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?

  • $\begingroup$ 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 $\Phi(f)(b)\circ a$. $\endgroup$ – Praphulla Koushik Mar 15 '18 at 5:46
  • 1
    $\begingroup$ 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. $\endgroup$ – Praphulla Koushik 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

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 $\mathsf{F}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.