Let $\mathcal{C}'$ and $\mathcal{D}'$ be categories so that $\mathcal{C} \subseteq \mathcal{C}'$ and $\mathcal{D} \subseteq \mathcal{D}'$ are full subcategories. Suppose the forgetful functors $F_{\mathcal{C}}:\mathcal{C} \to \mathcal{C}'$ and $F_{\mathcal{D}}: \mathcal{D} \to \mathcal{D}'$ have left adjoints $G_{\mathcal{C}}$ and $G_{\mathcal{D}}$. Then in general does an equivalence of categories between $\mathcal{C}'$ and $\mathcal{D}'$ induce an equivalence of categories between $\mathcal{C}$ and $\mathcal{D}$ by composing with the adjunctions? If this is not true in general are there any criteria to guarantee this?

The specific situation this came up in was trying to prove an equivalence of categories between sheaves on two different sites. I didn't want to deal with showing things were sheaves so I wanted to just prove an equivalence between the presheaf categories and say we can sheafify to get an equivalence on the sheaf level. I ended up proving that my functor sent sheaves to sheaves directly but I was still wondering if there was an answer to this question in general.

  • $\begingroup$ A single category can carry many different topologies, for all of these sites the category of presheaves is the same, but the categories of scheaves can be non-equivalent. So you definitely can't just "sheafify to get an equivalence on the sheaf level". $\endgroup$ Jun 27 '13 at 15:00

I think that in this generality the answer is "no". For example, take $\mathcal{C}'=\mathcal{D}'$ equal to some additive category, take the identity as the equivalence, take $\mathcal{C}=\mathcal{C}'$ and $\mathcal{D}=0$. This seems to satisfy your hypotheses.


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