# Adjoint functors for the power set monad

There is the power set functor, $T$, which gives raise to a monad: For a set $X$, we set $TX:=\mathcal P(X)$ and for $f:X\to Y$, we set $T(f):=S\mapsto f(S)$, where $f(S)$ denotes the direct image. The unit maps to the singleton $x\mapsto\{x\}$ and the second map gives a union, as in $\{\{a,b\},\{c\},\{\{d,e\},f\}\}\mapsto \{a,b,c,\{d,e\},f\}$. There is also the very similar list-monad, where $TX:=\bigcup_nX^n=X\cup (X\times X)\cup (X\times\dots$

What are the adjoint functor pairs $F,G$ for, so that $T=FG$ and what is the co-unit? I know there are at least "the two extremal solutions" - are they used somewhere?

• In the powerset monad that I'm familiar with, the contravariant powerset functor is both the $F$ and the $G$, and the counit is the unit in $\mathbf{Set^{op}}$. Sep 11 '14 at 18:03
• edit: For the list monad, it probably should be $X^0\cup X\cup(X\times\dots$. Sep 11 '14 at 20:20
• @MaliceVidrine: I'm confused: If both $F$ and $G$ have the power set operation as object map, isn't then $T$ the twice applied power set operation? Sep 11 '14 at 20:52
• Yes, though I've seen it referred to simply as the "powerset monad". It sounds like there are others I'm not familiar with, though. Sep 12 '14 at 1:02
• The "twice applied power set operation" cannot be given the structure of a monad: mathstat.dal.ca/mfps2018/preproc/paper_4.pdf So, it's impossible to have an adjunction where both left and right adjoint are the covariant power set functor. Jun 9 '18 at 18:20

I prefer to write $F$ for the left-adjoint. So we have $T = GF$. You can consider for $F$ the free functor from the category of sets to the category of complete semi-lattices or the free functor from the category of sets to the category of free complete semi-lattices. The functor $G$ is the forgetful functor that takes a complete semi-lattice (resp. a free complete semi-lattice) and forgets the structure to give the set of its elements.
• And, for the list-monad we get the free/forgetful adjuntion between $\bf{Set}$ and [free] semigroups. Sep 11 '14 at 19:00
• Yes, but it would be less ambiguous to write $TX = \bigcup_{n \geq 1} X^n$. Because if $TX = \bigcup_{n \geq 0} X^n$, we get the free/forgetful adjunction between Set and [free] monoids. It seems to me that usually the expression "list-monad" is used for the latter. Sep 11 '14 at 19:13
• So $F$ maps $\{a,b,c\}$ to $\langle\mathcal{P}(\{a,b,c\}),>\rangle$, where e.g. $\{a,c\}>\{c\}$, and then $G$ on that tuple projects out the first tuple? Sep 11 '14 at 20:27
• Yes, but it can be confusing to write $(\mathcal{P}(\{ a, b, c \} ), >)$ for an object of the category of complete semi-lattices, because morphisms have to preserve joins and not only to be monotone. Sep 11 '14 at 20:41
• Example : $A = \mathcal{P}(\{ a, b \})$ and $B = \mathcal{P}(\{ a, b, c \})$. The function $f$ from $A$ to $B$ defined by $f(\emptyset) = \emptyset$, $f(\{ a \}) = \{ a \}$, $f(\{ b \}) = \{ a, b \}$ and $f(\{ a, b \}) = \{ a, b, c \}$ preserves $\subseteq$ but not $\bigcup$ : it is not a morphism of the category of complete semi-lattices. Sep 11 '14 at 21:17