Latest revision as of 17:07, 24 September 2010

Week Four

This is the page with specific information for Week 4 of our Algebraic Geometry Graduate Reading Course

Discussion Leader: Amanda

For 9/22: Be prepared to discuss 3.4
- Hand in 6 written up problems
For 9/24: Read & be prepared to discuss 3.5
For 9/27: Read & be prepared to discuss 3.5, 3.6, 3.7.
- Meeting with faculty: Bring questions from all of Chapter 3.

Why does sheafification being left-adjoint to the forgetful functor "explain" why you don't need to sheafify when taking kernel, and why you need to sheafify when taking cokernel (see 3.4.K)?
- If $f:A \to B$ is a morphism of sheaves, then we really have a morphism of presheaves $F(f): F(A) \to F(B)$, where $F$ denotes the forgetful functor. Since $F$ is right-adjoint to sheafification, it commutes with kernels, so we "should" have, $ker_{pre} F(f) = F(ker_{sheaf} f)$. Since $F$ is just the forgetful functor, it doesn't really do anything. So taking the kernel in presheaves "should" actually be a kernel in sheaves. So (sh, F) being adjoint explains why we don't need to sheafify with kernels.

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 == Questions ==
 * Why does sheafification being left-adjoint to the forgetful functor "explain" why you don't need to sheafify when taking kernel, and why you need to sheafify when taking cokernel (see 3.4.K)?
+** If $f:A \to B$ is a morphism of sheaves, then we really have a morphism of presheaves $F(f): F(A) \to F(B)$, where $F$ denotes the forgetful functor. Since $F$ is right-adjoint to sheafification, it commutes with kernels, so we "should" have, $ker_{pre} F(f) = F(ker_{sheaf} f)$. Since $F$ is just the forgetful functor, it doesn't really do anything. So taking the kernel in presheaves "should" actually be a kernel in sheaves. So (sh, F) being adjoint explains why we don't need to sheafify with kernels.
 == Comments ==