1
$\begingroup$

I'm learning about fibrations and I read that the functor $dom: C^{\rightarrow} \rightarrow C$ is one for arbitrary category $C$. I can't see it.

I need to show that any morphism in $C^{\rightarrow}$ is (for now just weakly) cartesian wrt $dom$. According to Zhang, it should follow merely from the fact that morphisms in $C^{\rightarrow}$ are commutative squares.

Letting $s \in C^{\rightarrow}[b, a]$ be such a morphism with $dom(s) = f$. I need to show that if $s' \in C^{\rightarrow}[b', a]$ is another such morphism with $dom(s') = f$, then there is a unique morphism $t \in C^{\rightarrow}[b', b]$ with $dom(t) = 1$ and $s' = s \circ t$. Where does the arrow $cod(t)$ in $C$ come from?

enter image description here

$\endgroup$

2 Answers 2

3
$\begingroup$

It's crucial to remember that fibrations over $C$ are exactly the Grothendieck constructions of contravariant (pseudo)functors from $C$ to $\mathrm{Cat},$ with the fibers of the fibration corresponding to the values of the functor. Indeed, this is really how people ought to define fibrations, with the usual definition as a characterization, since the usual definition is so hard to grasp at first.

So, what is the fiber of $\mathrm{dom}$ over some $c\in C$? It's the class of arrows with domain $c$ and commutative squares whose domain component is $\mathrm{id}_c.$ That is, the fiber is precisely the slice category $c/C.$ And we have a very simple pseudofunctor out of $C^{\mathrm{op}}$ sending $c\mapsto c/C$ and $f:c\to c'$ to the functor $c/C\to c'/C$ given by precomposition with $f.$ In fact, this happens to be a strict functor into $\mathrm{Cat}.$

For my money, that's all you should realize to understand how $\mathrm{dom}$ is a fibration. You may well find it easier to work out the proof that the Grothendieck construction of an arbitrary functor into $\mathrm{Cat}$ is a fibration than to work it out in this particular case.

$\endgroup$
1
$\begingroup$

Partial answer. EDIT: the definitions used in the paper seem to be a bit weird, conflicting problematically with the definition given by nLab. When I have time I’ll update this to match the definitions on nLab.


$\newcommand{\C}{\mathsf{C}}\newcommand{\D}{\mathsf{D}}\newcommand{\dom}{\operatorname{dom}}\require{AMScd}$Copying definitions from the paper:

Given the context of a functor $p:\C\to\D$, we say an arrow $f:a\to b$ in $\C$ is $p$-Grothendieck cartesian if for all arrows $f':a'\to b$ with $p(f)=p(f')$ there is a unique $g:a'\to a$ with $p(g)=1$ making $f\circ g=f'$.

We say $p$ is a Grothendieck fibration if composition of $p$-Grothendieck cartesian arrows in $\C$ always gives another $p$-Grothendieck cartesian arrow and if for every $\varsigma\in\C$ and every arrow $\sigma:\partial\to p(\varsigma)$ in $\D$ there is a $p$-Grothendick cartesian arrow $\overline{\sigma}:a\to\varsigma$ with $p(\overline{\sigma})=\sigma$.

Ok, so we consider $p=\dom:\C^{\to}\to\C$. This is a partial answer because I can confirm the lifting property of $p$ but I'm unsure about the composition property.

Let's first classify what the $p$-Grothendieck cartesian arrows are in $\C^{\to}$; they'll be commutative squares: $$\begin{CD}a@>f_1>>b\\@V\alpha VV@VV\beta V\\a'@>>f_2>b'\end{CD}$$Where for every commutative square: $$\begin{CD}c@>f'_1>>b\\@V\gamma VV@VV\beta V\\c'@>>f'_2>b'\end{CD}$$Satisfying $p((f'_1,f'_2))=p((f_1,f_2))$ i.e. satisfying $f'_1=f_1$ so that $c=a$, there is a unique: $$\begin{CD}a@>g_1>>a\\@VV\gamma V@VV\alpha V\\c'@>>g_2>a'\end{CD}$$With $f_1\circ g_1=f'_1=f_1$, $f_2\circ g_2=f'_2$ and $g_1=p((g_1,g_2))=1$.


Diagram


Let's check the lifting criterion first. Say we have some $\beta:b\to b'$ an object in $\C^{\to}$ and an arrow $\sigma:a\to p(\beta)=b$ in $\C$. One way to construct a lift is $\overline{\sigma}=(\sigma,1)$: $$\begin{CD}a@>\sigma>>b\\@V\beta\sigma VV@VV\beta V\\b'@>>1>b'\end{CD}$$It remains to check this arrow is $p$-Grothendieck cartesian.

So, suppose there is a square: $$\begin{CD}a@>\sigma>>b\\@V\gamma VV@VV\beta V\\c@>>f'>b'\end{CD}$$

We need to find a unique $g=(g_1,g_2)$ in: $$\begin{CD}a@>g_1>>a\\@V\gamma VV@VV\beta\sigma V\\c@>>g_2>b'\end{CD}$$

That also satisfies:

  • $\sigma\circ g_1=\sigma$
  • $1\circ g_2=f'$
  • $p((g_1,g_2))=1$

The bullet points force $g_1=1,g_2=f'$ to be chosen. So, we have uniqueness, if it works. However, it's obvious that $g_1=1$ and $g_2=f'$ does work, so we are done.


We also should check the composite of $p$-Grothendieck cartesian arrows is again $p$-Grothendieck cartesian. I confess that I am not sure about this yet.


This edit from the OP seems to depend on a different definition than the one used by the author.

OP edit:

to see this condition suppose (below) that $f, f'$ are cartesian, then we want $f \circ f'$ cartesian. So for another square $g$, we want the square $h$ pictured to the left below. Applying first cartesianness of $f$ (middle) then of $f'$ (right) we get the required components $h_1, h_2$

enter image description here

$\endgroup$
5
  • 1
    $\begingroup$ Thanks for showing this in the spirit of the paper, especially for not requiring the Grothendieck construction. $\endgroup$
    – Mark
    Commented Jul 10, 2023 at 10:26
  • 1
    $\begingroup$ @Mark I’m not sure about your edit. Notice that $p(f)=p(f’)$ is required; so in the first image on the left, we should require $d=c$ and $q_1=f_1f’_1$. But then in the second image, to use the $p$-Grothendieck cartesianness of $(f_1,f_2)$ in the way that you want, we’d have to have $q_1=f_1$ and $d=b$ which is a priori not necessarily the case. So I don’t think it works. I suspect I misunderstood what the author of the paper meant with that since they sometimes say ‘Cartesian’ and sometimes say ‘$p$-Grothendieck Cartesian’ and it’s also the case that ‘Cartesian’ is used for other purposes. $\endgroup$
    – FShrike
    Commented Jul 10, 2023 at 11:15
  • 1
    $\begingroup$ In fact I think your author has incorrectly written the definition. Looking at the nLab, I find your author has defined Grothendieck fibration a little more flexibly, in a problematic way. When I have time I will have to edit the post accordingly $\endgroup$
    – FShrike
    Commented Jul 10, 2023 at 11:21
  • $\begingroup$ I see what you mean. I can also see that if $p$ is a weak fibration (by nLab definition) and composition of weak Cartesian arrows in E is also weak Cartesian, then $p$ is strong Cartesian (as nLab says). Is this the problematic part you mean? Then I think your original answer wasn't partial but was actually complete $\endgroup$
    – Mark
    Commented Jul 15, 2023 at 4:25
  • $\begingroup$ @Mark The problematic part is that your author’s definition only tests the lifting criterion against special cases where $p(f)=p(f’)$ but the nLab one tests the lifting criterion against a more general situation, we can have $p(f)=p(f’)\circ g$ or something like that. I haven’t checked but I believe it’s probably true that dom is a Grothendieck fibration à la nLab and that Cartesian arrows compose nicely $\endgroup$
    – FShrike
    Commented Jul 15, 2023 at 7:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .