Summary from last time Let f : X Y be a map of Our goal was to - PowerPoint PPT Presentation

Summary from last time ◮ Let f : X → Y be a map of Our goal was to construct a functor f ! : D + (Sh( Y )) → D + (Sh( X )) so that there are natural isomorphisms Hom D + (Sh( Y )) ( Rf ! F , G ) = Hom D + (Sh( X )) ( F , f ! G ) for all F ∈ D + (Sh( X )), G ∈ D + (Sh( Y )). ◮ We assume that R i f ! = 0 for all i > n . From that we construct a resolution 0 → Z X → K 0 → K 1 → ... → K n → 0 where K i are f -soft and flat (i.e. having torsion free stalks). ◮ When F is a sheaf (in Sh( X ) rather than D + (Sh( Y ))) we then have 0 → F → F ⊗ K 0 → ... → F ⊗ K n → 0. We prove this to be again a f -soft resolution. ◮ Let us write K U := j ! j ∗ K for K ∈ Sh( X ) and j : U ֒ → X open. ◮ For G ∈ Sh( Y ), we define f ! G to be the complex of sheaves given as: ( f ! G ) − a ( U ) = Hom Sh( Y ) ( f ! ( K a U ) , G ), for any open j : U ⊂ X .

Proof of Verdier duality, V ( f ! G ) − a ( U ) = Hom Sh( Y ) ( f ! ( K a U ) , G ) ◮ Why is this even a sheaf? ◮ Well, Cheng-Chiang made a mistake, and it isn’t. The reason is that H om ( − , G ) rarely give a sheaf unless G is injective. ◮ So I mean, let us suppose G is injective. Now let’s prove this. ◮ Lemma The functor F �→ f ! ( F ⊗ K a ) from Sh( X ) to Sh( Y ) is exact. ◮ Proof. We have seen last time that F ⊗ K a are f -soft. Then f ! is exact on f -soft sheaves. ◮ So for an open covering U i of V and U ij = U i ∩ U j , we begin with � � Z ij → Z i → Z V → 0 . then we have � f ! ( Z U ij ⊗ K a ) → � f ! ( Z U i ⊗ K a ) → f ! ( Z V ⊗ K a ) → 0 . then we have 0 → Hom Sh( Y ) ( f ! ( K a V ) , G ) → Hom Sh( Y ) ( ⊕ f ! ( K a U i ) , G ) → Hom Sh( Y ) ( f ! ( ⊕K a U ij ) , G ) . This proves ( f ! G ) − a to be a sheaf when G is injective.

Proof of Verdier duality, VI ( f ! G ) − a ( U ) = Hom Sh( Y ) ( f ! ( K a U ) , G ) ◮ Lastly, we have to prove that when G is injective, there is a natural isomorphism α Hom Sh( Y ) ( f ! ( F ⊗ K a ) , G ) → Hom Sh( X ) ( F , f ! G − a ) − ◮ That is to say given any φ ∈ Hom Sh( Y ) ( f ! ( F ⊗ K a ) , G ), for any open U ⊂ X we want to have a natural (i.e. functorial) and unique α φ ∈ Hom( F ( U ) , Hom Sh( Y ) ( f ! ( K a U ) , G )) = Hom( F ( U ) , Hom Sh( Y ) ( f ! ( K a U ) , G )). ◮ That is, for any s ∈ F ( U ) and s ′ V ∈ f ! ( K a U )( V ) for any V ⊂ Y open, we have to construct ( α φ ( s ))( s ′ V ) ∈ G ( V ). ◮ This is evident; we just tensor s with s ′ V and feed it to φ ( V ). ◮ It’s easy to check all these are functorial (w.r.t. both F and G ), i.e. α indeed induces natural transformations. ◮ Now we have to go back to prove that α is an isomorphism. ◮ When F = Z U for some U ⊂ X open, one checks this readily. ◮ In general, we can resolve F via such. That F �→ f ! ( F ⊗ K a ) and Hom Sh( Y ) ( − , G ) are exact then gives us what we want.

Proof of Verdier duality, VII Hom Sh( Y ) ( f ! ( F ⊗ K a ) , G ) α → Hom Sh( X ) ( F , f ! G − a ) − ◮ For G ∈ D + (Sh( Y )) in general, we assume it is a complex of injective sheaves and define f ! G ′ to be the single complex associated to the double complex ( f ! G b ) a . ◮ We check that f ! so defined sends homotopic objects to homotopic objects, so that this is well-defined on D + (Sh( Y )). ◮ We claim that each ( f ! G b ) a ∈ Sh( X ) is injective. Indeed, for F ∈ Sh( X ) we have proved α Hom Sh( X ) ( F , ( f ! G b ) a ) ∼ Hom Sh( Y ) ( f ! ( F ⊗ K a ) , G b ) ← − is exact in F . ◮ We have to prove for F ∈ D + (Sh( X )): Hom D + (Sh( Y )) ( Rf ! F , G ) = Hom D + (Sh( X )) ( F , f ! G ) Again, we may assume F is a complex of injective sheaves. ◮ In this case, the Rf ! F in the left hand side can be computed via the f -soft resolution F ⊗ K a , or rather the single complex associated to the double complex F b ⊗ K a . Things are then reduced to the pre-derived level, namely the isomorphism α we proved above.

Sheaf Hom Hom D + (Sh( Y )) ( Rf ! F , G ) = Hom D + (Sh( X )) ( F , f ! G ) ◮ Since the Verdier duality (a.k.a. adjointness) characterize f ! . We have ( g ◦ f ) ! = f ! ◦ g ! . ◮ It is desirable to have Verdier duality in fully derived and/or sheaf version. Let’s prepare the notation. ◮ To begin with, for F , F ′ ∈ Sh( X ) we may construct H om X ( F , F ′ ) to be H om X ( F , F ′ )( U ) = Hom Sh( U ) ( F| U , F| U ′ ). ◮ This is a sheaf. But warning: for the stalks H om X ( F , F ′ ) x → Hom( F x , F ′ x ) can be neither injective nor surjective! ◮ It seems a good time to compare with F ⊗ F ′ , which is defined to be the sheafification of F ( U ) ⊗ F ( U ′ ). ◮ One readily checks that ( F ⊗ F ′ ) x = F x ⊗ F ′ x . ◮ Another remark: the usual adjoint property for H om X reads: H om Y ( F , f ∗ G ) = f ∗ H om X ( f ∗ F , G ) ◮ And maybe likewise (?) f ∗ ( F ⊗ G ) = f ∗ F ⊗ f ∗ G .

Derived Hom Hom D + (Sh( Y )) ( Rf ! F , G ) = Hom D + (Sh( X )) ( F , f ! G ) ◮ Viewing Hom( F , − ) : Sh( X ) → Ab , one may right derive it to RHom Sh( X ) ( F , − ) : D + (Sh( X )) → D + ( Ab ). ◮ In fact, it fits into a more general scheme of derived bifunctors, so that we have RHom Sh( X ) ( − , − ) : D − (Sh( X )) × D + (Sh( X )) → D + ( Ab ). We refer interested readers to [KS, § 1.10]. ◮ Likewise, we have H om ( − , − ) : D − (Sh( X )) × D + (Sh( X )) → D + (Sh) R ◮ Now our preferred version of Verdier duality reads RHom Sh( Y ) ( Rf ! F , G ) = RHom Sh( X ) ( F , f ! G ) ∈ D + ( Ab ) . H om X ( F , f ! G ) ∈ D + (Sh( Y )) . H om Y ( Rf ! F , G ) = f ∗ R R