2
$\begingroup$

(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.)

$\endgroup$

1 Answer 1

1
$\begingroup$

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 $(\textbf{Alg}_A)^\textrm{op}$.

$\endgroup$
2
  • $\begingroup$ 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? $\endgroup$ Sep 7, 2012 at 19:59
  • $\begingroup$ 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. $\endgroup$
    – Zhen Lin
    Sep 8, 2012 at 2:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.