We have a natural identification
\[ \left\{ \begin{gathered} \text{Comonads in}\\ \text{$\boldsymbol {\mathsf{Rel}}$ on $X$} \end{gathered} \right\} \cong \left\{ \text{Subsets of $X$}\right\} . \]
8.5.5 Internal Comonads
Let $X$ be a set.
Proof of Proposition 8.5.5.1.1.
A comonad in $\boldsymbol {\mathsf{Rel}}$ on $X$ consists of a relation $R\colon X\mathrel {\rightarrow \kern -9.5pt\mathrlap {|}\kern 6pt}X$ together with maps
\begin{align*} \Delta _{R} & \colon R \subset R\mathbin {\diamond }R,\\ \epsilon _{R} & \colon R \subset \Delta _{X} \end{align*}
making the diagrams
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1.
For each $x,y\in X$, if $x\sim _{R}y$, then there exists some $k\in X$ such that $x\sim _{R}k$ and $k\sim _{R}y$.
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2.
For each $x,y\in X$, if $x\sim _{R}y$, then $x=y$.
The second condition implies that $R\subset \Delta _{X}$, so $R$ must be a subset of $X$. Taking $k=y$ in the first condition above then shows it to be trivially satisfied. Conversely, any subset $U$ of $X$ satisfies $U\subset \Delta _{X}$, defining a comonad as above.
Let $f\colon A\to B$ be a function.
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1.
The density comonad $\operatorname {\mathrm{Lan}}_{f}(f)\colon B\mathrel {\rightarrow \kern -9.5pt\mathrlap {|}\kern 6pt}B$ is given by
for each $b\in B$. Thus, it corresponds to the image $\mathrm{Im}(f)$ of $f$ as a subset of $B$.
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2.
The dual density comonad $\operatorname {\mathrm{Lift}}_{f^{\dagger }}(f^{\dagger })\colon A\mathrel {\rightarrow \kern -9.5pt\mathrlap {|}\kern 6pt}A$ is given by
for each $b\in B$. Thus, it also corresponds to the image $\mathrm{Im}(f)$ of $f$ as a subset of $B$.
