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In the book "Spectral Agebraic Geometry" that Jacob Lurie is currently writing, he gives a construction (11.1.5.1), which describes the Serre functor: it has already been shown that any proper $A$-linear category $C$ over a ring spectrum (or just ring, I don't think anything about the construction is inherently $\infty$-categorical) $A$ is compactly generated and therefore dualizable. Let $e:C\otimes C^\vee\rightarrow Mod_A$ be the duality datum. Then this functor has a right adjoint $e^R:Mod_A\rightarrow C\otimes C^\vee$ which is also $A$-linear and can be identified with an $A$-linear functor $S_C:C\rightarrow C$, that can be more explicitly described as the composition

$C\simeq Mod_A\otimes C\overset{e^R\otimes id}{\rightarrow} C\otimes C^\vee \otimes C\overset{id\otimes e}\rightarrow C\otimes Mod_A\simeq C$.

Furthermore he describes in the appendices (D.7.2) that $C^\vee$ can be identified with $Ind(C_c^{op})$, where $C_c$ denotes the full subcategory of $C$ on the compact objects and that the duality datum $e$, restricted to the compact objects can be identified with the enriched $hom$ functor $\underline{hom}:C_c\otimes C_c\rightarrow Mod_A$.

This would certainly a very beautiful description of the Serre functor of such a category, alas, I fail to see why the described thing is indeed this functor. Given that he only only offhandedly mentions that, for $F,G\in C_c$ it induces an equivalence $\underline{hom}(F,S_C(G))\simeq \underline{hom}(G,F)^\vee$ and therefore those two coincide, I am probably missing something basic, maybe some well-known fact about the interaction of duals of categories and adjoints, but no matter how long I stare at it, I just cannot see it. Anything that might help me understand why those this abstract category-theoretic construction gives us indeed the Serre functor would be greatly appreciated.

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    $\begingroup$ The Serre functor is uniquely characterized by $\hom(F,S(G))\cong\hom(G,F)^\vee$. Is it this statement that you cannot prove? Or that his construction fits into such isomorphism? Or is it something else? $\endgroup$ Commented Jul 11, 2018 at 16:34
  • $\begingroup$ Yes, , I don't understand how to get from the the described construction to the isomorphism you mentioned. $\endgroup$
    – John Doe
    Commented Jul 11, 2018 at 16:47

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