Let $R\colon A\mathrel {\rightarrow \kern -9.5pt\mathrlap {|}\kern 6pt}B$ be a relation.
-
(a)
The left Kan lift
\[ \operatorname {\mathrm{Lift}}_{R}\colon \mathbf{Rel}(X,B)\to \mathbf{Rel}(X,A) \]along $R$ exists.
-
(b)
The relation $R$ admits a right adjoint in $\boldsymbol {\mathsf{Rel}}$.
-
(c)
The relation $R$ is of the form $f^{-1}$ (as in Definition 8.2.3.1.1) for some function $f$.
- 1Specifically for $R$ and $S$, not $\operatorname {\mathrm{Lift}}_{S}$ the functor.
8.5.16 Internal Left Kan Lifts
Proof of Proposition 8.5.16.1.1.
Item 1: Non-Existence of All Internal Left Kan Lifts in $\boldsymbol {\mathsf{Rel}}$
By Item 2, it suffices to take a relation that doesn’t have a right adjoint.
Item 2: Characterisation of Relations Admitting Left Kan Lifts Along Them
This proof is dual to that of Item 2 of Proposition 8.5.15.1.1, and is therefore omitted.
Given a function $f\colon A\to B$, the left Kan lift
\[ \operatorname {\mathrm{Lift}}_{f^{\dagger }}\colon \mathbf{Rel}(X,A)\to \mathbf{Rel}(X,B) \]
along $f^{\dagger }$ exists by Item 2 of Proposition 8.5.16.1.1. Explicitly, given a relation $R\colon X\mathrel {\rightarrow \kern -9.5pt\mathrlap {|}\kern 6pt}A$, the left Kan lift
\begin{align*} [\operatorname {\mathrm{Lift}}_{f}(R)](x) & = [\operatorname {\mathrm{Gr}}(f)\mathbin {\diamond }R](a)\\ & = \bigcup _{a\in R(x)}f(a)\end{align*}
for each $x\in X$.
Given relations $S\colon A\mathrel {\rightarrow \kern -9.5pt\mathrlap {|}\kern 6pt}X$ and $R\colon A\mathrel {\rightarrow \kern -9.5pt\mathrlap {|}\kern 6pt}B$, is there a characterisation of when the left Kan lift1
\[ \operatorname {\mathrm{Lift}}_{S}(R)\colon X\mathrel {\rightarrow \kern -9.5pt\mathrlap {|}\kern 6pt}A \]
exists in terms of properties of $R$ and $S$?
This question also appears as [Emily, Existence and characterisations of left Kan extensions and liftings in the bicategory of relations II].
