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.



Can Inequivalent Topologies Have Same Sheaves/Cohomology?
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.


