1
$\begingroup$

I am reading Andreas Gathmann's notes on Algebraic geometry,http://www.mathematik.uni-kl.de/~gathmann/class/alggeom-2002/main.pdf

Def 7.1.14(iv)says the following

As usual, a sequence of sheaves and morphisms $$ \cdots \rightarrow F_{i−1} \rightarrow F_i \rightarrow F_{i+1} \rightarrow \cdots $$ is called exact if ker$( F_i \rightarrow F_{i+1}) = $im$( F_{i−1}\rightarrow F_i ) $ for all $i$.

However,even every arrow is a morphism of sheaves, the image of a sheaf will not necessarily be a sheaf, so what does this definition really mean? I have two guesses.

One is that if one of the image is not a sheaf, then the sequnce is never exact. The other is that the notation im$( F_{i−1}\rightarrow F_i ) $ stands for the sheafification of the presheaf image.

Which one is true? (or both are wrong?)

$\endgroup$
  • $\begingroup$ It depends on whether you're working in the category of sheaves or the category of presheaves. Also, in the category of sheaves, images (in the categorical sense) of morphisms generally can't be computed pointwise. $\endgroup$ – user14972 Jun 13 '15 at 7:07
  • $\begingroup$ This definition ceratinly works for presheaves.....I have no question about that. $\endgroup$ – Z. Jiang Jun 13 '15 at 7:12
  • 4
    $\begingroup$ Your second guess is the correct one. $\endgroup$ – Zavosh Jun 13 '15 at 7:16
  • 1
    $\begingroup$ The presheaf image is a sub-presheaf of the codomain sheaf. The sheafification of the image-presheaf can still be assumed to be a subsheaf of the codomain sheaf. The obstruction for the image presheaf to be a sheaf is that you might have a cover $\{U_i\}$ of a set $U$, with sections $s_i$ on each $U_i$ that agree on any intersection, but the glued section in the codomain sheaf does not exist in the image presheaf. The sheafification adds such sections to the presheaf to make it a sheaf. $\endgroup$ – Arthur Jun 13 '15 at 8:25
1
$\begingroup$

Sheaves of abelian groups form an abelian category, and the definition given is the usual definition of the exactness of a sequence in an abelian category. Image here means the categorical image in an abelian category, which is equivalently either the kernel of the cokernel or the cokernel of the kernel; either way, to compute it you have to compute a cokernel, and that cokernel can be computed as a presheaf cokernel and then sheafified.

$\endgroup$
1
$\begingroup$

In any Abelian category, the image of a morphism is the kernel of the cokernel.

We can use the relationship bewteen sheaves and presheaves to compute the image, however: if $i,a$ are the inclusion and sheafification functors, then if $f$ is any morphism:

$$ \begin{align}\mathop{\mathrm{im}} f &= \ker \mathop{\mathrm{coker}} f \\&\cong \ker ai(\mathop{\mathrm{coker}} f) \\&\cong a(\ker \mathop{\mathrm{coker}}if) \\&\cong a(\mathop{\mathrm{im}} (if)) \end{align}$$

using the fact that $ai \cong 1$, $a$ is left exact, and $i$ is right exact. Thus, the image of a sheaf morphism is indeed the sheafification of its image when viewed as a presheaf morphism.

Note, for example, $\mathop{\mathrm{im}} (f)$ means the image as a sheaf morphism, and $\mathop{\mathrm{im}} (if)$ means the image as a presheaf morphism. Allow me to emphasize that the result of this calculation literally is the image of a morphism in the category of sheaves of abelian groups.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.