Sorry, I guess my mistake is that I've got the adjunction in the wrong direction: it is a left Kan extension: [CxC,2](Lan_Delta P, Q) =~ [C,2](P, Q o Delta) And equality is given by Lan_Delta T : CxC -> 2. Does it always exist? And what was the right Kan extension in my previous mail when it exists? I wrote:

Thank you all for your replies. If I understood you well, the adjunction is as follows:

 [C,2](P o Delta, Q) =~ [CxC,2](P, Ran_Delta Q)

where Ran_Delta Q is the right Kan extension if Q along Delta. (I write [C,D] for the functor category, and 2 for the category with only two objects True and False and identities). If T is the constant functor that sends every object to True, then Ran_Delta T : CxC -> 2 is the equality predicate, right?

Now I write this adjunction in logical form:

   P(x,x) |- Q(x) ============== P(x,y) |- Q(x) /\ x=y

But in Pitts it was reversed:

  Q(x) |- P(x,x) ============== Q(x) /\ x=y |- P(x,y)

Where is my mistake?

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