# Existence and characterisations of left Kan extensions and liftings in the bicategory of relations I

The bicategory $$\mathsf{Rel}$$ of sets, relations, and inclusions of relations has right Kan extensions and right Kan lifts¹, however I believe it does not have all left Kan extensions/lifts.

1. Is it indeed the case that $$\mathsf{Rel}$$ does not have all left Kan extensions/lifts?
2. (Moved to a sequel question)
3. Is there an if and only if characterisation of the stronger condition of having $$\mathrm{Lan}_S\colon\mathsf{Rel}(A,X)\to\mathsf{Rel}(B,X)$$ exist for a relation $$S\colon A⇸B$$ in terms of properties of $$S$$?

¹I'll add a reference here with full proofs for this in the future.

• Question 1 is "easy" to evade: a 1-cell in a bicategory has a left adjoint if and only if $Lan_F1$ exists (and it turns out to be an absolute extension). So, take a relation that doesn't have a right adjoint. I am not aware of a sharp theorem for 2 or 3, but certainly some Lan's exist and some don't. I'd start assuming something on $S$ (it is total, univalued, functional...) and see what happens. (I am ashamed I never thought about this before!) Commented Dec 19, 2023 at 16:49
• Note that the hom-categories of $Rel$ are co/complete lattices, and the composition functor preserves colimits in each variable. So by the adjoint functor theorem, right Kan extensions / lifts exist as you indicate, and left Kan extensions / lifts exist in the sense of your (3) if and only if the relevant composition functors preserve limits (which doesn't happen in general). Commented Dec 19, 2023 at 19:18
• @fosco oooh I forgot about the $Lan_F1$ characterisation, that's super useful! Commented Dec 19, 2023 at 22:30

Here is a partial answer, addressing (3). Given a relation $$S : A⇸B$$, composition with $$S$$ is a cocontinuous order-preserving map $$Rel(B,C) \to Rel(A,C)$$. These hom-posets are powerset lattices $$2^{B\times C}$$ and $$2^{A \times C}$$, which are complete. By the adjoint functor theorem, this map has a right adjoint giving the right Kan extension as you indicated in the question.
In order for a left adjoint to exist, by the adjoint functor theorem it is necessary and sufficient for composition with $$S$$ to preserve limits. If $$C = \emptyset$$, this is always the case. Otherwise, $$C$$ admits $$1$$ as a retract, and we reduce to this case. So for left Kan extensions to exist when $$C \neq \emptyset$$, it is necessary and sufficient for $$S$$ to admit a left adjoint in $$Rel$$, i.e. for $$S$$ to be the graph of a function $$f : A \to B$$. Dually, for all left Kan lifts to exist when $$C \neq \emptyset$$, it is necessary and sufficient for $$S$$ to be the dual graph of a function $$f : B \to A$$.
• Hi Tim, thank you so much for the answer! Do you happen to know an explicit description for left Kan extensions (resp. lifts) along $S=\mathrm{Gr}(f)$ (resp. along the relation $b\mapsto f^{-1}(b)$)? Commented Dec 19, 2023 at 22:30