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At “Definition as a 2-functor”, the article says
For a proarrow H:B→D and ordinary arrows f:A→B and g:C→D, we write H(g,f) for the composite D(g,1)∘H∘B(1,f); it is a proarrow from A to C. We also write UA, A(1,1), or simply A for the identity proarrow A↛A.
In seems that the nLab has settled on the convention that a profunctor H:B↛D is a functor H:Dop×B→Set. Therefore, in the expressions H(g,f) and B(1,f) the morphism f is in the covariant slot, which means that either f should be a morphism out of B or H should be a profunctor into A.
What am I missing?
It seems right to me. The covariant slot of H is the category that it is “from” as a profunctor, and we are substituting a value of f into that slot; so it’s correct that H is a profunctor from B to something else, and that the values of f belong to B, i.e. that f is a morphism into B.
In that case, D and g are the problem. Thinking of H as a set of heteromorphisms from B to D, we can precompose with f to get a set of heteromorphisms from A to D, but to postcompose, we need g:D→C, not g:C→D. That is,
H(f:A→B,g:D→C)(h:B→D)=g∘h∘f:A→C.f is not a morphism in B, it’s a functor with codomain B, and similarly g is not a morphism in D, it’s a functor with codomain D. We’re not describing the action of the functor H on its elements/heteromorphisms; we’re describing its restriction along a pair of functors.
It’s like the multiplication of three matrices (the outer two being special in deriving from functions), if that helps. Or the composition of relations.
Mike: Then shouldn’t it be D(1,g)∘H∘B(1,f), that is, f followed by H followed by the taking the preimage under g? The expression D(g,1) doesn’t make sense; since B is the codomain of f, having g in the other slot means we’d want the domain of g. If we want the codomain of g, it needs to be in the same slot that f is.
The D in D(g,1) represents the hom-functor of D, which is a profunctor from D to D. Thus, g being a functor with codomain D, it can be plugged into either slot of this profunctor.
Although in the context of the quote in question, D(1,g) and D(g,1) are just notations for the two proarrows associated to the arrow g. The reason for using that notation is as I said: in the case of categories and profunctors, they are (x,y)↦HomD(x,g(y)) and (y,x)↦HomD(g(y),x).
OK, so in that notation, both slots are covariant. I’ll add a note to that effect. Thanks!
I think I have some idea of what you mean, but I don’t think “both slots are covariant” is really the correct way to say it. H is still a functor from Dop×B to Set, so in that sense the first slot is contravariant. And if the question is what kind of a functor H(g,f) is as a functor of g and f, then it’s a functor from K(A,B)op×K(C,D) to M(A,C), so in that sense also the first slot is contravariant.
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