Does the concept of a basis for a topology on a category exist?

2010-06-22T22:40:15Z

If we want to define a sheaf F on a topological space X and we have a basis B for the topology of X, what we can do is to define objects and restrictions for guys in B, check that they satisfy the "B-sheaf axioms" and then use

Theorem 1: the B-sheaf extends uninquely to the whole of X.

I was wondering if there's a similar thing for more general sites, and actually not just for sheaves on a given site but for stacks.

The question I'm really interested in is the following:

If one has a fibred category over Schemes (say Schemes over some fixed field with the fppf topology) and one wants to check that descent is effective, would it be sufficient to check it on some subcategory of schemes (using perhaps some vague analogue of Theorem 1)?

Thanks.

EDIT

For example one might want to construct the stack M of coherent sheaves on some scheme X. One way to do it is to define the functor which associates with each scheme S the groupoid of coherent sheaves on $S\times X$ flat over S $$ M(S) = { E \in Coh\ S\times X,\ E \text{ flat over S} }.$$

Let's say I want to use a different characterization of $M(S)$, perhaps using Lemma 3.31 on page 82 of Huybrechts' Fourier-Mukai book.

My ignorance prevents me from knowing if the that lemma is valid for a general scheme S (no matter how nice my X might be). This is why I'd like to work over some nice subcategory of schemes (where the lemma is valid) and then extend.

The stack I'd be interested in defining would be a stack of perverse sheaves on X, where the matters would be a bit worse.

Answer by David Carchedi for Does the concept of a basis for a topology on a category exist?

2010-06-23T00:52:16Z

Let $S$ be your Grothendieck site. What you want is a subcategory $j:B \hookrightarrow S$ such that the Grothendieck topology of $S$ restricts to $B$ in the sense that every covering sieve of $b \in B$ can be refined by one coming from a family $(U_i \to b)_i$ with each $U_i \in B$, AND such that $j^*:Sh(B) \to Sh(S)$ is an equivalence of topoi. This is exactly what makes Theorem 1 work.

Concretely, you want the topology to restrict and every $s$ in $S$ to have a covering family $(b_i \to s)$ with $b_i \in B$, so that you can then say:

$$F(s):=\varprojlim \left(\prod_{i}F(b_i)\rightrightarrows \prod_{i,j}F(b_i \times_{s} b_j)\right).$$

However, you need to make sure this doesn't depend on the covering family you chose.

Suppose only that every $s$ in $S$ to have a covering family $(b_i \to s)$ with $b \in B$ and that the Grothendieck topology on $S$ restricts to $B$ in the sense described above. Then, since the Grothendieck topology is subcanonical, $s$ is a colimit of $b_i$s, hence $s \mapsto Hom(blank,s)$ embeds $S$ into $Sh(B)$ (note that the left-Kan extension of this embedding is precisely $j^*$, which is literally restriction).

I claim $j^*$ is fully-faithful. This is essentially because $Hom(j^*F,j^*G)$ for two sheaves on $S$ determines $Hom(F,G)$ since the value of $F(s)$ is determined by the value of $F$ on $b_i$s by the cover $(b_i \to s)$, by descent.

Now, if $F$ is a $B$-sheaf (i.e. an element of $Sh(B)$), then $j_*F(s)=Hom(j^*s,F).$ Hence, $$j_*j^*(F)(s)\cong Hom(s,j^*j_*F)\cong Hom(j^*s,j^*F)\cong Hom(s,F)\cong F(s),$$

by Yoneda, adjointness, and full an faithfulness.

Note also that $j^*j_*(G) \cong G$ for all $G \in Sh(B)$ pretty much by definition. Hence the adjoint pair $j_*,j^*$ is an equivalence.

So, what does this mean concretely? You need to find a subcategory $B$ of schemes such that

1.)every cover in the fppf topology of an element of $b \in B$ can be refined by one with domains in $B$ (at least you need to be able to find a family of morphisms whose SIEVE is in the topology GENERATED by the fppf pretopology)

2.) Every scheme can be covered by elements of $B$.

I'll leave it to you to find such a subcategory, as I don't know much AG.

P.S., everything I said will hold for stacks as well.

EDIT: Condition 2.) implies condition 1.), so this becomes simpler:

You just need a category subcategory $B$ of schemes such that every scheme can be covered by elements of $B$.

Does the concept of a basis for a topology on a category exist? - MathOverflow

Does the concept of a basis for a topology on a category exist?

Answer by David Carchedi for Does the concept of a basis for a topology on a category exist?