The problem as stated is to find the unit for the adjunction: ((- x A), (-)^A x A) The latter functor takes an arrow f to (f . -) x id_A and does the obvious thing for objects. The co-unit diagram is given as: B^A x A ---- eval_AB ----> B ^ ^ | | | | curry(g) x id_A g : C x A -> B | | | | | | C x A --------------------+ This diagram is somewhat puzzling, because it seems the second functor has turned into (-)^A ! Continuing in with that, we get a unit diagram like this: C ---- magic ----> (C x A)^A | | | | | | curry(g) (g . -) | | | | | v +------------------> B^A So what is the magic? It is an arrow that takes a C to an arrow that takes an A and makes the product C x A. I want to write curry(C x A) but that is ridiculous looking. What's the right notation for this thing? -- _jsn