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If necessary for other users, I will write down the definitions for pseudo functor and/or fibered category, otherwise they can find them in page $42,44$ of Angelo vistoli's part Grothendieck topologies, fibered categories and descent theory in book FGA explained.

Given a fibered category $\mathcal{F}\rightarrow \mathcal{C}$ with a cleavage fixed, we associate a pseudo functor $\Phi$ on $\mathcal{C}$. I understand this.

It says given any pseudo functor $\Phi$ on $\mathcal{C}$ we can associate a fibered category, with a cleavage.

He does that assuming $\Phi$ is actually a functor $\Phi:\mathcal{C}^{op}\rightarrow (Cat)$ to category of categories.

The idea is to define a category $\mathcal{F}$ with a functor $\mathcal{F}\rightarrow \mathcal{C}$ such that the fiber $\mathcal{F}(U)$ is canonically equivalent to the category $\Phi(U)$.

It says an element of $\mathcal{F}$ is a pair $(\xi,U)$ where $U$ is an object of $\mathcal{C}$ and $\xi\in \Phi(U)$. So, objects are pairs.

Let $(\xi,U),(\eta,V)$ be two objects in $\mathcal{F}$. We have to define what are the morphisms $(a,f):(\xi,U)\rightarrow (\eta,V)$ between these. We expect morphisms to be pairs as objects are also pairs. Choice for $f$ is more or less obvious i.e., a morphism $f:U\rightarrow V$ in $\mathcal{C}$. This $f$ gives map $\Phi(f):\Phi(V)\rightarrow \Phi(U)$ and $\eta\in \Phi(V)$ mapsto an element $\Phi(f)(\eta)\in \Phi(U)$. So, we have two elements $\xi,\Phi(f)(\eta)$ and we pick a morphism $\xi\rightarrow \Phi(f)(\eta)$ and call it $a$. Thus, morphism set is also clear.

Now, we need to define what does it mean to say composition between two morphisms. Coniser $$(\xi,U)\xrightarrow{(a,f)} (\eta,V)\xrightarrow{(b,g)}(\zeta,W)$$ This composition should give a map $(\xi,U)\xrightarrow{(c,h)} (\zeta,W)$ and one obvious choice for $h:U\rightarrow W$ is just the composition $g\circ f:U\rightarrow V\rightarrow W$ and $c$ is expected to be morphism $$\xi\rightarrow \Phi(g\circ f)(\zeta)$$ that is an arrow in $\Phi(U)$.

But author writes something else for $c$. He writes $$(b,g)\circ (a,f)=(\Phi(f)(b)\circ a,g\circ f)$$ This is not clear for me.

We have $a:\xi\rightarrow \Phi(f)(\eta)$. $\Phi(f):\Phi(V)\rightarrow \Phi(U)$.

As $b:\eta\rightarrow \Phi(g)(\zeta)$ is an arrow in $\Phi(V)$, $\Phi(f)(b)$ gives an arrpw in $\Phi(U)$ we have $$\Phi(f)(b):\Phi(f)(\eta)\rightarrow \Phi(f)(\Phi(g)(\zeta))=\Phi(g\circ f)(\zeta)$$ (This is because $\Phi$ is a functor).

So, we have $a:\xi\rightarrow \Phi(f)(\eta)$ in $\Phi(U)$ and $\Phi(f)(b):\Phi(f)(\eta)\rightarrow \Phi(g\circ f)(\zeta)$ and their composition $$\Phi(f)(b)\circ a:\xi\rightarrow \Phi(g\circ f)(\zeta)$$ which is the same thing as I have expected but what I do not understand is what is the point of writing down as $\Phi(f)(b)\circ a$ when you can just write $\xi\rightarrow\Phi(g\circ f)(\zeta)$?

Is it just because we can not write $\Phi(g\circ f)=\Phi(f)\circ\Phi(g)$ if $\Phi$ is just a pseudo functor or Is there any other reason that I am missing?

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  • $\begingroup$ Huh? $\Phi(g\circ f)(\zeta)$ is the codomain of $\Phi(f)(b)\circ a$. They certainly aren't equal, since they're not the same kind of thing... $\endgroup$ – Kevin Carlson Mar 15 '18 at 5:39
  • $\begingroup$ @KevinCarlson I mean to say $\xi\rightarrow \Phi(g\circ f)(\zeta)$ is same as $\Phi(f)(b)\circ a$. $\endgroup$ – user537667 Mar 15 '18 at 5:46
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    $\begingroup$ That doesn't make sense. Is $\mathbb{R}\to\mathbb{R}$ "the same as" the exponential function? You have to specify which morphism you're talking about, not just its domain and codomain. $\endgroup$ – Kevin Carlson Mar 15 '18 at 5:48
  • $\begingroup$ It makes sense, I mean what you are saying makes sense. Though for me it is the map $\Phi(f)(b)\circ a$ when I am writing $\xi\rightarrow \Phi(g\circ f)(\zeta)$ there will be other maps $\xi\rightarrow \Phi(g\circ f)(\zeta)$. So, writing $\xi\rightarrow \Phi(g\circ f)(\zeta)$ will only cause confusion and it is wrong in some sense. Thanks. $\endgroup$ – user537667 Mar 15 '18 at 5:52
  • $\begingroup$ Yes. If you've somehow gotten into the habit of naming your morphisms solely by their domains and codomains, you should get out of that habit right now. Glad to help. $\endgroup$ – Kevin Carlson Mar 15 '18 at 5:54
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I think, I understand what your problem is. To start off, the definition of the composition he gives is the correct one. Assume that someone starts with a functor $\Phi: \mathsf{C}^{{\rm op}} \to \mathsf{Cat}$. Then this functor-"like" object conists of a data, which fulfils some compatibility conditions (you can find them in his notes somewhere). Assume now that you have constructed the corresponding fibered category over $\mathsf{C}$, say $\mathsf{F}$. Pick out three arbitrary objects, $(\xi, V), (\eta, W),(\zeta,U)$ and think of the following diagram $$(\xi,U)\xrightarrow{(a,f)} (\eta,V)\xrightarrow{(b,g)}(\zeta,W),$$ where the pairs of morphisms conist of the following: $f:U \to V$, $g:=V \to W$, are morphisms in $\mathsf{C}$, while $a: \xi \to \Phi(f)(\eta)$, $b: \eta \to \Phi(g)(\zeta)$, are mophisms in $\Phi(U)$ and $\Phi(V)$ correspondingly. Now since the composition (whatever it is) must give morphisms of the same form, we should end up with a morphism $(\tau,h): (\xi, U) \to (\zeta, W)$. Since clearly $h=g \circ f$,the second morphism must be something like $\tau:\xi \to \Phi(g \circ f)(\zeta)$.

We define now the composition to be $(b,g) \circ (a,f)=(\Phi(f)(b) \circ a, g \circ f)$. As I said, the composition $g \circ f$ is obvious. Regarding its first part, notice that $\Phi(f): \Phi(V) \to \Phi(U)$ from the definition of pseudo-functor is a functor and $b$ is a morphism in the category $\Phi(V)$, therefore we obtain $\Phi(f)(b):\Phi(f)(\eta) \to \Phi(f)(\Phi(g))(\zeta)=\Phi(g \circ f)(\zeta)$. So defining $\Phi(f)(b) \circ a$ gives a morphism in $\Phi(U)$, hence our choice is well-defined. Clearly this morphism has domain $\xi$, and codomain $\Phi(g \circ f)(\zeta)$. Thus this definition of composition is totally consistent with the definition of a morphism in $\mathsf{F}$.

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