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    • CommentRowNumber1.
    • CommentAuthorHarry Gindi
    • CommentTimeAug 28th 2010
    • (edited Aug 28th 2010)

    Let p:CDp: C\to D be a functor, and let f:yxf:y\to x be a morphism of CC. Then applying the following (perhaps incorrect) specialization of the definition for quasicategories, we want to show that the canonical map (Cf)(Cx)× (Dp(x)(Dp(f))(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:zxg: z\to x and h:p(z)p(y)h:p(z)\to p(y) such that p(g)=p(f)hp(g)=p(f) \circ h. Now, if we look at the definition of the natural model structure on CatCat, the trivial fibrations are the surjective (on objects) equivalences of categories. This means that if we require that (Cf)(Cx)× (Dp(x)(Dp(f))(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 (Cf)(Cx)× (Dp(x)(Dp(f))(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?

    • CommentRowNumber2.
    • CommentAuthorUrs
    • CommentTimeAug 29th 2010
    • (edited Aug 29th 2010)

    • CommentRowNumber3.
    • CommentAuthorHarry Gindi
    • CommentTimeAug 29th 2010
    • (edited Aug 29th 2010)

    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.

    • CommentRowNumber4.
    • CommentAuthorUrs
    • CommentTimeAug 29th 2010
    • (edited Aug 29th 2010)

    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:XYp : X \to Y the map and f:x 1x 2f : x_1 \to x_2 a morphism in XX, an object in the pullback category X/x 2× Y/p(x 2)Y/p(f)X/{x_2} \times_{Y/p(x_2)} Y/p(f) is a compatible pair

    ( a x 1 f x 2, b p(x 1) p(f) p(x 2)) \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/fX/f over this? I.e. can there be a non-identity morphism

    a h a x 1 f x 2 \array{ && a \\ &\swarrow & \downarrow^{\mathrlap{h}} & \searrow \\ && a' \\ \downarrow & \swarrow && \searrow & \downarrow \\ x_1 &&\stackrel{f}{\to}&& x_2 }

    in X/fX/f for h:aah : a \to a' a non-identity morphism in XX mapping to the identity morphism on the above object in the pullback category?

    It cannot, because a non-identity hh would map to a non-identity morphism

    a h a x 2 \array{ && a \\ && \downarrow^{\mathrlap{h}} & \searrow \\ && a' \\ & && \searrow & \downarrow \\ &&&& x_2 }

    in X/x 2X/{x_2}, hence to a non-identity morphism in the pullback category.

    • CommentRowNumber5.
    • CommentAuthorUrs
    • CommentTimeAug 29th 2010

    I turned this discussion into a formal proof at Cartesian morphism – Definition – In categories – Reformulations. Have a look.