0
$\begingroup$

Let $\mathcal{C}= {\bf{Set}}$, and let $X= \mathbb{C}$, and the category of open sets of $X$ equal $X$. We can define a presheaf on this space by the contravariant functor $\mathcal{F}: \mathbb{C} \to {\bf{Set}}$. By Axioms (a) and (b) in the first page of http://ocw.mit.edu/courses/mathematics/18-726-algebraic-geometry-spring-2009/lecture-notes/MIT18_726s09_lec03_sheaves.pdf we can define this to become the sheaf of open complex sets. One can see that this presheaf is also a sheaf. Does this hold in general, i.e. when is a presheaf a sheaf?

$\endgroup$
  • 3
    $\begingroup$ First, a presheaf is a contravariant functor, not a covariant one. Second, there are tons of notes on sheaf theory, why don't you google for them? $\endgroup$ – Alexei Averchenko Dec 1 '12 at 16:34
  • 1
    $\begingroup$ A sheaf is by definition a presheaf that satisfies certain additional properties... $\endgroup$ – user314 Dec 1 '12 at 16:57
  • 2
    $\begingroup$ I'm not sure what you are asking. Are you asking if every presheaf is a sheaf? The answer is no. Are you asking what the conditions are for a presheaf to be a sheaf? It's written in the PDF you linked. $\endgroup$ – Najib Idrissi Dec 1 '12 at 17:21
3
$\begingroup$

Here is an example of a pre-sheaf that is not in general a sheaf:

The constant pre-sheaf: Let $X$ be a topological space, and $S$ a set, then define a pre-sheaf $\mathcal{F}$ as follows:

$\mathcal{F}(U) = S$ for all non empty open sets $U$.

$\mathcal{F}(\emptyset) = \{e\}$ where $\{e\}$ is just some one point set.

The last condition is a technicality because sheaf axioms force the sections over an empty set to be the final object in the category, which in our case is the category of sets.

Verify that $\mathcal{F}$ is a pre-sheaf with obvious restriction maps, but that it is not a sheaf in general.

Hint: to show it is not a sheaf consider $X$ to be a two point set with discrete topology and $S$ with more than one element, and show that sheaf gluability fails.

A more natural example would be to show that the pre-sheaf of bounded functions on $\mathbb{R}$ with restriction of functions as restriction maps is not a sheaf (again this fails to satisfy gluability).

$\endgroup$

Your Answer

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