TITLE: Can we always make a strictly functorial choice of pullbacks/re-indexing?
QUESTION [10 upvotes]: $\newcommand{\C}{\mathbf{C}} \newcommand{\D}{\mathbf{D}}$ Let $\C$ be a category with pullbacks.  Taking any choice of pullbacks gives us re-indexing functors $f^* \colon \C /Y \to \C/X$, and these will be functorial in $f$ up to natural isomorphism, in that $g^* \cdot f^* \cong (f \cdot g)^*$.  However, these will usually not be strictly functorial in $f$; that is, $g^* \cdot f^*$ and $(f \cdot g)^*$ will not be literally equal.  Strict functoriality also requires that $1_X^* = 1_{\C/X}$; while this typically does hold on the nose, it’s still not automatic.
My main question: Is there always some choice of pullbacks that make re-indexing strictly functorial?  I believe the answer should be “no”, but I don’t know any counterexample.  Even in the case of $\mathbf{Set}$, it’s not obvious whether there’s a choice that works.
An equivalent phrasing of the question is: can the codomain fibration $\mathrm{cod} \colon \C^\rightarrow \to \C$ be equipped with a splitting?  It can always be replaced by an equivalent split fibration over $\C$; but splitting the codomain fibration itself seems hard.

REPLY [3 votes]: Here's an argument (assuming global choice) that the codomain fibration will always be split in "natural" cases. Perhaps it's well-known, or perhaps it has an important shortcoming (such as the use of choice?). Or perhaps there's an error -- strictification is tricky!

Claim: Let $C$, $D$ be categories with pullbacks, and suppose there is a functor $p: C \to D$ with the following properties:

$p$ preserves pullbacks.
$p$ is an isofibration, and induces an isofibration on all slices (perhaps this is redundant?).
$p$ reflects identities (i.e. if $f: c \to c$ is an endomorphism in $C$, and $pf: pc \to pc$ is an identity, then so is $f$).
There is a cardinal $\kappa$ such that for every $f: x \to c \in C^{[1]}$, the isomorphism class $[f]$ of $f$ in the fiber category $p^{-1}(pf) \subseteq C/c$ has cardinality $\kappa$.

The claim is that if $D$ admits strict reindexing, then so does $C$, and $p$ preserves the strict reindexing.

Upshot: Most "natural" categories $C$ admit such a functor $p: C \to \mathsf{Set}$, with $\kappa$ being the cardinality of the universe (perhaps with an exception for objects over $\emptyset$, which often doesn't matter because there are few maps into $\emptyset$). I'm assuming that $\mathsf{Set}$ admits strict reindexing (right?). So most "natural" categories do as well.
EDIT: $\mathsf{Set}$ does indeed admit strict reindexing. To see this, as in the comment below, note that the fibration $\mathsf{Set}^{(-)} \to \mathsf{Set}$ admits a splitting, and the fibration $\mathsf{Set}/(-) \to \mathsf{Set}$ is isomorphic as a fibered caegory -- the fact that it's equivalent as a fibered category is standard, and each isomorphism class in each fiber category has the cardinality of the universe, in both fibrations, so an isomorphism of fibrations exists. So $\mathsf{Set}/(-) \to \mathsf{Set}$ also admits a splitting as desired.
So indeed, most "natural" categories admit a strict reindexing. Of course, it's completely undecideable to compute it because the choice function must solve the isomorphism problem!

Proof of Claim: Let $()^\ast$ be a system of strict reindexing in $D$. Fix enumerations of the isomorphism classes of the fibers of $p$ by $\kappa$ (using condition (4) and choice). Now, if $\gamma : c' \to c$ is a morphism in $C$ and $f: x \to c \in C/c$, take an arbitrary pullback of $f$ along $\gamma$. Because $p$ is an isofibration (condition (2))and $p$ preserves pullbacks (condition (1)), we can correct this choice to lie over the canonical pullback $(p\gamma)^\ast(pf)$. Moreover, we can choose a representative which has the same index under our enumeration as $f$ does; let's call it $\gamma^\ast(f)$. It remains only to nail down the lift of the other leg of the pullback square, but any two choices differ by an automorphism of $\gamma^\ast(f)$ which lies over the identity on $(p\gamma)^\ast(pf)$, so by condition (3) this choice is uniquely determined.
It's now obvious that the lifted operation $()^\ast$ is strictly functorial, because it preserves the indexing by our enumerations.