 Solution

A monad on a category $\mathcal{C}$ consists of an endofunctor $T$, a unit $\eta :1_{\mathcal{C}}\Rightarrow T$, and a multiplication $\mu :T^2\Rightarrow T$ satisfyingIf $F:\mathcal{C}\rightleftarrows \mathcal{D}:G$ is an adjunction with unit $\eta$ and counit $\epsilon$, thendefines the monad induced by an adjunction. The two triangle identities give the unit laws, while naturality of $\epsilon$ gives associativity.

Let $\mathcal{D}$ now be a full subcategory of $[\mathcal{C},\mathcal{C}]$ that contains the identity endofunctor and is closed under composition, and suppose $T$ is terminal in $\mathcal{D}$. There is exactly one natural transformationand exactly oneBoth sides of either unit law are endomorphisms of the terminal object $T$, so they equal $1_T$; both sides of associativity are maps $T^3\to T$, so they are equal as well. This gives a monad, and terminality also makes both structure maps unique. This is the monad structure on a terminal endofunctor.

For a set $A$, let $\mathcal{U}(A)$ be the set of ultrafilters on $A$. A function $f:A\to A^{\prime }$ induces the pushforwardwhich makes $\mathcal{U}$ the ultrafilter functor. There is no ultrafilter on the empty set. Moreover, every ultrafilter on $A\bigsqcup B$ contains exactly one of the complementary summands $A$ and $B$, and restriction gives a unique ultrafilter on that summand. Thereforeso $\mathcal{U}$ preserves finite coproducts.

Let $F:\mathbf{Set}\to \mathbf{Set}$ preserve finite coproducts. For $x\in F(A)$ defineThe decomposition $A=B\bigsqcup (A\setminus B)$ and preservation of coproducts say that $x$ lies in exactly one of the two corresponding images. Thus exactly one of $B$ and its complement belongs to ${\Phi }_A(x)$. Upward closure follows by factoring subset inclusions. If $B,C\in {\Phi }_A(x)$, decompose $A$ into the four disjoint Boolean cells determined by $B$ and $C$. The unique cell containing $x$ must lie inside both sets, so $B\cap C\in {\Phi }_A(x)$. Hence ${\Phi }_A(x)$ is an ultrafilter.

For $f:A\to A^{\prime }$, the decompositionsshow thatThus $\Phi :F\Rightarrow \mathcal{U}$ is natural.

It is the only such natural transformation. For $B\subseteq A$, let ${\chi }_B:A\to \{0,1\}$ be its characteristic function. Sinceand the two ultrafilters on $\{0,1\}$ are the principal ones, naturality with the two singleton inclusions forces any transformation $F\Rightarrow \mathcal{U}$ to send each summand to the corresponding principal ultrafilter. Naturality with ${\chi }_B$ then says that $B$ belongs to the image ultrafilter exactly when $x$ lies in the image of $F(B)\to F(A)$. Hence the transformation must be $\Phi$.

The ultrafilter functor is therefore the terminal finite-coproduct-preserving set endofunctor. The preceding terminal-object argument supplies its unique ultrafilter monad structure.