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Let $C$ be a fixed category, and let $T_1$ and $T_2$ be two Grothendieck (pre)topologies on $C$. We say $T_1$ is subordinate to $T_2$ if every covering in $T_1$ has a refinement in $T_2$. We say $T_1$ and $T_2$ are equivalent if $T_1$ is subordinate to $T_2$ and $T_2$ is subordinate to $T_1$. We know that if $T_1$ is subordinate to $T_2$, then any sheaf for the topology $T_2$ is also a sheaf for $T_1$. Is the converse true? In particular, if the categories of sheaves are the same, does that imply the topologies are equivalent? At least, if we know that $T_1$ is subordinate to $T_2$ but not equivalent to it, do we know that there is a sheaf for $T_1$ that is not a sheaf for $T_2$? How about if both topologies are subcanonical?

If $T_1$ is subordinate to $T_2$, we have a morphism of topologies $g: T_1 \to T_2$. If $g_*$ is exact (so the cohomology of all sheaves for $T_2$ is the same in the two topologies) does it imply that the two topologies are equivalent? Thank you.

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  • $\begingroup$ What do you mean by "the same" here? (There are two things this could mean, namely just isomorphic or literally the same sheaves.) $\endgroup$
    – Qiaochu Yuan
    Commented Apr 24, 2013 at 2:08
  • $\begingroup$ I mean, literally the same sheaves. For example, say I have a topology $T$ that is subcanonical and I can prove that there is a universal regular epimorphism (so a covering in the canonical topology) which does not have a refinement in $T$. Is there a specific sheaf in $T$ that is not representable, or at least a sheaf that is not a sheaf for the canonical topology? $\endgroup$
    – Igor Minevich
    Commented Apr 24, 2013 at 2:36
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    $\begingroup$ The short answer is, Grothendieck topologies are uniquely determined by the subcategory of sheaves. This is not so for pretopologies. $\endgroup$
    – Zhen Lin
    Commented Apr 24, 2013 at 7:14

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I would think so.

Two pretopologies are equivalent when they generate the same topology, that is, when the have the same sieves. If $A$ is an object of $C$, a sieve on $A$ is a subfunctor of the functor $h_A$ represented by $A$; and a subfunctor $S$ of $h_A$ is a sieve with respect to a topology $T$ if and only if the induced morphism $S^{\rm sh} \to h_A^{\rm sh}$ of the sheafifications is an isomorphism. Of course the category of sheaves determines the sheafifications; so if two topologies define the same sheaves, they admit the same sieves, so they are equivalent.

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  • $\begingroup$ Great! So, if the sheaves for $T_1$ and $T_2$ are the same but there is a covering in $T_2$ that does not have a refinement in $T_1$, then the sieve $S$ generated by that covering has $S^{sh}\cong h_A^{sh}$ in $T_2$ but not in $T_1$. But the sheafification of any presheaf is the same in the two topologies (assume the two sheafifications are sheaves in both topologies and use the universal property of sheafification in both topologies): a contradiction. $\endgroup$
    – Igor Minevich
    Commented Apr 24, 2013 at 20:22
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Does this hold water?

If the sheaves are the same, they have the same subobject classifier, which is the sheaf of closed sieves, so both topologies have the same notion of closed sieves, so the same notion of covers, hence they are equivalent.

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    $\begingroup$ I wouldn't expect anything involving sieves to hold water, but I think this does. $\endgroup$
    – Tom Goodwillie
    Commented Apr 24, 2013 at 11:04

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