es regular; de donde, los $$A_{n,\eta}:=\{f\in \kappa^\omega\mid fn=\eta\},$$ con $\eta<\kappa$, son elementos de $X$: el conjunto $\{\eta\}$ es cerrabierto de $\kappa$; más aún, los $A_{n,\eta}$ son subbásicos de $\kappa^\omega$. Tenemos que la familia $\{A_{n,\eta}\mid n<\omega,\eta<\kappa\}$ genera a $X$; en efecto, sea $V\in X$; entonces, $V$ es unión de intersecciones finitas de $A_{n,\eta}$ s, digamos, $V=\bigcup V_i$; de donde, como $V$ es abierto regular y $\bigvee V_i=\mathrm{IntCl}(\bigcup V_i)$ es el abierto regular más pequeño que contiene a $\bigcup V_i$, tenemos que $V=\bigvee V_i$.
     Ahora, la cardinalidad de $X$ es al menos $\kappa$, pues si $\eta<\eta'<\kappa$, entonces $A_{0,\eta}$ y $A_{0,\eta'}$ son distintos.
     Antes de seguir, notemos que si $B\subseteq\kappa^\omega$ y $B$ depende de un número finito de coordenadas, entonces $B$ es cerrabierto. En efecto, que $B$ dependa de un número finito de coordenadas significa que $\exists\,n_1,\ldots,n_m\in\mathbb{N}$ $$B=\{f\in\kappa^\omega\mid R(fn_1,\ldots,fn_m)\},$$ donde $R\subseteq\kappa^m$ y $m\in\mathbb{N}$. Tenemos que $R$ es cerrabierto de $\kappa^m$ y $B=p^{-1}R$ con $p:\kappa^\omega\rightarrow\kappa^m$ definida como $pf:=(fn_1,\ldots,fn_m)$; $p$ es continua.
     Prosigamos. Dados $n,m<\omega$, defínase $$B_{n,m}:=\{f\in\kappa^\omega\mid fm\leq fn\}.$$ Entonces, por la observación anterior, $B_{n,m}$ es cerrabierto luego abierto regular. Afirmamos que los $B_{n,m}$ generan a $X$. En efecto, sea $Y$ la subálgebra completa más pequeña de $X$ que contiene a $\{B_{n,m}\mid n,m<\omega\}$. Bastará demostrar que $A_{n,\eta}\in Y$ para todo $n<\omega$ y para todo $\eta<\kappa$; hagámoslo por inducción sobre $\eta$; supongamos entonces que para $m<\omega$ y $\xi<\eta$ se tiene que $A_{m,\xi}\in Y$. Ahora, sean $$D_{n,\eta}:=\{f\in\kappa^\omega\mid fn<\eta\}\quad\text{y}\quad E_{n,\eta}:=\{f\in\kappa^\omega\mid fn\leq\eta\}$$ para $n<\omega$ y $\eta<\kappa$. Lo que queremos hacer es ver que $D_{n,\eta},E_{n,\eta}\in Y$, ya que $$A_{n,\eta}=D_{n,\eta}\cap E_{n,\eta}=D_{n,\eta}\wedge E_{n,\eta}.$$ Como $D_{n,\eta}$ y $E_{n,\eta}$ dependen sólo de una coordeanada, son cerrabiertos; luego, son elementos de $X$. Por inducción, estamos suponiendo que $A_{n,\xi}\in Y$, así que como $D_{n,\eta}$ es abierto regular y $$D_{n,\eta}=\bigcup_{\xi<\eta} A_{n,\xi},$$ entonces $D_{n,\eta}=\bigvee_{\xi<\eta} A_{n,\xi}\in Y$.
     Por otro lado, dados $n,m<\omega$, $$C_{m,n}:=\{f\in\kappa^\omega\mid fn\leq fm\text{ o }fm<\eta\}$$ es cerrabierto, ya que su definición depende sólo de $m$ y $n$; luego, $C_{m,n}\in X$. Notemos que $$C_{m,n}=B_{m,n}\cup\bigvee_{\xi<\eta} A_{m,\xi}.$$ Nuevamente, como $C_{m,n}\in X$, se tiene que $C_{m,n}=B_{m,n}\vee\bigvee_{\xi<\eta} A_{m,\xi}$. De donde, $C_{m,n}\in Y$. Demostremos que $E_{n,\eta}=\bigwedge_{m<\omega} C_{m,n}$; es decir, que $E_{n,\eta}=\mathrm{Int}(\bigcap_{m<\omega} C_{m,n})$.
     Ahora, dados $n<\omega$ y $f\in\kappa^\omega$, los $$U(n,f):=\{h\in\kappa^\omega\mid\forall\,m\leq n\; hm=fm\}$$ forman una base para $\kappa^\omega$ (es claro que $U(n,f)\in\tau(\kappa^\omega)$). En efecto, sea $\langle U_{i_1},\ldots,U_{i_m}\rangle$ un básico de $\kappa^\omega$ y sea $f\in\langle U_{i_1},\ldots,U_{i_m}\rangle$. Sin pérdida de generalidad, supóngase que $i_1<\cdots < i_m$. Entonces, $U(i_m,f)\subseteq\langle U_{i_1},\ldots,U_{i_m}\rangle$.
     Demostremos que $E_{n,\eta}=\mathrm{Int}(\bigcap_{m<\omega} C_{m,n})$. Sea $g\in E_{n,\eta}$; es decir, $gn\leq\eta$. Entonces, $U(n,g)\subseteq\bigcap_{m<\omega} C_{m,n}$, pues si $h\in U(n,g)$, entonces dado $m<\omega$, si $hn\leq hm$ entonces $h\in C_{m,n}$ y si $hm< hn=gn$ entonces $hm<\eta$ y $h\in C_{m,n}$; luego, $h\in\bigcap_{m<\omega} C_{m,n}$.
     Recíprocamente, sea $g\in\mathrm{Int}(\bigcap_{m<\omega} C_{m,n})$ y supóngase que $gn>\eta$. Como $g\in\mathrm{Int}(\bigcap_{m<\omega} C_{m,n})$, existe $N\in\omega$ tal que $U(N,g)\subseteq\bigcap_{m<\omega} C_{m,n}$. Como $U(k,g)\subseteq U(N,g)$ para todo $k\geq N$, podemos suponer que $N\geq n$. Ahora defínase $h:\omega\rightarrow\kappa$ como $$hm:=\begin{cases} gm &\text{si $m\leq N$,}\\ \eta &\text{si $m > N$}. \end{cases}$$ Entonces, $h\in U(N,g)\subseteq\bigcap_{m<\omega} C_{m,n}$; de donde, $h\in C_{N+1}$; es decir, $hn< h(N+1)$ o $h(N+1)<\eta$. Sin embargo, $h(N+1)=\eta< gn=hn$ y $h(N+1)\geq\eta$ !! Luego, $g\in E_{n,\eta}$. Así que $E_{n,\eta}\in Y$.
     Por lo tanto, $X$, que tiene cardinalidad por lo menos $\kappa$, es generado por $\{B_{n,m}\mid n,m<\omega\}$, que es numerable.
     Así que tenemos el siguiente teorema.

Teorema (Solovay). Sea $\kappa$ un cardinal infinito. Si $\kappa$ tiene la topología discreta y $\kappa^\omega$ la topología producto, entonces $\mathbf{AR}(\kappa^\omega)$ es un álgebra boolena completa numerablemente generada con cardinalidad por lo menos $\kappa$.

Y tenemos el siguiente corolario.

Corolario. No hay álgebras booleanas completas libres sobre conjuntos infinitos.

Demostración. Sean $X$ un conjunto infinito, $FX$ el álgebra completa libre generada por $X$ y $\eta_X:X\rightarrow FX$ la función con la propiedad universal de álgebra libre. Sea $\kappa$ un cardinal mayor que $|FX|$. Entonces, por el teorema anterior, $\mathbf{AR}(\kappa^\omega)$ tiene un conjunto numerable $Y$ de generadores. Sea $f:X\rightarrow Y$ una función suprayectiva. Entonces, si $g:FX\rightarrow\mathbf{AR}(\kappa^\omega)$ es homomorfismo de álgebras booleanas completas y $g\circ\eta_X=f$, entonces $gFX$ es una subálgebra completa de $\mathbf{AR}(\kappa^\omega)$ que incluye a $Y$; de donde, $g$ sería suprayectiva y $\kappa\leq|\mathbf{AR}(\kappa^\omega)|\leq|FX|<\kappa$ !!