By Sze-Tsen Hu

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3; 3. 2 and 3. 3; 4. 2 and 4. 5; Exact sequences 6. 1 Definition. (i) If f : F - G is a morphism of presheaves, we define the (presheaf) image of f to be PIm(f) = Ker(G - PCok f). (ii) If f : F - G is a morphism of sheaves, we define the (sheaf) image of f to be SIm(f) = Ker(G - SCok f). 6. 2 Exercise. Formulate the universal property that you would like a concept of 'image' to satisfy, and verify that PIm and SIm do in the categories Presh and Shv. Exercise. Check that PIm(f) is a presheaf whose abelian group of sections over each open U is the image of f(U), while SIm(f) is a sheaf whose stalk at each x E X is the image of fX 6.

These bijections fit together to give a bijection 0 such that p = p1 0 0. If U is open in X and v E r(U, E), then $[a[Ull = a[UM. Hence 0 is open, and by 3. 5 it is also continuous; since this means that 0 is a homeomorphism. // is bijective The sheafification of a presheaf 2. 4 4. 1 Given a presheaf F over X we can construct the sheaf space LF and then obtain a sheaf rLF called the sheafification of F. Now we have a morphism of presheaves nF : F - TLF defined as follows: given U open in X and s E F(U), s defines the function s: x x as in 3.

Furthermore p is continuous with respect to this topology on LF, since for any open U of X p-1(U) = U {s[V]; s E F(V) with V s U open }, and p is a local homeomorphism since on s[U] it has the continuous inverse s. A presheaf morphism f : F -, F' gives a collection of stalk maps fx : Fx - F'x and so a map Lf : LF - LF' such that LF -> LF' commutes; also Lf[s[U]] = f(U(s)[U], so Lf is continuous by 3. 5(c). o g)=Lf o Lg Check the functorial properties rL(f l L(id)=id. Now it is natural to ask what happens when we do L, IF in succession.