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A superextensive site is a site whose covering families can be reduced to single covers combined with stable coproducts. Many naturally occuring sites are superextensive. The prefix ‘super-’ is opposite to ‘sub-’ as in ‘subset’ and indicates that the Grothendieck topology of the site contains the extensive topology; compare subcanonical site.
Let be a finitely extensive category. A Grothendieck topology on is (finitely) superextensive if it includes the (finitely) extensive topology (generated by families of injections into finite coproducts) and every covering family is generated by a finite one.
Likewise, if is infinitary-extensive, a topology is infinitary-superextensive if it includes the infinitary-extensive topology (generated by families of injections into small coproducts) and every covering family is generated by a small one. (Of course, the last condition is vacuous if is small.)
If is a single morphism in that generates a covering family, we call it a cover. The basic property of a superextensive site is that one can replace any (finite or small, as appropriate) covering family with the single map . This is a cover, since the original covering family factors through it, and the original covering family can be written as the composite of this cover and the family of coproduct injections .
Note that covers are stable under pullback, whenever such pullbacks exist. Even without pullbacks, they always satisfy the condition that if is a cover and is any morphism, then there is a cover such that factors through . For by the pullback condition on a Grothendieck topology, there is a covering family such that each factors through , but then is the required cover. Covers are also easily seen to be stable under composition, and generate another Grothendieck topology (which is a “singleton pretopology”). If the original topology was denoted , we denote the topology generated by single covers by .
The main conclusion is:
A superextensive Grothendieck topology on an extensive category is generated by the union of and the extensive topology .
By definition, and are sub-topologies of , so it suffices to observe that any covering family in is the composite of the family of coproduct injections with the cover .
Beware, however, of the following common misconception: is not identical with . That is, although a superextensive topology is determined by its singleton covers, in the sense that it can be recovered uniquely by knowing only them, it is not generated by its singleton covers as a Grothendieck topology; we need to also include in the generating covering families.
An easy way to see that and must be different is to notice that , like any singleton pretopology, is a locally connected site, so that all constant presheaves are sheaves for . But since the initial object is covered by the empty family in , no nontrivial constant presheaf can be a sheaf for .
In general, it is not true, for two Grothendieck topologies and on the same category, that -sheafification preserves -sheaves (and hence takes them to -sheaves). For example, let be the walking arrow , let be generated by the nonidentity morphism (so that -sheaves are those for which the image of is an isomorphism), and let be generated by the empty family covering (so that -sheaves are those for which the image of is terminal).
However, if is the extensive topology and for some superextensive topology (which is then recovered as ), and coproducts preserve covers, then it is true.
Let be a finitary superextensive site, and suppose that finite coproducts of covers in are again covers. If is a presheaf on which is a -sheaf, then the -sheafification of is also a -sheaf, and hence a -sheaf.
We show that if is a -sheaf, then so is , where is the reflection into -separated presheaves. Since is -sheafification, this will prove the theorem.
Since is generated by single covers, for any we have
where the colimit is over all covers with target , and consists of those such that , where is the kernel pair of .
The extensive topology is generated by the empty covering family of the initial object, and by the injections into binary coproducts. First consider the initial object . being a sheaf for the empty family on means that is terminal?. But since is extensive, its initial object is strict, so any morphism is an isomorphism. Therefore, .
Now consider a binary coproduct . being a sheaf for this means that is an isomorphism (since coproducts are disjoint and is terminal, there is no compatibility condition to worry about). We want to show that is also an isomorphism.
First suppose given . Since covers are stable under pullback and composition, we can represent and by some for the same cover . Now since is extensive and , decomposes as , and moreover and are the pullbacks of along the two inclusions . Thus, since covers are stable under pullback, we have covers and , and so the restrictions of and to and can be represented by the restrictions of and to and respectively.
Now if and become equal in and , there must be covers for such that in for and . Now since covers are stable under coproducts, the map is a cover, and moreover and can be obtained by first restricting to and then along the two coproduct injections.
But since is a -sheaf and for , we must have that and become equal in . Now since the composite is a cover, it follows that and are equal in . Thus, is injective, i.e. is -separated.
The case of surjectivity is easier. Suppose given , represented by for some covers . Then since covers are stable under coproduct, is a cover. And since is a -sheaf, we have a which restricts to for . Hence represents some which restricts to each , as desired. Thus, is a -sheaf.
Of course, there is an analogous result for infinitary superextensive sites. With a little more work we can also prove the same thing for stacks and stackification.
This result is especially interesting because sheaves and stacks for the extensive topology are easier to come by than those for some other topologies. For instance, if is an extensive coherent category, then any internal category in represents a functor which is a stack for the extensive topology, but not usually for the coherent topology. However, the coherent topology is superextensive and its covers are precisely the regular epimorphisms, i.e. is the regular topology. Thus, the stackification of an internal category in relative to the regular topology is still a stack for the extensive topology, and hence also for the coherent topology.
Some discussion about terminology in this entry is on the nForum here.