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Let $C$ be a category and $F\in[C^{op}, Cat]$ be a strong functor.

(1) There are functors

$$hom_C(c',c)\times F(c)\to F(c').$$

(2) The grothendieck construction gives a 2-equvalence

$$\int_C: [C^{op},Cat]\to Fib_C$$

(3) Fibrations over $C$ are categories over $C$ together with a left action of the morphism double category $Mor_C$. ($Mor_C$ is a 2-monoid in the 2-category of spans from $C$ to $C$; the 2-monoidal structure is given by the strong pullback. This induces a 2-monad on the 2-category of spans from $C$ to $pt$, the one-point category. Algebras for this monad are fibrations.)

(4) Let $\int_C^D$ denote the grothendieck construction for distributors. We have

$$\int_C^C hom_C = Mor_C$$


$$\int_C^{pt} F = \int_C F$$

Question: How do these parts fit together?

The maps from (1) should result in a transformation

$$hom_C\otimes_C F \to F$$

and the action of $Mor_C$ is given by a functor

$$\int hom \times_C \int F \to \int F.$$

Now: If the grothendieck construction was laxely (or even strongly) compatible with the the products involved - that is, A lax (or strong) bifunctor - these two actions would naturally correspond to each other.

Question 2: Is it? And if so: Can this be shown without using the explicit description of the grothendieck construction; only by using the characterisation as a weighted limit?

Related Question

Edit: I reformulated the question.

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I am not sure if I understand your questions. However, "distributorial" composition is generally far different from composition via pullbacks. – Michal R. Przybylek Apr 22 '13 at 21:31
Yeah; It seems i was a bit vauge. I reformulated the question to make it more comprehensible. – Garlef Wegart Apr 23 '13 at 7:05

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