# Kan extension along any functor implies adjoint exists

I trying to decide whether the existence of a, say, a left Kan extension along $$i: C \to D$$ for any functor $$F: C \to E$$ implies that $$i$$ has a right adjoint.

I have proven the converse, and I know that if left Kan extensions existed and were absolute, i.e. preserved by any functor $$G: E \to E'$$, then $$Ran_{F}id_C$$ would be the desired adjoint.

I initially looked for a counterexample, but this turned out to be unsuccessful. I really have no idea how to approach this unfortunately, so any help is appreciated!

• @Max maybe this is something you are interested in or can help with? – ThePuix Oct 6 '19 at 17:58
• Maybe this can help : math.stackexchange.com/questions/2314042/… . I don't know the answer at the moment, although note : if $i$ is in fact a left adjoint (= has a right adjoint), then the Kan extensions $Lan_i F$ are all absolute. In other words, if $i$ has all Kan extensions along it, but some of them aren't absolute, then $i$ can't be a left adjoint. (1/2) – Maxime Ramzi Oct 6 '19 at 18:33
• So two possible strategies are: find an example with non absolute extensions; or prove that with your hypotheses all extensions are absolute (specifically, you only need $Lan_i id_C$ to be absolute). [unrelated : I don't know if you @ was to me but I didn't get a notification, I just stumbled onto your question by chance ] (2/2) – Maxime Ramzi Oct 6 '19 at 18:35
• @Max Was trying to get a hold of you, but I guess it doesn't work like that. And thanks for the possible strategies :). Do you have any intuition for what the answer might be? – ThePuix Oct 6 '19 at 19:14
• My intuition about Kan extensions is not very good yet, but I would say that only imposing existence will not yield absoluteness, and so I would say there are counterexamples. But maybe some trick with some functor or comma category can yield the desired properties, so I'm not quite sure. – Maxime Ramzi Oct 6 '19 at 20:05