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Unformatted text preview: SOLUTIONS TO HARTSHORNE CHAPTER II & III WINTER 2008 – MATH 632 1. Section II.1 – Sheaves Problem 1.1. Let A be an abelian group, and define the constant presheaf associated to A on the topological space X to be the presheaf U 7→ A for all U 6 = ∅ , with restriction maps the identity. Show that the constant sheaf A defined in the text is the sheaf associated to this presheaf. Proof. [by Zachary Scherr ] Let F denote the constant presheaf defined in the problem, and define a map ϕ : F → A such that for each open U ⊆ X , p ∈ U and s ∈ F ( U ), ϕ ( U )( s )( p ) = s . ϕ is well-defined, since ϕ ( U )( s ) : U → A is continuous as A is given the discrete topology. It is also trivial to check that ϕ gives a morphism of presheaves. To show that A is in fact the sheafification of F , it suffices to show that ϕ induces an isomorphism on all stalks. Fix p ∈ X , then we have the induced homomorphism ϕ p : F p → A p . First we show that ϕ p is injective. Suppose that h U,s i ∈ Ker ϕ p , so that h U,ϕ ( U )( s ) i = 0. Then there exists an open set W ⊆ U so that h W,ϕ ( U )( s ) | W i = h W,ϕ ( W )( s | W ) i where ϕ ( W )( s | W ) is the zero map on W . By definition of ϕ however, this can only happen if s | W = 0, so h U,s i = h W,s | W i = 0, so indeed ϕ p is injective. Next we show that ϕ p is surjective. Suppose that h U,f i ∈ A p . Choose some t ∈ Im f ⊆ A , then as A is given the discrete topology, V = f- 1 ( t ) is open in U . Then h U,f i = h V,f | V i , where f | V ( V ) = t . Thus h U,f i = h V,f | V i = h V,ϕ ( V )( t ) i = ϕ p ( h V,t i ), which shows that ϕ p is surjective. Thus ϕ p is an isomorphism for each p ∈ X , so indeed A is the sheafification of F . 1 2 WINTER 2008 – MATH 632 Problem 1.2. (a) For any morphism of sheaves ϕ : F → G , show that for each point P , (ker ϕ ) P = ker( ϕ P ) and (im ϕ ) P = im( ϕ P ). (b) Show that ϕ is injective (respectively, surjective) if and only if the induced map on the stalks ϕ P is injective (respectively, surjective) for all P . (c) Show that a sequence ··· → F i- 1 ϕ i- 1 → F i ϕ i → F i +1 → ··· of sheaves and morphisms is exact if and only if for each P ∈ X the corresponding sequence of stalks is exact as a sequence of abelian groups. Proof. [by Kevin Wilson ] Part a. Suppose that the pair h U,s i is killed by ϕ P . Then we have that h U,ϕ ( U )( s ) i = 0. But that means that for some open V ⊆ U containing P we have that ϕ ( U )( s ) | V = ϕ ( V )( s | V ) = 0. As h U,s i = h V,s | V i , we may as well replace U with V and s with s | V . Now ϕ ( U )( s ) = 0 means that s ∈ ker( ϕ ( U )), which means by defi- nition that h U,s i ∈ ker( ϕ ) P ....
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