Let $\mathsf{C,D}$ be (small) categories. A functor $P: \mathsf{C} \to \mathsf{D}$ is a discrete fibration if for every arrow $g: d \to Pc$ in $\mathsf{D}$ there is a unique arrow $f: b \to c$ in $\mathsf{C}$ such that $Pf=g$.
I've read that (1) $P$ is a discrete fibration iff the following is a pullback square in $\mathsf{Set}$: $$ \require{AMScd} \begin{CD} \mathsf{C}^{\bullet \to \bullet} @>{P}>> \mathsf{D}^{\bullet \to \bullet} \\ @V {\rm codom} VV @VV{\rm codom}V \\ \mathsf{C} @>>P> \mathsf{D} \end{CD} $$
However, another source I've got claims that (2) $P$ is a discrete fibration iff the above square is a pullback square in $\mathsf{Cat}$.
Naively, I'm inclined to think that (1) and (2) are not equivalent. Also, I was able to prove (1) but not to prove (2). Therefore my questions are:
- is it true that (1) does not imply (2)?
- is (2) true at all? (I'm not looking for a proof; I'd like to work it out by myself, but I want to be sure that there is hope in proving it).
Later edit
Yes, (2) is true, I managed to exhibit a proof. It's surprising, however, that (1) and (2) are equivalent then.
It seems you have satisfied yourself that the claim is true, so I will explain how it fits into a bigger picture.
Given morphisms $f : A \to B$ and $g : C \to D$ in a category, we say that $f$ is left orthogonal to $g$, or $g$ is right orthogonal to $f$ if, for every commutative square of the form below, $$\require{AMScd} \begin{CD} A @>{h}>> C \\ @V{f}VV @VV{g}V \\ B @>>{k}> D \end{CD}$$ there is a unique morphism $l : B \to C$ such that $l \circ f = h$ and $g \circ l = k$. You should convince yourself this is the same as saying that the following is a pullback square: $$\begin{CD} \textrm{Hom} (B, C) @>{f^*}>> \textrm{Hom} (A, C) \\ @V{g_*}VV @VV{g_*}V \\ \textrm{Hom} (B, D) @>>{f^*}> \textrm{Hom} (A, D) \end{CD}$$ Thus, your (1) is the claim that a functor is a discrete fibration if and only if it is right orthogonal to the inclusion $\{ 1 \} \hookrightarrow \{ 0 \to 1 \}$.
But what other functors are left orthogonal to discrete fibrations? Well, it is a general fact that the class of morphisms left orthogonal to a given class is closed under isomorphisms, transfinite composition, retracts, and colimits. So, for example, the inclusion $\{ 0 \} \hookrightarrow \{ \cdots \to -2 \to -1 \to 0 \}$ is left orthogonal to discrete fibrations. Here is another general fact:
Proposition. In a cartesian closed category, assuming $\mathcal{L}$ is the class of morphisms left orthogonal to $\mathcal{R}$ and $\mathcal{R}$ is the class of morphisms right orthogonal to $\mathcal{L}$, the following are equivalent:
For every morphism $f : A \to B$ in $\mathcal{L}$ and every object $E$, $E \times f : E \times A \to E \times B$ is also in $\mathcal{L}$.
For every morphism $g : C \to D$ in $\mathcal{R}$ and every object $E$, $g^E : C^E \to D^E$ is also in $\mathcal{R}$.
For every morphism $f : A \to B$ in $\mathcal{L}$ and every morphism $g : C \to D$ in $\mathcal{R}$, the following is a pullback square: $$\begin{CD} C^B @>{C^f}>> C^A \\ @V{g^B}VV @VV{g^A}V \\ D^B @>>{D^f}> D^A \end{CD}$$
But this is something above and beyond the basic notion of orthogonality – there are examples of orthogonal classes in cartesian closed categories that do not have this property! – so in a sense you are right that it is surprising. I think it is not surprising, because for discrete fibrations it amounts to saying that if I have a natural transformation of diagrams in the base and a lift of the codomain diagram, then I can uniquely lift the natural transformation too. Obviously, the components of the lifted natural transformation are supplied by the standard definition of discrete fibration, so we only need to check that the relevant diagrams commute; and they do, because the lifts are unique.