Does adding "co"refinements to a Grothendieck pretopology change the topos? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T20:19:19Z http://mathoverflow.net/feeds/question/44023 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/44023/does-adding-corefinements-to-a-grothendieck-pretopology-change-the-topos Does adding "co"refinements to a Grothendieck pretopology change the topos? anon 2010-10-28T21:37:12Z 2010-10-31T23:30:57Z <p>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 <code>$\{V_\beta \to X\}$</code> such that there exists some $\tau$-cover <code>$\{U_\alpha \to X\}$</code> and a factorisation <code>$U_\alpha \to V_{\beta_\alpha} \to X$ for each $\alpha$</code>. 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?</p> <p>Edit: I could't read the relevant page in Elephant either, but Mike's answer lead me to the saturation section of <a href="http://ncatlab.org/nlab/show/coverage" rel="nofollow">http://ncatlab.org/nlab/show/coverage</a> 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.</p> http://mathoverflow.net/questions/44023/does-adding-corefinements-to-a-grothendieck-pretopology-change-the-topos/44029#44029 Answer by David Roberts for Does adding "co"refinements to a Grothendieck pretopology change the topos? David Roberts 2010-10-28T22:28:03Z 2010-10-28T23:36:23Z <p>Edit again: this answer is wrong, see the comments.</p> <hr> <p>The new set of families (for each object $X$) is called the <a href="http://ncatlab.org/nlab/show/sieve" rel="nofollow">sieve</a> generated by the existing covers of $X$. One term for a Grothendieck pretopology is a <em>basis for a Grothendieck topology</em>, and different bases can give rise to the same Grothendieck topology. All of them, and the topology they generate, have the same sheaves.</p> <p>See <a href="http://ncatlab.org/nlab/show/coverage#grothendieck_coverages_18" rel="nofollow">here</a> for example.</p> <p>Edit: Actually it is proposition C.2.1.9 in Johnstone's <em>Sketches of an Elephant</em> (<a href="http://books.google.com.au/books?id=TLHfQPHNs0QC&amp;pg=PA692&amp;lpg=PA692&amp;dq=johnstone+elephant+sieve&amp;source=bl&amp;ots=yqaqD8_tRt&amp;sig=ev0RFXyl0rDezGy-9RtXiJFmuLc&amp;hl=en&amp;ei=ygPKTLDhNoaMvQP6r40Q&amp;sa=X&amp;oi=book_result&amp;ct=result&amp;resnum=1&amp;ved=0CBkQ6AEwAA#v=onepage&amp;q=immediately%20deduce&amp;f=false" rel="nofollow">Google books</a> )</p> http://mathoverflow.net/questions/44023/does-adding-corefinements-to-a-grothendieck-pretopology-change-the-topos/44038#44038 Answer by David Carchedi for Does adding "co"refinements to a Grothendieck pretopology change the topos? David Carchedi 2010-10-29T00:44:26Z 2010-10-29T00:44:26Z <p>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:</p> <p>If $V \stackrel{f}{\rightarrow} X$ has the property that there exists a covering family $$\left(U_\alpha \stackrel{i_\alpha}{\rightarrow}X\right)_\alpha$$</p> <p>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. </p> <p>So if anything, you still have AT LEAST the same amount of sheaves as before.</p> <p>Conversely, if my topology can be generated by singletons, any singleton cover trivially satisfies your requirements.</p> <p>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.</p> http://mathoverflow.net/questions/44023/does-adding-corefinements-to-a-grothendieck-pretopology-change-the-topos/44058#44058 Answer by Mike Shulman for Does adding "co"refinements to a Grothendieck pretopology change the topos? Mike Shulman 2010-10-29T05:05:56Z 2010-10-29T05:05:56Z <p>The answer is yes. David Roberts had the right idea&mdash;adding those new covering families gives you a new pretopology which generates the same Grothendieck topology&mdash;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.</p> <p>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 <em>also</em> a sheaf.</p> http://mathoverflow.net/questions/44023/does-adding-corefinements-to-a-grothendieck-pretopology-change-the-topos/44390#44390 Answer by Buschi Sergio for Does adding "co"refinements to a Grothendieck pretopology change the topos? Buschi Sergio 2010-10-31T23:30:57Z 2010-10-31T23:30:57Z <p>I think that the topologies are the some:</p> <p>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:</p> <p>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).</p> <p>b) the natural morphism $\coprod_{i\in I} X_i \to X$ is Epi in $\widetilde{\mathscr{C} }$ </p> <p>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 $X_{i,j}:=X_i\times_X X_j$</p> <p>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.</p> <p>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.</p> <p>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.</p> <p>Excuse my for your time If I wrong.</p>