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It is well known that the codomain functor $$cod:\mathcal{C}^\to\to\mathcal{C}$$ from the arrow category of a category $\mathcal{C}$ to itself is a fibration iff $\mathcal{C}$ has binary pullbacks.

Is there another functor that always exists for which promotion to a fibration is equivalent to $\mathcal{C}$ having a terminal object?

I naively tried $!:\mathcal{C}\to{\bf 1}$ and $\{X\}:{\bf 1}\to\mathcal{C}$ without any luck.

This would be of interest for the obvious reason -- having binary pullbacks and a terminal object is equivalent to being finitely complete, so a functor answering the above question positively would allow for a 'fully fibrational' characterization of finite completeness. Further, the construction below yields a functor that always exists such that promotion to a fibration is equivalent to having all pullbacks -- this together with Alexanders answer below offers a fully fibrational characterization of completeness, as originally desired.

Thanks to Alexander and Andrej for pointing out in the comments that $cod$ as above being a fibration only yields binary pullbacks, not arbitrary ones.

EDIT: I was able to fix the following construction to actually yield a functor that always exists such that promoting this functor to a fibration is equivalent to $\mathcal{C}$ having all pullbacks. (The first correction worked; we want them all to be the same commutative square universally, so that all paths from the pullback object into the object we're pulling back over are equal.)

For a category $\mathcal{C}$, consider the category $Sink(\mathcal{C})$ whose objects are sinks in $\mathcal{C}$ indexed over ordinals $\alpha\in{\bf O_n}$ $$\{f_i:X_i\to X\}_{i<\alpha}$$ and whose arrows are only defined from sinks indexed over $1$ to arbitrary sinks $$\hat h:\{f:A\to B\}\to\{g_i:C_i\to D\}_{i<\alpha}$$ given by ordered pairs $$\hat h=\big(\{h_i:A\to C_i\}_{i<\alpha},h:B\to D\big)$$ such that all the obvious squares commute, so $$h\circ f=g_i\circ h_i$$ for all $i<\alpha$. We have an obvious functor $$cod:Sink(\mathcal{C})\to\mathcal{C}$$ $$\{f_i:X_i\to X\}_{i<\alpha}\mapsto X,$$ $$\hat h\mapsto h$$ and this functor is a fibration iff $\mathcal{C}$ has all pullbacks.

We can identify $\mathcal{C}^\to$ with the subcategory of $Sink(\mathcal{C})$ indexed over $1$.

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    $\begingroup$ N.B. A category has pullbacks and a terminal object iff it is finitely complete. $\endgroup$
    – Alexander Campbell
    Commented Feb 1, 2022 at 7:46
  • $\begingroup$ @AlexanderCampbell A category is complete iff it has all products and equalizers iff it has all pullbacks and a terminal object. For finite completeness we would only require finite pullbacks and a terminal object. $\endgroup$
    – Alec Rhea
    Commented Feb 1, 2022 at 8:10
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    $\begingroup$ @AlecRhea: You should have an opinion poll that finds out how many people think that "pullback" means the limit of $\bullet \to \bullet \leftarrow \bullet$. $\endgroup$
    – Andrej Bauer
    Commented Feb 1, 2022 at 9:05
  • $\begingroup$ @AlexanderCampbell Ah, I see the confusion thanks to Andrej -- when I say that a category 'has pullbacks', I mean arbitrary ones not binary ones. (limits of diagrams shaped like arbitrarily many points each with a unique morphism to some specific point) $\endgroup$
    – Alec Rhea
    Commented Feb 1, 2022 at 9:20
  • $\begingroup$ But now I'm wondering if the fibrational characterization of pullbacks only holds for binary ones; I think I need to be more careful here. $\endgroup$
    – Alec Rhea
    Commented Feb 1, 2022 at 9:21

1 Answer 1

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A category $C$ has a terminal object if and only if the canonical functor $C \star \mathbf{1} \to \mathbf{1} \star \mathbf{1} = \mathbf{2}$ is a fibration.

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    $\begingroup$ It's funny how $C \mapsto C \star \mathbf 1$ is more of a "colimit-y" construction whereas $C \mapsto C^{\mathbf 2}$ from the motivating example is more of a "limit-y" construction. $\endgroup$
    – Tim Campion
    Commented Feb 1, 2022 at 23:55
  • $\begingroup$ @TimCampion Seeing as $\star$ here is a coproduct (colimit) and $C^2\cong C\times C$ is a product (limit), that doesn't strike me as odd. $\endgroup$
    – Alec Rhea
    Commented Feb 4, 2022 at 5:25
  • $\begingroup$ @AlecRhea By $C^{\mathbf 2}$ I meant the morphism category of $C$, not the product of $C$ with itself. This is indeed a $Cat$-weighted limit, though. I'm not sure how $\star$ is a colimit. The more precise thing to say is that $C \mapsto C \star 1$ commutes with connected colimits and few limits, whereas $C \mapsto C^{\mathbf 2}$ commutes with limits and few colimits. I find it odd because these two constructions are playing analogous roles in the story here -- $C \star 1$ is to terminal objects as $C^{\mathbf 2}$ is to pullbacks. $\endgroup$
    – Tim Campion
    Commented Feb 4, 2022 at 11:17

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