《Hartshorne_代数几何解答》.pdfVIP

Chapter 2 2.1 1.1 Show that A has the right universal property. Let G be any sheaf and let F be the presheaf U → A, and suppose ϕ : F → G. Let f ∈ A (U ), i.e. f : U → A is a continuous map. Write U = Vα with Vα the connected components of U so f (Vα ) = aα ∈ A. Then we get bα = ϕVα (aα ) since F (U ) = A for any U , and since G is a sheaf we obtain b ∈ G(U ). We define : A → G by U (f ) = b. This map has the right properties. 1.2 a) Observe (ker ϕ)P = lim (ker ϕ)(U ) = lim ker ϕU is a subgroup of FP , as is ker ϕP , so we show −→U P −→U P equality inside FP . For x ∈ (ker ϕ)P pick (U, y) representing x, with y ∈ ker ϕU . Then the image of y in FP , i.e. x, is mapped to zero by ϕP . Conversely, if x ∈ ker ϕP there exist (U, y) with y ∈ F (U ) and ϕU (y) = 0 so x ∈ (ker ϕ)P . For im ϕ one proceeds similarly, noting only that (im ϕ)P = lim im ϕU since the presheaf −→U P “imϕ” and the sheaf im ϕ have the same stalks at every point. b) The morphism ϕ is inj. resp surj. iff (ker ϕ)P = 0 resp. (im ϕ)P = GP for all P . By part a), this holds iff ker ϕP = 0 resp. im ϕP = GP for all P , that is, iff ϕP is inj. resp. surj. i−1 i i−1 i−1 i i c) We have im ϕ = ker ϕ iff im ϕP = (im ϕ )P = (ker ϕ )P = ker ϕP . 1.3 a) By 1.2, ϕ is surjective iff ϕP is surj. for all P , that is, iff for all U and all s ∈ G(U ) there exist (U , t ) ∈ F with t ∈ F (U ) such that ϕ (t ) = s , or shrinking U if need be, iff for all P ∈ U we have i i P i i Ui i P P i ϕ (t ) = s with each U a nbd of P . The U cover U . Ui