The codensity monad of the small skeleton of finite sets is the ultrafilter monad

By [L1], an element of ${\mathrm{Ran}}_{J}J(X)$ is exactly a cone over the diagram $(X\downarrow J)\to \mathbf{F}\mathbf{i}\mathbf{n}\mathbf{O}\mathbf{r}\mathbf{d}\to \mathbf{S}\mathbf{e}\mathbf{t}$. Since an object of $(X\downarrow J)$ is a map $f:X\to [n]$, such a cone is exactly a family of chosen elements ${\alpha }_f\in [n]$, one for each $f:X\to [n]$, satisfying the compatibility condition ${\alpha }_{h\circ f}=h({\alpha }_f)$ for every map $h:[n]\to [m]$.

Given such a coherent family $\alpha$, define ${\mathcal{U}}_{\alpha }:=\{A\subseteq X:{\alpha }_{{\chi }_A}=1\}$, where ${\chi }_A:X\to [2]$ is the characteristic function of $A$. Let $!:X\to [1]$ be the unique map, and let $c_0,c_1:[1]\to [2]$ be the constant maps with values $0$ and $1$. Since ${\chi }_{\mathrm{\varnothing }}=c_0\circ !$ and ${\chi }_X=c_1\circ !$, coherence gives ${\alpha }_{{\chi }_{\mathrm{\varnothing }}}=0$ and ${\alpha }_{{\chi }_X}=1$, so $\mathrm{\varnothing }\mathrm{\notin }{\mathcal{U}}_{\alpha }$ and $X\in {\mathcal{U}}_{\alpha }$. If $\tau :[2]\to [2]$ swaps $0$ and $1$, then ${\chi }_{X\setminus A}=\tau \circ {\chi }_A$, so exactly one of $A$ and $X\setminus A$ lies in ${\mathcal{U}}_{\alpha }$ for every $A\subseteq X$. If $A,B\in {\mathcal{U}}_{\alpha }$, define $g:X\to [4]$ by $g(x)=2{\chi }_A(x)+{\chi }_B(x)$, let $p,q:[4]\to [2]$ recover the first and second bits, and let $m:[4]\to [2]$ send only $3$ to $1$. Then $p\circ g={\chi }_A$, $q\circ g={\chi }_B$, and $m\circ g={\chi }_{A\cap B}$, so coherence gives $p({\alpha }_g)=q({\alpha }_g)=1$ and hence ${\alpha }_{{\chi }_{A\cap B}}=m({\alpha }_g)=1$. Thus $A\cap B\in {\mathcal{U}}_{\alpha }$. Finally, if $A\in {\mathcal{U}}_{\alpha }$, $A\subseteq B$, and $B\mathrm{\notin }{\mathcal{U}}_{\alpha }$, then $X\setminus B\in {\mathcal{U}}_{\alpha }$, so $A\cap (X\setminus B)=\mathrm{\varnothing }$ lies in ${\mathcal{U}}_{\alpha }$, contradicting $\mathrm{\varnothing }\mathrm{\notin }{\mathcal{U}}_{\alpha }$. Therefore ${\mathcal{U}}_{\alpha }$ is a filter deciding every subset, hence an ultrafilter by [L2].

Conversely, let $\mathcal{U}$ be an ultrafilter on $X$. For each map $f:X\to [n]$, the fibres $f^{-1}(i)$ form a finite partition of $X$, so [L2] gives a unique index ${\alpha }_f^{\mathcal{U}}\in [n]$ with $f^{-1}({\alpha }_f^{\mathcal{U}})\in \mathcal{U}$. If $h:[n]\to [m]$, then $f^{-1}({\alpha }_f^{\mathcal{U}})\subseteq (h\circ f{)}^{-1}(h({\alpha }_f^{\mathcal{U}}))$, so upward closure of the ultrafilter puts $(h\circ f{)}^{-1}(h({\alpha }_f^{\mathcal{U}}))$ into $\mathcal{U}$; uniqueness of the selected partition cell therefore gives ${\alpha }_{h\circ f}^{\mathcal{U}}=h({\alpha }_f^{\mathcal{U}})$. Thus ${\alpha }^{\mathcal{U}}$ is a coherent family of the kind described in step 1.1.

The two constructions are inverse. Starting from $\alpha$, let ${\chi }_i:[n]\to [2]$ be the characteristic function of $\{i\}$. Then for any $f:X\to [n]$ and any $i\in [n]$, one has $f^{-1}(i)\in {\mathcal{U}}_{\alpha }$ exactly when ${\alpha }_{{\chi }_i\circ f}=1$, which by coherence is exactly when ${\chi }_i({\alpha }_f)=1$, that is, when $i={\alpha }_f$; so the ultrafilter ${\mathcal{U}}_{\alpha }$ selects precisely the fibre of ${\alpha }_f$, and step 2.2 recovers ${\alpha }_f$. Starting from an ultrafilter $\mathcal{U}$, the definition of ${\mathcal{U}}_{{\alpha }^{\mathcal{U}}}$ says $A\in {\mathcal{U}}_{{\alpha }^{\mathcal{U}}}$ exactly when the fibre $A$ is the cell selected by $\mathcal{U}$ in the partition $\{A,X\setminus A\}$, which is exactly the condition $A\in \mathcal{U}$. Hence ${\mathrm{Ran}}_{J}J(X)\cong \beta X$ naturally in $X$.

For $x\in X$, the family ${\alpha }_f^x:=f(x)$ is coherent, and the assignment $x\mapsto {\alpha }^x$ is natural in $X$. On a finite ordinal $[n]$ and an element $i\in [n]$, the counit of the right Kan extension evaluates ${\alpha }^i$ at the identity map of $[n]$, so it returns $i$; the defining equation $1_J=\epsilon \circ (\eta J)$ in [L3] therefore forces the codensity unit to be $x\mapsto {\alpha }^x$. Under step 3.1 this corresponds to the principal ultrafilter at $x$, since $A\in {\mathcal{U}}_{{\alpha }^x}$ exactly when ${\chi }_A(x)=1$.

Now let $\mathbb{W}$ be an ultrafilter on $\beta X$, and let $\mathrm{\Omega }$ be its coherent family from step 2.2. For each $f:X\to [n]$, define $\widehat{f}:\beta X\to [n]$ by sending $\mathcal{U}$ to the unique index $i$ with $f^{-1}(i)\in \mathcal{U}$; step 2.2 guarantees that this is well-defined. For the flattened ultrafilter of [L4], one has $f^{-1}(i)\in {\mu }_X(\mathbb{W})$ if and only if ${\widehat{f}}^{-1}(i)\in \mathbb{W}$, so the coherent family attached to ${\mu }_X(\mathbb{W})$ by step 2.2 takes the value ${\mathrm{\Omega }}_{\widehat{f}}$ at $f$. But ${\mathrm{\Omega }}_{\widehat{f}}$ is exactly the finite-set value produced by $T\epsilon$ in the defining equation $\epsilon \circ (\mu J)=\epsilon \circ (T\epsilon )$ of [L3]. Hence the codensity multiplication is identified with ultrafilter flattening. Therefore step 3.1 matches both the codensity unit and multiplication with the principal unit and flattening multiplication of [L4], so the codensity monad of $J$ is the ultrafilter monad.