$ \newcommand{\eps}{\epsilon} \newcommand{\op}{^\text{op}} \newcommand{\id}{\text{id}} \newcommand{\set}{\text{Set}} \newcommand{\xra}{\xrightarrow} $Can someone help me to understand how algebras over the monad $T:=PP\op$, with $P:\set\op\to\set$ being the powerset functor taking maps to inverse images, look like? Note that this is not a duplicate of understanding the algebras of the powerset monad, which concerns the monad $P':\set\to\set$ which takes maps to direct images. This question was also flagged as duplicate of what are the algebras of the double powerset monad, which is very close, although I am more interested in sets than in Boolean algebras.
This is what I got so far:
Let $P:\set\op\to\set$ be the powerset functor with left adjoint $P\op:\set\to\set\op$ where the unit $\eta(U):U\to PP\op U$ maps $x\in U$ to the set $\{X\subseteq U:x\in X\}$ and the counit $\eps$ is just the corresponding arrow in the opposite category. So we can construct a monad $T:=PP\op$ from this adjoint pair as usual with $\mu(U):TTU\to TU$ defined via $$TTU=PP\op PP\op U\xra{P\eps(P\op U)} PP\op U=TU$$ and I wonder how the algebras over this monad look like.
Recall that an algebra over $U$ is a map $a:TU\to U$ such that
- $U\xra{\eta(U)}TU\xra{a} U=\id(U)$ and
- $TTU\xra{\mu(U)}TU\xra{a}U=TTU\xra{T(a)}TU\xra{a}U$.
(1.) is relatively straight forward, the set of all $X$ containing $x$ must be mapped back to $x$ by $a$.
For (2.), note is that both, $\mu(U)$ and $T(a)$, are of the shape $Pf$ for some $f:P\op U\to P\op TU$, so given some $W\in TTU$, they return all $X\in P\op U$ such that $f(X)\in W$. This relation can further simplified by noting that if $a$ already satisfies (1.), $f$ is in both cases injective. For $\mu(U)$ we have $f_0:=e(P\op U)$ mapping $X$ to the principal filter containing all $Y\in TU$ with $X\in Y$, which is clearly injective, and for $T(a)$ we have $f_1:=P\op(a)$ which maps $X$ to the set of $Y$ with $a(Y)\in X$, which is also injective since $\eta(U)$ is injective and $a$ satisfies (1.). Let $W_i$ be the image of $P\op U$ under $f_i$ and let $Y_{0\cap 1}$ be the inverse image of $W_0\cap W_1$.
So $a$ is only allowed to care for elements in $Y_{0\cap 1}$, since the others can be independently added or removed under $f_0$ and $f_1$ by changing $W$ accordingly.
We can see that $W_0$ is what I call a disjoint free antichain, for $Z_0\neq Z'_0\in W_0$, let's say $Z_0$ is the principal filter of $X$ and $Z'_0$ the principal filter of $X'$, neither $Z_0\subset Z'_0$ nor $Z'_0\subset Z_0$ nor $Z_0\cap Z'_0=\emptyset$ since only $Z_0$ contains $\{X\}$, only $Z'_0$ contains $\{X'\}$ and both contain $\{X,X'\}$. On the other hand, if $X\subset X'$, clearly $f_1(X)\subseteq f_1(X')$ and by (1.) also $f_1(X)\subset f_1(X')$, a contradiction. Similar with $X\cap X'=\emptyset$, so it follows that $Y_{0\cap 1}$ must inherit the disjoint free antichain property from $W_0$.
But from (1.) we can also see that if $x,x'\in U$ with $x\neq x'$, we must have $X,X'\in Y_{0\cap 1}$ such that $X$ contains $x$ but not $x'$ and $X'$ contains $x'$ but not $x$, otherwise $\eta(U)x\cap Y_{0\cap 1}=\eta(U)x'\cap Y_{0\cap 1}$ and $a$ would not be able to differentiate the two sets.
So all in all we have this strange system of disjoint free antichains which can differentiate between all elements. I don't know if there is a special name for this property.
