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 $1$.

finitelycomplete. $\endgroup$1more comment