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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?

Edit: All that remains is the following question. 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.

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.

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?

A possibly equivalent question is the following. 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.

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?

Edit: All that remains is the following question. 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.

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?

A possibly equivalent question is the following. If every covering in $T_1$ belongs 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.

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?

A possibly equivalent question is the following. 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.