Grothendieck Topologies versus Pretopologies - MathOverflow

Grothendieck Topologies versus Pretopologies

2012-07-30T05:04:32Z

The wikipedia article(s) as well as the nlab article(s) about Grothendieck topologies and Grothendieck pretopologies are careful to differentiate the two very emphatically and to point out that distinct pretopologies can give rise to the same topology on a category. My (almost certainly trivial) questions are the following:

1) What are the key differences between Grothendieck topologies and Grothendieck pretopologies, and

2) What are some nice examples of distinct pretopologies that give rise to the same topology?

Answer by name for Grothendieck Topologies versus Pretopologies

2012-07-30T06:58:20Z

A) Let $\tau$ be a Grothendieck pretopology on any category with fiber products. Define a new Grothendieck pretopology $\tau'$ where a cover $\{U_i \to X\}$ is a $\tau'$ cover if and only if there exists a refinement $\{V_{ij} \to X\}$ (i.e. there exists for each $ij$ an $X$-morphism $V_{ij} \to U_i$) such that $\{V_{ij} \to X\}$ is a $\tau$ cover. Then $\tau$ and $\tau'$ generate the same topologies.

B) The operation (lets say $\Phi$) that takes topologies to pretopologies composed with the operation (say $\Psi$) that takes pretopologies to topologies gives back the same topology that you started with (i.e. $\Psi \Phi = id$). So for any pretopology $\tau$, the pretopology $\Phi\Psi \tau$ gives the same topology as $\tau$.

C) Related to (A), on the category of affine schemes, we can define an fppf cover as a family $\{U_i \to X\}$ which is jointly surjective, and such that each morphism is flat, and finitely presented. We could also include the requirement that in addition the morphisms be quasi-finite. Corollary 17.16.2 in EGA IV implies that these two pretopologies gives rise to the same topology.

In response to the first question, there are surely people more expert than me, but it seems to me that Grothendieck topologies in algebraic geometry are almost always defined via Grothendieck pretopologies. I think topologies are the nicer concept, but while the idea is suited to proving very general things about topoi, if you have a specific pretopology such as the étale, Zariski, Nisnevich, flat, cdh, envelopes (also called proper cdh - see 18.3 in Fulton's "Intersection theory" and the Mazza, Voevodsky, Weibel book), etc having actual scheme morphisms in your hands allows you to apply the strong algebr-geometric results which are largely responsible for making topologies such a powerful tool in algebraic geometry. Of course, I'm sure someone in logic would have a very different point of view.

In relation to this line of thought, Voevodsky - while working with the Nisnevich and cdh topologies - found it useful to further simplify the data leading to a topology in the notion of a cd structure (see the papers "Unstable motivic homotopy categories in Nisnevich and cdh-topologies" and "Homotopy theory of simplicial sheaves in completely decomposable topologies").

While I'm talking about Voevodsky's work, he also uses from time to time an idea which he calls "covers of normal form" in "Homology of schemes I", but the same idea is used implicitely in the appendix to "Singular homology of abstract algebraic varieties". The way I understand this phenomenon is the following. If you have two pretopologies $\sigma$ and $\rho$ on a category with fiber products, then the covers of the pretopology generated by $\sigma$ and $\rho$ are finite compositions of covers which are either a $\sigma$ cover or a $\rho$ cover. Lets denote the new pretopology by $\langle \sigma, \rho \rangle$. Many pretopologies $\tau$ in common use, are actually generated by two other pretopologies in the sense that $\langle \sigma, \rho \rangle$ and $\tau$ give rise to the same topology.

For example, the cdh pretopology is generated in this way (by definition) by the Nisnevich pretopology and the pretopology of envelopes. Voevodsky shows that the h-pretopology is generated like this Zariski and the pretopology whose covers are jointly surjective families of proper morphisms. The qfh pretopology is generated like this by étale and the pretopology whose covers are jointly surjective families of finite morphisms.

Answer by Zhen Lin for Grothendieck Topologies versus Pretopologies

2012-07-30T07:31:58Z

A Grothendieck topology by definition consists of sieves – what Johnstone calls a sifted coverage – whereas a Grothendieck pretopology in any non-trivial case will contain a non-sieve. (Recall that $\lbrace \textrm{id} : X \to X \rbrace$ is always a covering family for $X$, but it is a sieve if and only if there are no morphisms $Y \to X$ for any $Y \ne X$.) Thus, in any case of interest, no topology is a pretopology and no pretopology is a topology.

But siftedness is not the key difference between topologies and pretopologies. The key difference is saturation: as you are already aware, it is possible to add covering families to a pretopology without changing the category of sheaves. One can define the non-sifted analogue of a topology as a family of sinks satisfying the following axioms:

Any isomorphism constitutes a singleton covering family.
The composition of covering families is a covering family.
If a covering family factors through a given sink, then the sink is also a covering family.

One can show that every pretopology is contained in a unique such saturated coverage, and every saturated coverage contains a unique topology – just pick out the sieves!

Here's a reasonably "natural" example of a pair of pretopologies that generate the same topology. We consider the category $\textbf{Top}$ of topological spaces, or any full subcategory $\mathbf{T}$ thereof closed under pullbacks and open subsets.

A sink $\lbrace f_i : U_i \to X \rbrace$ is covering if and only if each $f_i$ is open and a homeomorphism onto its image, and the union of the images is the whole of $X$.
A sink $\lbrace f_i : Y_i \to X \rbrace$ is covering if and only if the induced map $f : \coprod_i Y_i \to X$ is a local homeomorphism.

Clearly, pretopology (1) is contained in pretopology (2). Conversely, given a covering family of type (2), we can obtain a covering family of type (1) by taking a suitable refinement. (For each point $y$ of $Y_i$, take an open neighbourhood $U_{i,j}$ that is mapped homeomorphically into $X$.) So the two pretopologies must generate the same topology.

Answer by David Roberts for Grothendieck Topologies versus Pretopologies

2012-07-30T07:54:01Z

For another example, the pretopologies on the category of finite dimensional smooth manifolds given by

~~open covers~~
maps of the form $\coprod U_i \to X$ for a given open cover $(U_i)$
surjective local diffeomorphisms
surjective submersions

all generate the same topology. The last three are nested, and the second is cofinal in the other two. ~~The second and the first are equivalent because of superextensivity.~~

If you are willing to weaken the concept of pretopology to that of coverage then the coverage of (EDIT: coproducts of) smooth good open covers on manifolds also generates the same topology.

Answer by Andreas Blass for Grothendieck Topologies versus Pretopologies

2012-07-30T13:39:39Z

I think of pretopologies (and coverages and other variations of this idea) as convenient ways of presenting a topology, just as groups (and other algebraic structures) can be conveniently presented by generators and relations, or (since I'm a set theorist) a complete Boolean algebra can be conveniently presented by giving a dense subposet (called a notion of forcing). Usually (though not always), the important thing is the bigger, more abstract object --- the Grothendieck topology, the group, or the complete Boolean algebra --- on which the interesting constructions can be based. Then the choice of a particular presentation is, in principle, irrelevant, but it can be very useful for working with the object and understanding it, especially because the presentation is often much smaller than the abstract object.