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Let $p: C\to D$ be a functor, and let $f:y\to x$ be a morphism of $C$. Then applying the following (perhaps incorrect) specialization of the definition for quasicategories, we want to show that the canonical map $(C\downarrow f) \to (C\downarrow x)\times_{(D\downarrow p(x)} (D\downarrow p(f))$ is a trivial fibration in the natural model structure. However, if we write out what the (strict 2-) pullback means, the objects are precisely the pairs of morphisms $g: z\to x$ and $h:p(z)\to p(y)$ such that $p(g)=p(f) \circ h$. Now, if we look at the definition of the natural model structure on $Cat$, the trivial fibrations are the surjective (on objects) equivalences of categories. This means that if we require that $(C\downarrow f) \to (C\downarrow x)\times_{(D\downarrow p(x)} (D\downarrow p(f))$ to only be a trivial fibration and not an honest isomorphism, we only have that the filler is unique up to a contractible groupoid of choices. That is, the arrow is not cartesian in the sense of Grothendieck, but it is homotopy cartesian.
Then the question: should we require that the map $(C\downarrow f) \to (C\downarrow x)\times_{(D\downarrow p(x)} (D\downarrow p(f))$ is an isomorphism of categories? Have I somewhere missed that uniqueness can actually be derived?
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By the way, the nLab page cartesian morphism has the same definition (that it should be a surjective equivalence), but when I asked on MO, Anton said that it should be an isomorphism of categories.
I had posted a reply but removed it when I saw I needed to improve something but had to go offline.
What Anton says in his MO-reply is effectively that requiring a surjective equivalence in this case implies that this is actually an isomorphism.
Here is what I wrote in the above reply:
using the notation at Cartesian morphism for $p : X \to Y$ the map and $f : x_1 \to x_2$ a morphism in $X$, an object in the pullback category $X/{x_2} \times_{Y/p(x_2)} Y/p(f)$ is a compatible pair
$\left( \array{ && a \\ &&& \searrow \\ x_1 &&\stackrel{f}{\to}&& x_2 } \;\;\;\;\;\;\,,\;\;\;\;\;\; \array{ && b \\ & \swarrow && \searrow \\ p(x_1) &&\stackrel{p(f)}{\to}&& p(x_2) } \right)$Can there be two different objects of $X/f$ over this? I.e. can there be a non-identity morphism
$\array{ && a \\ &\swarrow & \downarrow^{\mathrlap{h}} & \searrow \\ && a' \\ \downarrow & \swarrow && \searrow & \downarrow \\ x_1 &&\stackrel{f}{\to}&& x_2 }$in $X/f$ for $h : a \to a'$ a non-identity morphism in $X$ mapping to the identity morphism on the above object in the pullback category?
It cannot, because a non-identity $h$ would map to a non-identity morphism
$\array{ && a \\ && \downarrow^{\mathrlap{h}} & \searrow \\ && a' \\ & && \searrow & \downarrow \\ &&&& x_2 }$in $X/{x_2}$, hence to a non-identity morphism in the pullback category.
I turned this discussion into a formal proof at Cartesian morphism – Definition – In categories – Reformulations. Have a look.
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