TITLE: Can Inequivalent Topologies Have Same Sheaves/Cohomology?
QUESTION [6 upvotes]: 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.

REPLY [8 votes]: 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.

REPLY [5 votes]: 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.