# The domain functor is a fibration?

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?

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.

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 $$g_1=p((g_1,g_2))=1$$.

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 $$h_1, h_2$$

• Thanks for showing this in the spirit of the paper, especially for not requiring the Grothendieck construction.
– Mark
Commented Jul 10, 2023 at 10:26
• @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. Commented Jul 10, 2023 at 11:15
• 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 Commented Jul 10, 2023 at 11:21
• 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
– Mark
Commented Jul 15, 2023 at 4:25
• @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 Commented Jul 15, 2023 at 7:53