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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Sign up to join this communityThe 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.