Is $P(M)$ internally complete in the category of $M$-sets? Let $\textsf{MSet}$ be the category of $M$-sets. The powerset of $M$, $P(M)$, is an $M$-set with action given by division:


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*$iX = \{j \mid j\circ i \in X\}$


Suppose that $B$ is a Boolean algebra with an $M$-action that preserves the Boolean operations (such as $P(M)$). Say that $B$ is internally complete iff:


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*For any $M$-set $A$, the diagonal map $k:B\to B^A$ (the exponential transpose of $\pi_1$) has a left adjoint $\bigvee_A:B^A \to B$.


That is, for any equivariant $f:M\times A \to B$ and $b\in B$:


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*$\bigvee_A f \subseteq b$ if and only if, for all $j\in M, a\in A$, $f(j,a) \subseteq jb$


I'd like to know whether $P(M)$ is internally complete (whatever $M$ is), and if it is, what the explicit construction of $\bigvee$ is.
 A: Yes, and the construction is pretty much immediate from your explicit characterization.  Namely, given $f:M\times A\to P(M)$, define $c\in P(M)$ to be the intersection of all $b\in P(M)$ such that for all $j\in M$ and $a\in A$, $f(j,a)\subseteq jb$.  Since the action of $M$ on $P(M)$ preserves arbitrary intersections, it is clear that this $c$ is the least element of $P(M)$ satisfying this condition on $b$, so $c$ satisfies the definition of $\bigvee_A f$.
To prove that $f\mapsto\bigvee_A f$ preserves the action of $M$, let me first give a more concrete description of $\bigvee_A f$.  If $b\in P(M)$, let me write $bi^{-1}$ for what you have written $ib$.  This notation should be intuitive, since it is exactly the inverse image of $b$ under right-multiplication by $i$.  Let me also write $bj=\{xj:x\in b\}$ for the image of $b$ under right-multiplication by $j$.
We can now explicitly describe $\bigvee_A f$ as a set.  We want the smallest $b\in P(M)$ such that $f(j,a)\subseteq bj^{-1}$ for all $(j,a)$.  Note that $f(j,a)\subseteq bj^{-1}$ iff $f(j,a)j\subseteq b$, so the smallest such $b$ is simply the union $\bigcup_{j,a}f(j,a)j$.  Now note that if $x\in f(j,a)$, then $1\in f(j,a)x^{-1}=f(xj,xa)$, so if $y=xj\in f(j,a)j$ then $1\in f(xj,xa)$.  Conversely, if $1\in f(y,a)$ for some $a$, then $y\in f(y,a)y\subseteq\bigvee_f A$.  So we can also describe $\bigvee_A f$ as the set of all $y\in M$ such that $1\in f(y,a)$ for some $a\in A$.
Now let $g=if$ for some $i\in M$, so $g(j,a)=f(ji,a)$ for all $(j,a)$.  Then $\bigvee_A g$ is the set of all $y\in M$ such that $1\in g(y,a)=f(yi,a)$ for some $a\in A$.  But this is exactly the set of $y$ such that $yi\in \bigvee_A f$.  Thus $\bigvee_A g=\left(\bigvee_A f\right)i^{-1}$, as desired.
