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\begin{document}%Alpha, Beta, Gamma, Delta, Epsilon, Zeta, Eta, Theta, Iota, Kappa, Lambda, Mu, Nu, Xi, Omicron, Pi, Rho, Sigma, Tau, Upsilon, Phi, Chi, Psi, Omega

\centerline{\Huge Natural ways of mapping subsets to subsets}\vskip20pt

\centerline{Pierre-Yves Gaillard}\vskip10pt

{\footnotesize If $X$ is a set, $M$ its monoid of self-maps and $P$ its power set, then $P$ can be viewed as a left $M$-set ${}_MP$ or as a right $M$-set $P_M$. We compute the monoids $\End{}_MP$ and $\End P_M$.}\vskip10pt

Let $X$ be a set, $M=X^X$ its monoid of self-maps (that is, $M=\{f:X\to X\}$) and $P$ its power set (that is, $P=\{A\ |\ A\subset X$\}). Then $P$ has a left $M$-set structure given by 
$$
fA=f_*A=\{fa\ |\ a\in A\}
$$ 
and a right $M$-set structure given by 
$$
Af=f^*A=f^{-1}A=\{x\in X\ |\ fx\in A\}.
$$ 

We denote these two $M$-sets by ${}_MP$ and $M$-set $P_M$ respectively. Our purpose is to compute the monoids $\End{}_MP$ and $\End P_M$. In the sequel we denote $fA$ and $Af$ by $f_*A$ and $f^*A$ respectively.

Define the maps $\alpha,\beta,\gamma,\delta:P\to P$ by the formulas 
$$
\alpha A=A,\quad\beta A=\varnothing,\quad\gamma A=X\setminus A,\quad\delta A=A.
$$ 
(Here $\varnothing$ is the empty set and $X\setminus A$ the complement of $A$ in $X$.)

\begin{thm}\label1
We have $\End{}_MP=\{\alpha,\beta\}$ and $\End P_M=\{\alpha,\beta,\gamma,\delta\}$.
\end{thm} 

It suffices to show $\End{}_MP\subset\{\alpha,\beta\}$ and $\End P_M\subset\{\alpha,\beta,\gamma,\delta\}$. Indeed, the converse inclusions are clear. Moreover, to prove Theorem~\ref{1} we can, and do, assume that $X$ has at least two elements.

\section{The monoid $\End{}_MP$}

The $M$-sets considered in this section are \textbf{left} $M$-sets.  

\begin{lem}\label{1.1} 
If $\ee:P\to P$ is a morphism of $M$-sets, then $\ee\in\{\alpha,\beta\}$. 
\end{lem} 

\begin{proof} 
This will follow immediately from the four steps below. 

Step~1: We have $\ee\oo=\oo$. Proof: The equalities $\ee\oo=\ee f_*\oo=f_*\ee\oo$ hold for all $f$ in $M$. This implies $\ee\oo=\oo$. 

Note: In view of Step~1 it suffices to show that we have either $\ee A=\oo$ for all $A$ in $P$, $A\ne\oo$, or $\ee A=A$ for all $A$ in $P$, $A\ne\oo$. 

Step~2: We have $\ee X\in\{\oo,X\}$. Proof: Since $\ee f_*X=f_*\ee X$ for all $f$ in $M$, we get $\ee X=f_*\ee X$ for all surjection $f:X\twoheadrightarrow X$, and thus $\ee X\in\{\oo,X\}$.

Step~3: If $\ee X=\oo$, then $\ee=\beta$. Proof: We have $\ee f_*X=f_*\oo=\oo$ for all $f$ in $M$. This entails $\ee A=\oo$ for all $A$ in $P$, $A\ne\oo$, hence $\ee=\beta$ by the Note.

Step~4: If $\ee X=X$, then $\ee=\alpha$. Proof: We have $\ee f_*X=f_*X$ for all $f$ in $M$. This implies $\ee A=A$ for all $A$ in $P$, $A\ne\oo$, hence $\ee=\alpha$ by the Note.
\end{proof} 

\section{The monoid $\End P_M$}

The $M$-sets considered in this section are \textbf{right} $M$-sets. 

Let $\ee:P\to P$ be a morphism of $M$-sets. We must show: 

\begin{lem}\label{2.1} 
If $\ee:P\to P$ is a morphism of $M$-sets, then $\ee\in\{\alpha,\beta,\gamma,\delta\}$. 
\end{lem} 

\begin{proof} 
We generalize slightly the notation used so far. Set $2:=\{0,1\}$. For all set $Y$ write $Y^*$ for the set of subsets of $Y$ and identify $Y^*$ to the set $2^Y$ of all maps $Y\to2$ by attaching to $A\subset X$ the map $f$ defined by $fx=1$ if and only if $x\in A$. Moreover we associate with a map $g:Z\to Y$ the map $g^*:Y^*\to Z^*$ defined by $g^*A=g^{-1}(A)$. Note that, if $f:Y\to2$ is the map attached to $A$ described above, then the map $Z\to2$ attached to $g^*A$ is $f\circ g$, so that it is natural to denote this map by $g^*f$. 

Claim: If $A$ is a nonempty proper subset of $X$, then $\ee A\in\{\oo,A,X\setminus A,X\}$. 

Proof of the claim. Let $A$ be as above and $f:X\to2$ the map attached to $A$. Since for any $B\subset2$ we have $f^*B\in\{\oo,A,X\setminus A,X\}$, it suffices to show that $\ee A$ is of the form $f^*B$ with $B\subset2$. Pick $x_0$ in $X\setminus A$ and $x_1$ in $A$, and define $g:2\to X$ by $g(i)=x_i$. Then $f\circ g$ is the identity of $2$, and we get 
$$
\ee A=\ee f=\ee(f\circ g\circ f)=\ee((g\circ f)^*(f))=(g\circ f)^*(\ee f)=f^*(g^*\ee f).
$$ 
This proves the claim. 

Recall that $A$ is a nonempty proper subset of $X$. Let $C$ be any subset of $X$, define $h\in M$ by $hx=x_1$ if $x\in C$ and $hx=x_0$ if $x\in X\setminus C$, and observe the equalities $h^*A=C$ and $\ee C=\ee h^*A=h^*\ee A$. The claim implies $\ee A\in\{\oo,A,X\setminus A,X\}$. If $\ee A=\oo$ then $\ee C=h^*\oo=\oo$. If $\ee A=A$ then $\ee C=h^*A=C$. The cases $\ee A=X\setminus A$ and $\ee A=X$ are similar. This shows $\ee\in\{\alpha,\beta,\gamma,\delta\}$, as desired. 
\end{proof} 

Now Theorem~\ref1 follows from Lemmas~\ref{1.1} and \ref{2.1}.

%If X is a set, M its monoid of self-maps and P its power set, then P can be viewed as a left M-set L or as a right M-set R. We compute the monoids End L and End R.

\end{document}%

pierre.yves.gaillard@gmail.com

algebra

Natural ways of mapping subsets to subsets

Pierre-Yves Gaillard

If X is a set, M its monoid of self-maps and P its power set, then P can be viewed as a left M-set L or as a right M-set R. We compute the monoids End L and End R.

2 pages