Re: Equality as an adjunction

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? [For admin and other information see: http://www.mta.ca/~cat-dist/ ]