superextensive site

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 $C$ be a finitely extensive category. A Grothendieck topology on $C$ 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 $C$ 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 $C$ is small.)

- Well known examples of infinitary superextensive sites include Top and Diff with their usual topologies, generated by the open covers.
- The coherent topology on any (finitely or infinitary) extensive coherent category is (finitely or infinitary) superextensive. In particular, this includes (in the infinitary case) the canonical topology on a Grothendieck topos.

If $f$ is a single morphism in $C$ 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 $(j_i: U_i \to B)_i$ with the single map $[j_i]_i: \coprod_i U_i \to B$. 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 $U_i \to \coprod_i U_i$.

Note that covers are stable under pullback, whenever such pullbacks exist. Even without pullbacks, they always satisfy the condition that if $f\colon U\to B$ is a cover and $g\colon A\to B$ is any morphism, then there is a cover $\bar{f}\colon V\to A$ such that $g \bar{f}$ factors through $f$. For by the pullback condition on a Grothendieck topology, there is a covering family $(\bar{f}_j\colon V_j\to A)$ such that each $g \bar{f}_j$ factors through $f$, but then $[\bar{f}_j]\colon \coprod_j V_j \to A$ 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 $T$, we denote the topology generated by single covers by $T_{cov}$.

The main conclusion is:

A superextensive Grothendieck topology $T$ on an extensive category $C$ is generated by the union of $T_{cov}$ and the extensive topology $T_{ext}$.

By definition, $T_{cov}$ and $T_{ext}$ are sub-topologies of $T$, so it suffices to observe that any covering family $(j_i: U_i \to B)_i$ in $T$ is the composite of the family of coproduct injections $(j_i: U_i \to \coprod_j U_j)_i$ with the cover $[j_i]_i: \coprod_j U_j \to B$.

Beware, however, of the following common misconception: $T$ is *not* identical with $T_{cov}$. 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 $T_{ext}$ in the generating covering families.

An easy way to see that $T$ and $T_{cov}$ must be different is to notice that $T_{cov}$, like any singleton pretopology, is a locally connected site, so that all constant presheaves are sheaves for $T_{cov}$. But since the initial object is covered by the empty family in $T_{ext}$, no nontrivial constant presheaf can be a sheaf for $T_{ext}$.

In general, it is not true, for two Grothendieck topologies $T_1$ and $T_2$ on the same category, that $T_2$-sheafification preserves $T_1$-sheaves (and hence takes them to $T_1\cup T_2$-sheaves). For example, let $C$ be the walking arrow $0 \to 1$, let $T_1$ be generated by the nonidentity morphism (so that $T_1$-sheaves are those for which the image of $0\to 1$ is an isomorphism), and let $T_2$ be generated by the empty family covering $0$ (so that $T_2$-sheaves are those for which the image of $0$ is terminal).

However, if $T_1$ is the extensive topology and $T_2 = T_{cov}$ for some superextensive topology (which is then recovered as $T_1\cup T_2$), and coproducts preserve covers, then it is true.

Let $C$ be a finitary superextensive site, and suppose that finite coproducts of covers in $C$ are again covers. If $X$ is a presheaf on $C$ which is a $T_{ext}$-sheaf, then the $T_{cov}$-sheafification of $X$ is also a $T_{ext}$-sheaf, and hence a $T$-sheaf.

We show that if $X$ is a $T_{ext}$-sheaf, then so is $X^+$, where $(-)^+$ is the reflection into $T_{cov}$-separated presheaves. Since $X^{++}$ is $T_{cov}$-sheafification, this will prove the theorem.

Since $T_{cov}$ is generated by single covers, for any $c\in C$ we have

$X^+(c) = colim_{f\colon d \to c} gl(X,f)$

where the colimit is over all covers with target $c$, and $gl(X,f)\subseteq X(d)$ consists of those $x\in X(d)$ such that $r^*(x) = s^*(x)$, where $(r,s)\colon d\times_c d \rightrightarrows d$ is the kernel pair of $f$.

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 $0$. $X$ being a sheaf for the empty family on $0$ means that $X(0)=1$ is terminal?. But since $C$ is extensive, its initial object is strict, so any morphism $f\colon d\to 0$ is an isomorphism. Therefore, $X^+(0) = X(0) = 1$.

Now consider a binary coproduct $c = c_1 + c_2$. $X$ being a sheaf for this means that $X(c) \to X(c_1)\times X(c_2)$ is an isomorphism (since coproducts are disjoint and $X(0)$ is terminal, there is no compatibility condition to worry about). We want to show that $X^+(c) \to X^+(c_1)\times X^+(c_2)$ is also an isomorphism.

First suppose given $x,x'\in X^+(c)$. Since covers are stable under pullback and composition, we can represent $x$ and $x'$ by some $\bar{x}, \bar{x'}\in X(d)$ for the same cover $f\colon d\to c$. Now since $C$ is extensive and $c=c_1+c_2$, $f$ decomposes $d$ as $d = d_1+d_2$, and moreover $d_1$ and $d_2$ are the pullbacks of $f$ along the two inclusions $c_1 \to c \leftarrow c_2$. Thus, since covers are stable under pullback, we have covers $f_1\colon d_1\to c_1$ and $f_2\colon d_2\to c_2$, and so the restrictions of $x$ and $x'$ to $c_1$ and $c_2$ can be represented by the restrictions of $\bar{x}$ and $\bar{x'}$ to $d_1$ and $d_2$ respectively.

Now if $x$ and $x'$ become equal in $X^+(c_1)$ and $X^+(c_2)$, there must be covers $g_i\colon b_i \to d_i$ for $i=1,2$ such that $g_i^*(\bar{x}) = g_i^*(\bar{x'})$ in $X(b_i)$ for $i=1$ and $2$. Now since covers are stable under coproducts, the map $g_1+g_2\colon b_1+b_2 \to d_1+d_2 = d$ is a cover, and moreover $g_i^*(\bar{x})$ and $g_i^*(\bar{x'})$ can be obtained by first restricting to $b_1+b_2$ and then along the two coproduct injections.

$\array{b_i & \overset{}{\to} & b_1+b_2\\
^{g_i}\downarrow && \downarrow^{g_1+g_2}\\
d_i& \overset{}{\to} & d\\
\downarrow && \downarrow\\
c_i & \underset{}{\to} & c}$

But since $X$ is a $T_{ext}$-sheaf and $g_i^*(\bar{x}) = g_i^*(\bar{x'})$ for $i=1,2$, we must have that $\bar{x}$ and $\bar{x'}$ become equal in $X(b_1+b_2)$. Now since the composite $b_1+b_2 \to d \to c$ is a cover, it follows that $x$ and $x'$ are equal in $X^+(c)$. Thus, $X^+(c) \to X^+(c_1)\times X^+(c_2)$ is injective, i.e. $X^+$ is $T_{ext}$-separated.

The case of surjectivity is easier. Suppose given $x_i\in X^+(c_i)$, represented by $\bar{x_i}\in X(d_i)$ for some covers $f_i\colon d_i\to c_i$. Then since covers are stable under coproduct, $f_1+f_2 \colon d_1+d_2 \to c_1+c_2 = c$ is a cover. And since $X$ is a $T_{ext}$-sheaf, we have a $\bar{x}\in X(d)$ which restricts to $\bar{x_i}$ for $i=1,2$. Hence $\bar{x}$ represents some $x\in X^+(c)$ which restricts to each $x_i$, as desired. Thus, $X^+$ is a $T_{ext}$-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 $C$ is an extensive coherent category, then any internal category in $C$ represents a functor $C^{op}\to Cat$ 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. $T_{cov}$ is the regular topology. Thus, the stackification of an internal category in $C$ relative to the regular topology is still a stack for the extensive topology, and hence also for the coherent topology.

- Chapter 1 of David Roberts’ thesis uses superextensive sites in the study of anafunctors. The material is covered and updated in the paper
*Internal categories, anafunctors and localisations*.

Some discussion about terminology in this entry is on the nForum here.

Revised on June 7, 2014 03:11:52
by Urs Schreiber
(89.204.130.218)