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.



Does adding "co"refinements to a Grothendieck pretopology change the topos?
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.


