Lemma 7.14.10. Let $f : \mathcal{D} \to \mathcal{C}$ be a morphism of sites given by the functor $u : \mathcal{C} \to \mathcal{D}$. Given any object $V$ of $\mathcal{D}$ there exists a covering $\{ V_ j \to V\}$ such that for every $j$ there exists a morphism $V_ j \to u(U_ j)$ for some object $U_ j$ of $\mathcal{C}$.

Proof. Since $f^{-1} = u_ s$ is exact we have $f^{-1}* = *$ where $*$ denotes the final object of the category of sheaves (Example 7.10.2). Since $f^{-1}* = u_ s*$ is the sheafification of $u_ p*$ we see there exists a covering $\{ V_ j \to V\}$ such that $(u_ p*)(V_ j)$ is nonempty. Since $(u_ p*)(V_ j)$ is a colimit over the category $\mathcal{I}^ u_{V_ j}$ whose objects are morphisms $V_ j \to u(U)$ the lemma follows. $\square$

Comment #3433 by Remy on

The categories $\mathcal I_{V_j}^u$ may be empty. This happens for example if $\mathcal C = \varnothing$ is the empty site, but it could also just happen that none of the $V_j$ admit a map to any $u(U)$.

Comment #3434 by on

Dear Remy, I think the lemma (and the proof) is OK. The key is that the definition of a "morphism of sites" forces exactness of the functor $f^{-1} = u_s$. For example, if $\mathcal{C}$ is the empty site, then $\textit{Sh}(\mathcal{D})$ has to be the empty topos (because the final and initial object of it have to be the same as the pullback of such on $\mathcal{C}$ by an exact functor). And this in turn means that every object $V$ of $\mathcal{D}$ has an EMPTY covering! Thus the lemma holds in this case. OK?

Comment #3437 by Remy on

You're right; I just realised the same thing.

There are also:

• 1 comment(s) on Section 7.14: Morphisms of sites

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).