Amalgamation of matching family for empty cover

Question

Let $\mathcal{J}$ be a Grothendieck topology on a small category $\mathbb{C}$. If $C$ is an object of $\mathbb{C}$ for which $\mathcal{J}(C)$ contains the empty cover/sieve, then if $F : \mathbb{C}^{\mathsf{op}} \to \mathsf{Set}$ is any $\mathcal{J}$-sheaf, it is my understanding that $F(C)$ must be a singleton, since any element of $F(C)$ will be an amalgamation for the (empty) matching family for the empty cover of $C$, and such amalgamations are unique. Is it a 'common' situation for an object to have an empty cover in a Grothendieck topology? Is it possible to assume without loss of generality that no object is covered by the empty sieve, in the sense that there is another site $(\mathbb{D}, \mathcal{K})$ with this property for which the topos of $\mathcal{J}$-sheaves is equivalent to the topos of $\mathcal{K}$-sheaves?

Answer 1

There is already an answer about how common empty covers are. I will focus on your last question. In short the answer is yes. You can simply leave out all the objects from $\mathbb{C}$ that are covered by the empty sieve (in the Grothendieck topology $\mathcal{J}$). This will give you a full subcategory $\mathbb{D}$ of $\mathbb{C}$ with an induced Grothendieck topology $\mathcal{K}$, which consists of the same sieves as in $\mathcal{J}$ but only for the objects in $\mathbb{D}$ of course. Then the categories of sheaves on $(\mathbb{C}, \mathcal{J})$ and $(\mathbb{D}, \mathcal{K})$ are equivalent.

This is in a simple application of the comparison lemma. That lemma is much stronger, so it is probably overkill and you could probably also write a simple direct proof using the insight you had (about $F(C)$ being a singleton).

Answer 2

It's extremely common for there to be objects $C$ for which $\mathcal{J}(C)$ contains the empty sieve. Take the Grothendieck topology arising from any topological space in the standard way: then the empty set has an empty cover.

You have the same situation with the join Grothendieck topology on a distributive lattice with all joins and meets - a partial order with all joins will have a least element, and this element will have an "empty cover" since it is the join of the empty set. Not surprising, given that this is a generalisation of the above example.

I'm not sure whether it is always possible to assume WLOG that there is no empty covering of any element.