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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

$$\int_C^C hom_C = Mor_C$$

and

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

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.

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

$$\int_C^C hom_C = Mor_C$$

and

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

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.

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

$$\int_C^C hom_C = Mor_C$$

and

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

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?

Related Question

Edit: I reformulated the question.

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

$$\int_C^C hom_C = Mor_C$$

and

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

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.

Let $X,Y,Z$ be categories and

$D:X^{op}\times Y\to Cat$ and

$E:Y^{op}\times Z\to Cat$

be $Cat$-valued distributors; (strong/pseudo)-functors.

Their composite can then be described as a coend or as a coequalizer. The corresponding lax- or oplax colimits can be calculated as grothendieck constructions on the category with two parallel arrows. But what about the strong coend?

Question 1: Is there a nice and explicit description of the composite?

The reason i am asking is that the grothendieck construction for a distributor gives a 2-equivalence between between $Cat$-vaulued distributors and spans that are fibrations on one side and opfibrations on the other. My first idea of showing this would be by using the explicit description of the grothendieck construction and some calculations.

On the other hand, such spans can be described as spans with left and right actions of the morphism-double-categories respectively (they are 2-monoids in the 2-category of endo-spans). Now i have a formula in mind that should look like

$$\int D\otimes_Y E = \left(\int D\right) \times_Y \left(\int E\right)$$

(Here $\times_Y$ denotes the strong pullback).

Then - using $\int hom = Mor$ - we would get

$$\int hom\otimes_X E = Mor_X \times_X \int E$$

and the operation

$$hom\otimes_X E \to E$$

should directly induce an operation

$$Mor_X\times_X \int E \to E$$

that exposes $\int E$ as fibered over $X$.

Question 2: What's the story here? Is my guess correct? And can this be shown without reference to the explicit description of grothendieck construction?

Remark: "$=$" is of course to be read as an equivalence.

EDIT: Related Question

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

$$\int_C^C hom_C = Mor_C$$

and

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

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?

Related Question

Edit: I reformulated the question.