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HartshorneSolutions - PROBLEM SET 1 DUE TUESDAY JANUARY 18...

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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 A for all U = , 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 U, s Ker ϕ p , so that U, ϕ ( U )( s ) = 0. Then there exists an open set W U so that W, ϕ ( U )( s ) | W = W, ϕ ( W )( s | W ) 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 U, s = W, s | W = 0, so indeed ϕ p is injective. Next we show that ϕ p is surjective. Suppose that U, f 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 U, f = V, f | V , where f | V ( V ) = t . Thus U, f = V, f | V = V, ϕ ( V )( t ) = ϕ p ( V, t ), 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
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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 U, s is killed by ϕ P . Then we have that U, ϕ ( U )( s ) = 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 U, s = V, s | V , 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 U, s ker( ϕ ) P . In the other direction, if U, s ker( ϕ ) P then for some open V U containing P we have that ϕ ( V )( s | V ) = ϕ ( U )( s ) | V = 0. So in particular, U, s ker( ϕ P ). Now we must show that im( ϕ P ) = im( ϕ ) P . This follows from the fact that for any presheaf F and its associated sheaf F + that F P = F + P . For consider the map θ P : F P F + P . This is certainly surjective. For if U, f F + P , there exists an open neighborhood V U of P U and t F ( V ) such that f ( Q ) = t Q for all Q V . Then in particular, θ ( t ) = f | V by definition of θ . Then U, f = V, θ ( t ) = θ P ( V, t ).
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