6
$\begingroup$

First of all, I give my definition of weak sheaves:

By a weak sheaf on a topological space $ X $, we mean a presheaf $F$ such that for all open covering $\{ U_i\}_{i\in I} $ of $X$ sheaf condition holds; that is, the diagram
\begin{equation*} F(X)\stackrel{}{\rightarrow} \prod_{i\in I}F(U_i)\stackrel{}{\rightrightarrows}\displaystyle\prod_{(i,{i'})\in I\times I}F(U_i\times_XU_{i'}) \end{equation*} is an equalizer.

Notice that this condition for other open subsets of $X$ is not necessary. While for a sheaf this condition holds for every open subset $U$ of $X$ and all open coverings $\{ U_i\}_{i\in I} $ of $U$.


Update 1: Clearly, every sheaf is a weak sheaf, but not conversely. As an example, if $S=\{0,1\}$, then the constant presheaf with the value $S$ on a non-empty space is a weak sheaf but not a sheaf.


Consider weak sheaves as a full subcategory of the category of presheaves.

Now my questions are

(1) Does the category of "weak sheaves" constitute an (elementary) topos? or at least which properties of the category of sheaves are true for that of "weak sheaves"?

(2) Is the category of "weak sheaves" of abelian groups an abelian category?

In sheaf theory, these properties are obtained by a tool called sheafification. So I think we should have a reflection from presheaves to weak sheaves. But I don't know if there is such a reflection. Maybe another idea works as well!


Update 2: I don't know if it is possible to discuss a similar situation over a site, (say on a final object of the site, if it exists), or as David Roberts mentioned in comments, consider a fixed object $X$ in the site and only $X$ itself has nontrivial covering families (probably).

One motivation for this question is that you are still able to survey local problems by weak sheaves, mainly about the whole space. For example, local solutions of differential equations in de Rham cohomology by the Poincare lemma. But what is the need for these solutions to open submanifolds, as a sheaf does? I am aware that this gives rise to a resolution of the constant sheaf $\underline{\mathbb{R}}$, and then isomorphisms between de Rham cohomology and singular, Cech, Alexander-Spanier cohomologies on manifolds. But I don't believe these are admissible answers.

In fact, I want to know why we need to strengthen "weak sheaves" to the notion of sheaves.


Update 3:

Every sheaf is an assignment of a weak sheaf to each open subset of the space.


I am interested in a clear explanation.

$\endgroup$
26
  • 2
    $\begingroup$ What happens if you take the open cover to consist of all open sets of $X$? I'm pretty sure a weak sheaf as you define it is a sheaf, but I'd like to see what you are taking as the definition of a sheaf. $\endgroup$
    – David Roberts
    Commented May 20, 2019 at 5:56
  • 7
    $\begingroup$ What's the motivation behind this definition? The sheaf condition has to hold because sheaf theory is about solving local problems, not simply because people wanted an abelian category for no reason. Do you have natural examples of weak sheaves that are not sheaves (by natural, I mean an example that you thought of before writing down this definition, for example). $\endgroup$ Commented May 20, 2019 at 9:11
  • 2
    $\begingroup$ Oh, I see. So I don't think there is a site structure on the lattice of opens of $X$ such that only $X$ itself has nontrivial covering families as in the condition, but all other sets have no covering families. So weak sheaves are probably not a reflective subcategory of presheaves, else you'd get such a site structure, I think. $\endgroup$
    – David Roberts
    Commented May 20, 2019 at 10:47
  • 2
    $\begingroup$ Such a reflection is equivalent to a Lawvere-Tierney topology on the category of presheaves, which I'm pretty such is equivalent to a site structure on the category arising from the lattice of opens. $\endgroup$
    – David Roberts
    Commented May 21, 2019 at 8:02
  • 2
    $\begingroup$ there can be many L–T topologies on a given topos. The identity reflector always gives one, for instance. Just as there can be different Grothendieck topologies on a given category (yes, even when that is the lattice of opens of a space, if the space is mathoverflow.net/questions/331973/… enough). $\endgroup$
    – David Roberts
    Commented May 22, 2019 at 1:46

0

You must log in to answer this question.

Browse other questions tagged .