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Suppose we have a Grothendieck pretopology $\tau$ on a category C with fibered products. Now define a new Grothendieck pretopology $\tau'$ consisting of all families of morphisms refinable by $\tau$-covers. That is, the new covers are the families $\{V_\beta \to X\}$ such that there exists some $\tau$-cover $\{U_\alpha \to X\}$ and a factorisation $U_\alpha \to V_{\beta_\alpha} \to X$ for each $\alpha$. This new set of families is also a Grothendieck pretopology and the question is: do they give the same topos? That is, is a presheaf a $\tau$-sheaf if and only if it is a $\tau'$-sheaf?

Edit: I could't read the relevant page in Elephant either, but Mike's answer lead me to the saturation section of http://ncatlab.org/nlab/show/coverage after which I worked out how to prove it myself. If someone explains to me how to typeset diagrams, I'll write up the answer.

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When you want to use set brackets {} you need to surround the display in backticks. –  Harry Gindi Oct 29 '10 at 1:16

4 Answers 4

up vote 4 down vote accepted

The answer is yes. David Roberts had the right idea—adding those new covering families gives you a new pretopology which generates the same Grothendieck topology—but not because it's a sieve completion, rather because there is an additional saturation condition in the definition of Grothendieck topology (in addition to saying that it consists of sieves) which essentially gives you this property.

It's not hard to check that any presheaf which is a sheaf for your original pretopology must also be one for the new one you define. You can find it as C2.1.6 in the Elephant. Note what this does not say: it's not necessarily true that if you have just a pair of families with the same codomain one of which corefines the other, that a sheaf for one of them is necessarily a sheaf for the other. The proof uses the assumption that the first covering family is part of a pretopology, and in particular can be pulled back along any morphism to another covering family, for which your presheaf is also a sheaf.

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Ah yes - saturation. I was going to mention that at some point, but Google books wouldn't let me look at that page of the Elephant to check my intuition. –  David Roberts Oct 29 '10 at 5:33

Edit again: this answer is wrong, see the comments.


The new set of families (for each object $X$) is called the sieve generated by the existing covers of $X$. One term for a Grothendieck pretopology is a basis for a Grothendieck topology, and different bases can give rise to the same Grothendieck topology. All of them, and the topology they generate, have the same sheaves.

See here for example.

Edit: Actually it is proposition C.2.1.9 in Johnstone's Sketches of an Elephant (Google books )

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Hi David, thanks for your reply. From what I can tell, I would get the sieves if in my question I had written $V_{\beta_\alpha} \to U_\alpha \to \to X$ in the place of $U_\alpha \to V_{\beta_\alpha} \to X$. –  anon Oct 28 '10 at 23:14
    
Sorry for the double arrow, that shouldn't be there. –  anon Oct 28 '10 at 23:15
    
Using the language of sieves, my question is equivalent to asking if the sieve on $X$ generated by $\{V_\beta \to X\}$ is equal to a sieve generated by some $\tau$-cover (possibly a different $\tau$-cover to $\{U_\alpha \to X\}$). –  anon Oct 28 '10 at 23:17
    
For some context, the first sentance in Section 1 of Goodwillie and Lichtenbaum's paper on the $h$-topology claims that the answer to my question is yes, the way I read it, but I don't see why. –  anon Oct 28 '10 at 23:20
    
Hmm, yes, you are right. I'll think about this some more. –  David Roberts Oct 28 '10 at 23:34

I think you get the same sheaves if and only if your topos of sheaves can be expressed as as sheaves on some singleton pretopology:

If $V \stackrel{f}{\rightarrow} X$ has the property that there exists a covering family $$\left(U_\alpha \stackrel{i_\alpha}{\rightarrow}X\right)_\alpha$$

and for $\alpha$ a map $\lambda_\alpha:U_\alpha \to V$ such that $f \circ \lambda_\alpha=i_\alpha$, that implies that $ay(f):ay(V) \to ay(X)$ is an epimorphism of representable sheaves, where $a$ is sheafification and $y$ is Yoneda. By Corollay 7 p.144 of Sheaves in Geometry and Logic, this means that the sieve generated by the singleton $f$ is a covering sieve.

So if anything, you still have AT LEAST the same amount of sheaves as before.

Conversely, if my topology can be generated by singletons, any singleton cover trivially satisfies your requirements.

So, in summary, I think what you are describing is some sort of "singelton completion", which seems to be a way of making your topos locally-connected.

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Note that anon isn't talking about a 'singleton completion', but taking general covering families. I think I have an argument why the singleton pretopology may not have the same sheaves as the original pretopology, in the case that the original pretopology isn't superextensive, over at the nForum: math.ntnu.no/~stacey/Mathforge/nForum/… –  David Roberts Oct 29 '10 at 1:00
    
I know he wasn't TRYING to take a singleton covering family, but the point is, if you have a family of maps $f_\beta$ which satisfy the requirements he asked, each map $f_\beta$ is already a singleton cover by my argument. –  David Carchedi Oct 29 '10 at 1:06
    
But in that case, if C has an initial object, as you need pullbacks of covers to be covers, this could make the map 0 -> X a cover. Also, if C is extensive, then you'd want the singleton covers generated by {U_i -> X} to be the same as those generated by \coprod U_i -> X, which would exclude f_\beta. –  David Roberts Oct 29 '10 at 1:20
    
How is $0 \to X$ a cover? You'd need a cover $U_\alpha$ of $X$ such that each $U_\alpha \to X$ factored through $0 \to X$. I'm just not seeing that. –  David Carchedi Oct 29 '10 at 1:23
    
I shouldn't have used $X$ in $0 \to X$. Work with Top to keep things concrete. Let X be a space, covered by open sets $\{j_a:U_a \to X\}$, and let $Z \hookrightarrow X$ be a subspace disjoint from some $U_b$, and let $f_a = j_a$ for each $a$. $f_b$ can't be a cover, because if it were, $Z\times_X U_b = 0 \to Z$ should be a cover. I'm just saying that if C doesn't have lots of coproducts, the singleton pretopology generated by the given pretopology could be very different to the one outlined in the question. –  David Roberts Oct 29 '10 at 2:12

I think that the topologies are the some:

1) Let $\widetilde{\mathscr{C} }$ the topos of $\tau$-sheaves. Give a family $g_i: X_i\to X\ i\in I $ the follow are equivalent:

a) $\cup_{i\in I} Image(g_i) = X $ in $\widetilde{\mathscr{C} }$ (for simply notation all $\mathscr{C}$ objects and situations are traslated in $\widetilde{\mathscr{C}} $ by Yoneda imbedding and associate sheaf functor).

b) the natural morphism $\coprod_{i\in I} X_i \to X$ is Epi in $\widetilde{\mathscr{C} }$

c) The natural diagram $\coprod_{i,j\in I}X_{i,j}\rightrightarrows\coprod_{i,j\in I}X_{i}\to X$ in $\widetilde{\mathscr{C} }$ is a Coker (where $X_{i,j}:=X_i\times_X X_j$

d) For any $F\in \widetilde{C} $ the natural diagram $F(X)\to \prod_{i\in I}F(X_i) \rightrightarrows \prod_{i,j} F(X_{i, j})$ is a Ker.

PROOF: Only observe that in $\widetilde{\mathscr{C} } $ any Epi is a coequalizer, then the equalizer of its Ker-couple, and the coprodocts are disjoint i.e. are coherent (commutate) by pullback’s.

2) We have that the (old) $\tau$-coverings are also $\tau’$-coverings (consider trivial factorization by first morphism as identity). Only we have to prove that for any $F\in \widetilde{\mathscr{C}} $ and for any $\tau’$-coverings $g_i: X_i\to X\ i\in I $ the diagram in (d) is exat (i.e. a Ker diagram), or equivalently that (a) is true, but from the factorizations $U_\alpha \to V_{\beta_\alpha }\to X$ follow $\coprod_\alpha U_\alpha \to \coprod_\beta V_\beta \to X $ and this composition is Epi, then $\ \coprod_\beta V_\beta \to X $ is Epi.

Excuse my for your time If I wrong.

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