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A coverage $J$ on a category $C$ assigns to an object $U$ of $C$ a set of covering families $J(U)$. The covering families are required to be stable under pullback, which amounts to requiring that for every arrow $f: V \to U$ in $C$, there is a function $J(U) \to J(V)$ which sends a covering family $\{x_i: U_i \to U\}_{i \in I}$ of $U$ to a covering family $\{y_j: V_j \to V\}_{j \in J}$ of $V$ such that each composite $fy_j$ factors through some $x_i$. This apparently defines a presheaf on $C$, as the functor laws seem to hold.

I am interested in this presheaf and its properties. Does it have a name? For example, for every presheaf on a site we may ask whether it is a sheaf. One might want to know what are the requirements for a coverage, when considered as a presheaf as above, to be a sheaf w.r.t. itself.

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In general, for a given $f$, there will be many options for the map $J(U)\to J(V)$. It's not immediately clear to me that the choices can always be made in a way that produces a functor $J$. –  Andreas Blass Jul 20 '12 at 18:10
    
What if the category has all pullbacks? Then I guess the base-change functor will do, won't it? –  Ali Lahijani Jul 20 '12 at 18:33
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If you add the additional assumption that covering families are preserved by pullback (which is not part of the definition of coverage), then pullback will give you a pseudofunctor of some sort. Pullbacks are rarely strictly associative. You could strictify it to get an actual functor, but then the question of whether it is a sheaf will depend on the strictification chosen. –  Mike Shulman Jul 20 '12 at 19:05

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First of all, Andreas' comment is right: a coverage gives no specified way to "pull back" a covering family of $U$ to a covering family of $V$. However, if you consider what Sketches of an Elephant calls "sifted" coverages, meaning that all covering families are sieves, then there is a canonical choice: the pullback of a sieve $R$ on $U$ along $f:V\to U$ is the sieve $f^*(R)$ consisting of all $h:W\to V$ such that $f h\in R$.

For an arbitrary sifted coverage, this pullback sieve may not be a covering family, but it always contains a covering family and thus lies in the "saturation" of the coverage. If a sifted coverage $T$ is closed under pullback of sieves, in this sense, then it does yield a presheaf on $C$, which is in fact a sub-presheaf of the subobject classifier in the presheaf topos $[C^{\mathrm{op}},\mathrm{Set}]$ (which is defined by $\Omega(U) = $ the set of all sieves on $U$). (If $T$ is a Grothendieck topology, then this sub-presheaf $T$ is the classifier of dense sub-presheaves.) See also C2.1.10 in the Elephant.

I claim that this sub-presheaf $T\subseteq \Omega$ is $T$-separated iff the coverage $T$ contains at most one covering sieve of every object. (In particular, if $T$ is a Grothendieck topology, then it must be the trivial topology.) This condition is clearly sufficent; for necessity, suppose $R$ and $S$ are two $T$-covering sieves of an object $U$. Then for any $f:V\to U$ in $R$, the pullback sieve $f^*(S)$ is covering. It follows that any $T$-separated presheaf is also separated for the sieve generated by all composites $f h$ with $h\in f^*(S)$ (this sieve lies in the saturation of $T$ to a Grothendieck topology). But this sieve is precisely $R\cap S$.

Thus, if $T\subseteq \Omega$ is $T$-separated, it is also separated for $R\cap S$ for any $R,S\in T$. However, for any $f:V\to U$ in $R\cap S$, we have $f^*(R) = f^*(S)$ being the maximal sieve on $V$. Thus, since $T$ is separated for $R\cap S$, we must have $R=S$; hence $U$ admits at most one $T$-covering sieve.

Now assuming $T$ satisfies this condition so that $T$ is $T$-separated, then $T$ is a $T$-sheaf whenever if $R$ is a (the) covering sieve of $U$ and for each $f:V\to U$ in $R$ we have a (the) covering sieve $S_V$ of $V$, then there is another covering sieve of $U$ (which, of course, must also be $R$) such that $f^* R = S_V$. But when $f\in R$, then $f^*R$ is the maximal sieve on $V$, so this means that the domain of every morphism in a covering sieve is covered only by its own maximal sieve --- which is already implied by $T$ being a functor. Such objects are called $T$-irreducible (C2.2.18 in the Elephant).

Thus there are three classes of objects in $C$: the irreducible ones, which are covered by their maximal sieve; those that are covered by some non-maximal sieve whose domains are all irreducible; and those that are not covered by any $T$-sieve. The irreducible objects are themselves a sieve in $C$, by functoriality, as are the objects that are covered by any $T$-sieve at all. The objects that are not covered by any $T$-sieve will be covered only by their maximal sieve in the Grothendieck topology generated by $T$, so they will be irreducible there.

In particular, the Grothendieck topology generated by $T$ is rigid in the sense of C2.2.18: every object is covered by the family of morphisms out of irreducible objects. It follows that the category of $T$-sheaves is equivalent to the category of presheaves on the irreducible objects for this topology (which are those that are $T$-covered by the maximal sieve or that are not covered by any $T$-sieve).

This is perhaps not a complete answer to your question, but it shows that the condition of a sifted coverage being a sheaf for itself is very restrictive.

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Excellent answer, thank you! Is there any relationship between $T \subseteq \Omega$ and the subobject classifier of the topos of sheaves for $T$? –  Ali Lahijani Jul 20 '12 at 19:36
    
Explicitly, is the classifying map $j: \Omega \to \Omega$ of $T \subseteq \Omega$ the same as the local modality defining the sheaf topos? I guess this is the meaning of the phrase 'classifier of dense sub-presheaves'. –  Ali Lahijani Jul 20 '12 at 19:42
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Yes, if $T$ is a Grothendieck topology, then when regarded as a sub-presheaf of $\Omega$, its classifying map is precisely the local modality (= Lawvere-Tierney topology) that defines the subtopos. This is C2.1.10. The subobject classifier in the topos of sheaves is different -- it is the classifier of closed sub-presheaves, which is the equalizer of $j:\Omega\to\Omega$ and $1_\Omega$. –  Mike Shulman Jul 20 '12 at 23:09
    
Very informative. Thank you. –  Ali Lahijani Jul 21 '12 at 11:43

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