Tensor For F , F Sh( X ) we have F F Sh( X ). Since tensor - PowerPoint PPT Presentation

Tensor ◮ For F , F ′ ∈ Sh( X ) we have F ⊗ F ′ ∈ Sh( X ). Since tensor commutes with taking stalk, the non-exactness comes from that F x might not torsion-free, and this can be taken care by a simple resolution 0 → P ′ → P → F → 0 where P is e.g. direct sum of Z U indexed by ( U , s ) with U ֒ → X open, s ∈ F ( U ). ◮ Hence we can derive ⊗ : we will continue to denote it by · ⊗ · : D ∗ (Sh( X )) × D ∗ (Sh( X )) → D ∗ (Sh( X )) where ∗ = ∅ , + , − or b . ◮ Lemma. For F ∈ D + (Sh( X )), G ∈ D + (Sh( Y )), and f : X → Y a map of locally compact Hausdorff spaces with R k f ! = 0 for k ≫ 0, we have Rf ! ( F ⊗ f ∗ G ) = Rf ! F ⊗ G . ◮ This is because · ⊗ f ∗ G preserves f -softness. ◮ Lemma. For G , G ′ ∈ D + (Sh( Y )), there is a morphism f ! G ⊗ f ∗ G ′ → f ! ( G ⊗ G ′ ) ◮ This follows from Rf ! ( f ! G ⊗ f ∗ G ′ ) = Rf ! f ! G ⊗ G ′ → G ⊗ G ′ and adjunction again.

f ! for smooth morphisms ◮ Last time, we proved the Poincar´ e-Verdier duality that f ! Z Y = Z X [ n ] for a topological submersion of (real) dimension n assuming Y is locally compact Hausdorff and locally contractible. ◮ Corollary. We also have f ! G = O Y / X ⊗ f ∗ G [ n ] for general G ∈ D + (Sh( Y )). ◮ For general f : X → Y the functor f ! is local on X : Indeed, for j : U ֒ → X open we have Hom Sh( X ) ( j ! F ′ , F ) = Hom Sh( U ) ( F ′ , j ∗ F ). Since j ! is exact this gives j ! = j ∗ . The same can be checked for topological submersion of dimension 0, giving n = 0 case of above. ◮ We have ( f ! G ) | U = j ! f ! G = ( f ◦ j ) ! G , which is what we want. ◮ Another preparation: for any Cartesian diagram p X ′ X f ′ f q Y ′ Y we have a natural transformation p ∗ f ! → ( f ′ ) ! q ∗ .

f ! for smooth morphisms, II f ! G = O Y / X ⊗ f ∗ G [ n ] for a topological submersion of (real) dimension n assuming ... ◮ We have the natural map f ! Z Y ⊗ f ∗ G → f ! G . Since we know f ! Z Y = O Y / X [ n ], it remains to prove this is an (quasi-)isomorphism. ◮ Since f ! is local on X , we may assume X = Y × R n . Consider the Cartesian diagram R n X p f Y pt ◮ Take U ⊂ R n and V ⊂ Y open we have R Γ( U × V ; f ! G ) RHom Sh( X ) ( Z U × V , f ! G ) = RHom Sh( Y ) ( Rf ! Z U × V , G ) = RHom Sh( Y ) ( H ∗ c ( U ; Z ) ⊗ Z V , G ) = R Γ( U ; p ! Z ) ⊗ R Γ( V ; G ) = RHom Ab ( H ∗ c ( U ; Z ) , Z ) ⊗ RHom Sh( Y ) ( Z V , G ) = Since p ! Z is just a shift of Z R n , the last item sheafifies to R Γ( U × V ; f ! Z Y ⊗ f ∗ G ) and we have proved the corollary.

Computation of Rf ! ◮ It relies on the computation of Rf ! locally on X . Thanks to proper base change it is reduced to � 0 , if k � = n H k c ( R n ; Z ) = Z , if k = n ◮ At the same time, even earlier we assumed that f ! has finite cohomological dimension. Again by proper base change this reduces to that H k c ( f − 1 ( y ); F ) = H k c ( X ; i ! F ) = 0 ∀ k ≫ 0 where i : f − 1 ( y ) → X and i ! = i ∗ . ◮ That is, we can reduce to the fiber and then again reduce to H ∗ c of some sheaf on the whole space.

Distinguished triangles ◮ Recall that for j : U ֒ → X open and F ∈ Sh ( X ) we put F U := j ! j ∗ F = j ! j ! F . ◮ Now for U , V ⊂ X open we have Mayer-Vietoris 0 → F U ∩ V → F U ⊕ F V → F U ∪ V → 0 where the natural maps can be realized by adjunction j ! j ! → id. ◮ This extends from Sh( X ) to C (Sh( X )) and short exact sequence in C (Sh( X )) gives distinguished triangles in K (Sh( X )) and D (Sh( X )). Since derived functor preserves distinguished triangles, we have +1 H ∗ c ( U ∩ V ; F ) → H ∗ c ( U ; F ) ⊕ H ∗ c ( U ; F ) → H ∗ c ( U ∪ V ; F ) − − → ◮ Likewise, for closed subset E 1 , E 2 ⊂ X using id → i ∗ i ∗ = i ! i ∗ , we have 0 → F E 1 ∪ E 2 → F E 1 ⊕ F E 2 → F E 1 ∩ E 2 → 0 +1 H ∗ c ( E 1 ∪ E 2 ; F ) → H ∗ c ( E 1 ; F ) ⊕ H ∗ c ( E 2 ; F ) → H ∗ c ( E 1 ∩ E 2 ; F ) − − → ◮ One more: Suppose j : U ֒ → X is open and i : ֒ → Z ֒ → X is the complement. We have +1 0 → F U → F → F Z → 0 and H ∗ c ( U ; F ) → H ∗ c ( X ; F ) → H ∗ c ( Z ; F ) − − →

Compute H ∗ ([0 , 1]; − ) ◮ Prop. For any F ∈ Sh([0 , 1]), we have H k c ([0 , 1]; F ) = H k ([0 , 1]; F ) = 0 for k ≥ 2. Also H 1 ([0 , 1]; Z (0 , 1) ) = Z , i.e. H ∗ c ( R 1 ; Z ) = Z [1]. ◮ Suppose s ∈ H k ([0 , 1]; F ) is non-zero for some k > 1. For any t ∈ (0 , 1) we have the Mayer-Vietoris 0 = H k − 1 ( { a } ; F ) → H k ([0 , 1]; F ) → H k ([0 , a ]; F ) ⊕ H k ([ a , 1]; F ) Hence the restriction of s to either [0 , t ] or [ t , 1] is non-zero. We can decompose closed interval like this forever, and they have to “converge” to some point x . ◮ But for some injective resolution 0 → F → I 0 → ... → I k − 1 → I k → I k +1 → .. , this s is represented by some ker( I k ([ a , b ]) → I k +1 ([ a , b ]) which has to come from I k − 1 ([ a , b ]) when a , b → x . This shows s | [ a , b ] = 0 for [ a , b ] → x and thus we have a contradiction.

Compute H ∗ c H k − 1 ( { a } ; F ) δ → H k ([0 , 1]; F ) → H k ([0 , a ]; F ) ⊕ H k ([ a , 1]; F ) − δ ◮ When k = 1 and F = Z (0 , 1) , we have 0 → H 0 ( { a } ; Z ) − → H k ([0 , 1]; Z (0 , 1) ) → H k ([0 , t ]; Z (0 , 1) ) ⊕ H k ([ t , 1]; Z (0 , 1) ). By the same trick we can show the last morphism is zero and the middle morphism is an isomorphism, hence the assertion. ◮ With careful bookkeeping, one verifies that when [0 , 1] is reversed (i.e. t �→ 1 − t ), there is a − sign on δ . ◮ Given the proposition, suppose we have any X that we find “ n -dimensional” so that we can map X → [0 , 1] with “( n − 1)-dimensional” fiber (or, decompose X using Mayer-Vietoris and have each part with this propety). Proper base change then allows us to do induction and prove H k c ( X ; F ) = 0 for all F ∈ Sh( X ), k > n . ◮ We have H ∗ c ( R n ; Z ) = Z [ n ] by induction. And again careful bookkeeping shows that any linear transformation T on R n induces c ( R n ; Z ). sgn( T ) on H ∗