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Let be a functor, and let be a morphism of . Then applying the following (perhaps incorrect) specialization of the definition for quasicategories, we want to show that the canonical map 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 and such that . Now, if we look at the definition of the natural model structure on , the trivial fibrations are the surjective (on objects) equivalences of categories. This means that if we require that 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 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 the map and a morphism in , an object in the pullback category is a compatible pair
Can there be two different objects of over this? I.e. can there be a non-identity morphism
in for a non-identity morphism in mapping to the identity morphism on the above object in the pullback category?
It cannot, because a non-identity would map to a non-identity morphism
in , 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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