# Coverages that are not pretopologies

A coverage on a category $C$ is a collection of covering families $\{u_i \to a\}$ for each object $a$ of $C$ such that for each arrow $b\to a$ there is a covering family for $b$ which fits into a commutative square with the original covering family.

A pretopology is an example of a coverage, but much stronger, in that the weak pullbacks are required to be pullbacks, covering families need to be closed under composition, and families consisting of a single isomorphism are covering.

In order to define sheaves, it is enough to start with a coverage, and indeed there is a very useful coverage on the category $Diff$ of finite dimensional smooth manifolds consisting of good open covers. Also, given a topological space, and a basis for the topology on it, one can define a coverage which is not a pretopology using these subsets.

However, the coverage of good open covers on $Diff$ does satisfy the 'covers compose' axiom of a pretopology, but not the other two, so is not in a sense completely general. What I would like to see is examples of coverages which satisfy zero, one or two but not all of the pretopology axioms, in various combinations. And, most importantly, are not contrived finite examples, but on categories which can be considered 'geometric' (probably should be extensive, or similar), like $Ring^{op}$, $Top$, $Diff$ and so on.

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The natural coverage arising from the local ring axioms is a pullback-stable coverage on $\textbf{CRing}^\textrm{op}$, but not a pretopology. (Every covering family is a pair of localizations, so there's no chance of satisfying the transitivity axiom.) There's an obvious pretopology containing it, which is the usual Zariski pretopology. –  Zhen Lin Mar 15 '12 at 12:29
Slight correction: the "pullbacks" in a coverage are not even weak pullbacks; they are just commutative squares. –  Mike Shulman Mar 16 '12 at 5:09
Ah, yes. Just cones, then. I'll edit. –  David Roberts Mar 17 '12 at 0:51
on schemas category coverings maked by affine open immersion that make a topologically covering too, we have the composition axiom only. –  Buschi Sergio Mar 17 '12 at 14:36