1
$\begingroup$

Does anyone know a good example where the cokernel of a sheaf morphism is not a sheaf?

$\endgroup$
3
  • $\begingroup$ An additional example where the image of a sheaf is not a sheaf would be appreciated. $\endgroup$
    – Arthur
    Jun 25, 2014 at 17:40
  • 3
    $\begingroup$ For the sake of precision, you mean to compute its cokernel as a morphism of presheaves (which would give a presheaf that might not be a sheaf), rather than as a morphism of sheaves (which would always give a sheaf). $\endgroup$
    – user14972
    Jun 25, 2014 at 17:45
  • $\begingroup$ Search for "exponential exact sequence". $\endgroup$
    – Hoot
    Jun 25, 2014 at 18:42

2 Answers 2

6
$\begingroup$

Here is an example. For $X=\mathbb C$,let $F=(\mathcal O,+)$ be the sheaf of holomorphic functions (with addition) and $G=(\mathcal O^{*},\cdot)$ be the sheaf of nowhere zero holomorphic functions (with multiplication). Then there is a map of sheaves of abelian groups defined by $$exp:\mathcal O(U) \rightarrow\mathcal O^{*}(U),f\mapsto exp(f).$$ We can show that $coker^{pre}(exp)$ is not a sheaf.

Let $U = \mathbb C \setminus\{0\}$, which is covered by the open sets $U_{1}=\mathbb C\setminus [0,+\infty]$ and $U_2 = \mathbb C \setminus (−\infty, 0]$. By complex analysis, we know that the function $g=z\in\mathcal O^{*}(U)$ cannot be written as the exponential of some other holomorphic function $f\in\mathcal O(U)$. Thus $[g]\not=0 \in coker(exp(U))$.

On the other hand, the open sets $U_1,U_2$ are simply connected,so every nowhere zero function $g\hat{~}$ on them can be written as $exp(f\hat{~}$). Thus $coker(exp(U_i)) = 0$ for $i = 1,2$. But thus $coker^{pre}(exp)$ cannot be a sheaf, because the restriction of $g$ to the open cover $U_1, U_2$ of $U$ is zero on both sets, but globally nonzero.

$\endgroup$
2
  • $\begingroup$ What complex analysis do I need to know that $g = z$ cannot be written as the exponential of some other holomorphic function? $\endgroup$ Aug 7, 2020 at 21:23
  • $\begingroup$ @Siddharth Bhat, exponentials of holomorphic functions are never equal to $0$. $\endgroup$ Jun 18, 2021 at 1:13
3
$\begingroup$

Here's a simple example: consider shaves on the unit circle, which I will write as $\mathbf{R} / 1$. Let $F$ be the sheaf of sections of the bundle $\mathbf{R} \to \mathbf{R}/1$ given by the projection.

The cokernel of the "add 1" map is the sheaf $G$ of sections of the bundle $\mathbf{R}/1 \to \mathbf{R}/1$ given by the identity.

$F$ has no global sections but $G$ does: therefore $G$ is not the cokernel of the "add 1" map when computed as a map of presheaves.

EDIT: This is actually an example for the coequalizer of sheaves of sets, rather than a cokernel of abelian sheaves. (I'm talking about the coequalizer of the identity and the add 1 maps)

For another example in the same spirit, consider the abelian sheaves $F$ of continuously differentiable functions of period $1$ and $G$ of continuous functions of period $1$. What is the cokernel of the derivative map? On any proper open subset, $F(U) \to G(U)$ is surjective, so the cokernel sheaf is zero. However, the cokernel of $F(\mathbf{R}/1) \to G(\mathbf{R}/1)$ is $\mathbb{R}$, so the cokernel presheaf is nonzero.

The projection to the cokernel, incidentally, is

$$ f \mapsto \int_{0}^1 f(t) \, dt $$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .