Coverage, itself considered as a presheaf - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T00:51:33Z http://mathoverflow.net/feeds/question/102753 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/102753/coverage-itself-considered-as-a-presheaf Coverage, itself considered as a presheaf Ali Lahijani 2012-07-20T17:30:59Z 2012-07-20T19:03:01Z <p>A <a href="http://www.ncatlab.org/nlab/show/coverage" rel="nofollow">coverage</a> $J$ 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.</p> <p>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.</p> http://mathoverflow.net/questions/102753/coverage-itself-considered-as-a-presheaf/102762#102762 Answer by Mike Shulman for Coverage, itself considered as a presheaf Mike Shulman 2012-07-20T19:03:01Z 2012-07-20T19:03:01Z <p>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 <em>Sketches of an Elephant</em> 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 $f h\in R$.</p> <p>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 $T$ <em>is</em> 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 <em>all</em> 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.</p> <p>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 <code>$f^*(S)$</code> is covering. It follows that any $T$-separated presheaf is also separated for the sieve generated by all composites $f h$ with <code>$h\in f^*(S)$</code> (this sieve lies in the saturation of $T$ to a Grothendieck topology). But this sieve is precisely $R\cap S$.</p> <p>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 <code>$f^*(R) = f^*(S)$</code> 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.</p> <p>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 <code>$f^* R = S_V$</code>. But when $f\in R$, then <code>$f^*R$</code> 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).</p> <p>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.</p> <p>In particular, the Grothendieck topology generated by $T$ is <em>rigid</em> 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).</p> <p>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.</p>