5
$\begingroup$

In proposition 2.7. of the condensed notes of professors Scholze and Clausen it is said that the category of extremally disconnected sets is a site, but in the definition of a site in the Stacks Project (https://stacks.math.columbia.edu/tag/00VH) it is necessary for a site to have fibre products sometimes (Axiom (3) in the definition) and extremally disconnected sets don't have all fibre products.

The category of extremally disconnected sets is a site using the definition from the Stacks Project?

$\endgroup$
1
  • 2
    $\begingroup$ The Stacks Project is a work in progress and doesn't always have The Very Best things category theory has to offer (eg non-small sites that nonetheless have a good theory of sheaves!) See for now the awful temporary link: nlab-pages.s3.us-east-2.amazonaws.com/nlab/show/coverage and the definition of site there, which uses a coverage and comes from Johnstone's Sketches of an Elephant. $\endgroup$ Jan 24 at 10:00

1 Answer 1

8
$\begingroup$

Usage varies. Let's at least stipulate that "site" is synonymous with "category equipped with a Grothendieck topology".

Some, but not all, authors, require a site to have pullbacks, because this assumption simplifies the definition a bit. But e.g. the nlab gives the definition which doesn't assume one has pullbacks. Apparently this version of the definition goes back at least to SGA 4.

$\endgroup$
3
  • 2
    $\begingroup$ It should be noted that a size condition should be added in the absence of pullbacks… but the condition is automatically satisfied for essentially small sites. $\endgroup$
    – Zhen Lin
    Jan 24 at 0:15
  • $\begingroup$ @ZhenLin : although in the condensed formalism (as opposed to the pyknotic one), the site in question is not essentially small $\endgroup$ Jan 24 at 9:11
  • 1
    $\begingroup$ Indeed, that's why I mentioned it. To be explicit, the size condition I am speaking of is, I think, due to Shulman and is used to ensure that the category of small sheaves is closed under finite limits. $\endgroup$
    – Zhen Lin
    Jan 24 at 9:28

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.