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Here are the necessary notations and the statement of the proposition. I don't understand why the underlined sentence is true. Are the horizontal morphisms Cartesian fibration?

I would appreciate it if someone could give me some hints.

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1 Answer 1

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The two vertical maps are fully faithful, because $D\subset D'$ is, and it preserves $\mathcal K$-indexed colimits.

Therefore, the pullback of the right vertical morphism is also a full subcategory $E$ of $Fun_\mathcal K(P_\mathcal R^\mathcal K(C),D')$, and by the commutativity of the diagram, $Fun_\mathcal K(P_\mathcal R^\mathcal K(C),D)\subset E$ as full subcategories.

The statement that the square is cartesian is equivalent to the statement that this inclusion be an equivalence, i.e. that any object of $E$ be in the image of $Fun_\mathcal K(P_\mathcal R^\mathcal K(C),D)$. But objects of $E$ are exactly objects of $Fun_\mathcal K(P_\mathcal R^\mathcal K(C),D')$ whose restriction along $j$ lands in $D$, so this is equivalent to what Lurie claims.

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  • $\begingroup$ Sorry, Could you elaborate more on why this square being Cartesian implies it’s homotopy Cartesian? $\endgroup$
    – XiaYu
    Aug 12 at 8:23
  • $\begingroup$ Sorry, when I wrote cartesian I meant homotopy cartesian. $\endgroup$ Aug 12 at 8:36
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    $\begingroup$ Ok, I might get it. Since fully faithfulness is invariant under equivalence so at the start we can reduce to the case where the right vertical morphism is a fibration and still make sure the two vertical arrows induce full-subcategory relationships. thanks. $\endgroup$
    – XiaYu
    Aug 12 at 9:03

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