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(I find it easier to think in terms of Grothendieck pretopologies, instead of topologies. If this annoys any experts, please forgive me.)

Suppose that $C$ is a (full) subcategory of a category $D$. Furthermore, let $J$ be a Grothendieck pretopology on $C$ and let $K$ be a pretopology on $D$ (i.e., for an object $X$ of $C$, $J(X)$ is the collection of covering families for $X$, and similarly for $K$). I am interested in the following situation:

For any object $X$ of $C$, every covering family $\{U_i \to X\}$ for $J$ is also a covering family for $K$ (i.e., $J(X) \subseteq K(X)$).

Is there much precedent for this situation? Am I lucky enough that it can be named using standard terminology?

(Please notice that I am not assuming that the inclusion functor $C \to D$ preserves pullbacks. I realize that this is may create many complications when considering the two topologies. In particular, I don't think that it makes sense to say that the topology generated by $K$ "restricts" to a topology on $C$. This suggests that the answer to my question may not be as simple as saying that the topology from $J$ is subordinate to some restricted topology from $D$, to use terminology from Vistoli's notes, Definition 2.47.)

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I would just say that the inclusion preserves covering families (in the naïve sense). You don't need Grothendieck pretopologies to make sense of this – just plain coverages (in the sense of Johnstone; see [Sketches of an elephant, §C2.1]) are enough. It's not a good notion though – a better one would say something like "for each covering family $\mathfrak{U}$ in $\mathcal{C}$, there is a covering family in $\mathcal{D}$ that is subordinate to $\mathfrak{U}$". Notice that this condition depends only on the Grothendieck topologies on $\mathcal{C}$ and $\mathcal{D}$ and not the particular choice of coverage. (For an extreme example, let $\mathcal{C} = \mathcal{D}$, let $\mathcal{D}$ have any non-saturated Grothendieck pretopology, and put a strictly larger Grothendieck pretopology on $\mathcal{C}$ that has the same sheaves.)

It is true that when the inclusion $\mathcal{C} \to \mathcal{D}$ doesn't preserve pullbacks, the (pre)coverage induced by a Grothendieck pretopology on $\mathcal{D}$ isn't necessarily a Grothendieck pretopology on $\mathcal{C}$. It may even fail to be a coverage. Oddly enough, things don't improve even if we replace the Grothendieck pretopology on $\mathcal{D}$ with the Grothendieck topology it generates. I posted some counterexamples here. It doesn't really matter though – we can always replace a not-quite-coverage with the coverage it generates, or even the Grothendieck topology it generates if we are that way inclined.

Cover-presering inclusions arise in nature occasionally. If we take the full subcategory $\mathcal{B}$ of the frame of open subsets of $\operatorname{Spec} A$ spanned by the principal open subsets, we obtain a full subcategory of $(\textbf{Alg}_A)^\textrm{op}$, and the inclusion $\mathcal{B} \to (\textbf{Alg}_A)^\textrm{op}$ preserves covers. In fact, the topology on $\mathcal{B}$ is precisely the restriction of the Zariski topology on $(\textbf{Alg}_A)^\textrm{op}$.

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Thanks for your nice, informative answer. Your better property, using subordinate covering families, seems useful. Can you think of a concise way to state it, similar to the inclusion preserving covering families? –  Manny Reyes Sep 7 '12 at 19:59
    
I would also say it "preserves covering families" – after all, if $\mathcal{D}$ is equipped with a saturated coverage then it is equivalent to "preserves covering families in the naïve sense". The former is what Johnstone means by "cover-preserving" in the Elephant, so it is not without precedent. –  Zhen Lin Sep 8 '12 at 2:15
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