# Proof without sieves: Equivalent grothendieck topologies have the same sheaves

I'm currently learning about sheaf theory with Angelo Vistoli’s 2007 Notes on Grothendieck topologies, fibered categories and descent theory. And in page 35, there is the following definition of a refinement and a subordinate grothendieck topology:

Refinement: Let $$C$$ be a category, $$\{U_i\xrightarrow{\phi_i} U\}_{i\in I}$$ a set of arrows. A refinement $$\{Va\xrightarrow{\psi_a} U\}_{a\in A}$$ is a set of arrows such that for each index $$a\in A$$ there is some index $$i\in I$$ such that $$V_a\xrightarrow{\psi_a} U$$ factors through $$U_i\xrightarrow{\phi_i} U$$.

Subordinate Grothendieck Topology: Let $$C$$ be a category, $$\mathcal{T}$$ and $$\mathcal{T'}$$ two topologies on $$C$$. We say that $$\mathcal{T}$$ is subordinate to $$\mathcal{T'}$$, and write $$\mathcal{T}\prec\mathcal{T'}$$, if every covering in $$\mathcal{T}$$ has a refinement that is a covering in $$\mathcal{T'}$$. If $$\mathcal{T}\prec\mathcal{T'}$$ and $$\mathcal{T'}\prec\mathcal{T}$$, we say that $$\mathcal{T}$$ and $$\mathcal{T'}$$ are equivalent, and write $$\mathcal{T}\equiv\mathcal{T'}$$.

Now we have the following main-proposition: Let $$\mathcal{T}$$ and $$\mathcal{T'}$$ be two Grothendieck topologies on the same category $$C$$. If $$\mathcal{T}$$ is subordinate to $$\mathcal{T'}$$, then every sheaf in $$\mathcal{T'}$$ is also a sheaf in $$\mathcal{T}$$ . In particular, two equivalent topologies have the same sheaves.

Vistoli proved this proposition with sieves and I questioned myself: Can it be proven 'easier' without sieves? What do I mean with 'easier'? The prove with sieves in Vistolis paper uses several statements (Cor. 2.40, Prop. 2.42, Lemma 2.43,Prop. 2.46, Prop. 2.48) and with 'easier' I mean with less theory.

This part can be ignored, because it is using a false statement. My first idea was using the following statement: Let $$F:C^{op}\rightarrow Set$$ a presheaf. Then $$F$$ is a sheaf if and only if the following diagram is an equalizer for all coverings $$\{U_i\xrightarrow{\phi_i}U\}_{i\in I}$$ in $$C$$: $$F(U)\rightarrow \prod_{i}F(U_i) \rightrightarrows \prod_{i,j}F(U_i\times_UU_j)$$ where the function $$F(U) → \prod_i F(U_i)$$ is induced by the restrictions $$F(U)\xrightarrow{\phi_i^*} F(U_i)$$ and $$pr_1^*:\prod F(U_i) \rightarrow \prod_{i,j}F(U_i\times_UU_j)$$ and $$pr_2^*:\prod F(U_i) \rightarrow \prod_{i,j}F(U_i\times_UU_j)$$. So using that statement I want to proof the following main-lemma: Let $$C$$ be a category, $$\mathcal{T}$$ and $$\mathcal{T'}$$ two topologies on $$C$$ with $$\mathcal{T}\prec\mathcal{T'}$$ and $$F:C^{op}\rightarrow Set$$ a sheaf in $$\mathcal{T}$$. Let $$\{U_i\xrightarrow{\phi_i} U\}_{i\in I}$$ be a covering in $$\mathcal{T}$$ and $$\{Va\xrightarrow{\psi_a} U\}_{a\in A}$$ the refinement of $$\{U_i\xrightarrow{\phi_i} U\}_{i\in I}$$, then the diagram $$F(U)\rightarrow \prod_{a}F(V_a) \rightrightarrows \prod_{a,b}F(V_a\times_UV_b)$$ is an equalizer. Unfortunately, I don't know a way to prove this lemma.

Question 1: Is the way of proving the main-proposition without sieves even 'easier' (pretopology)? Answer: No, see Answer of Marc Hoyois.

Question 2: How do I prove the main-lemma? Answer: The main-lemma is wrong, because 'refinements' are not well-defined.

• Equivalent Grothendieck topologies are in fact equal, so they tautologically have the same sheaves. That is the point of using sieves in the definition of Grothendieck topology... Sieves tend to simplify proofs significantly. I assume your question is really about pretopologies, which are equivalent iff they induce the same topology. The simplest and most useful way to prove the statement is then to show that sheaves for a pretopology are the same as sheaves for the induced topology. Aug 10, 2022 at 17:55
• Thank you Marc. That cleared some questions for me. First of all, yes I mean pretopologies. But then what does induced topology mean (I'm new with the notion of that topic :D)? If we have a Grothendieck Pretopologie, the induced topology is the Grothendieck topology with sieves? Aug 10, 2022 at 21:30
• If you have access to it, section C2.1 of Sketches of an Elephant has a nice comprehensive discussion of pretopologies (or, rather, even weaker structures called "coverages") and their relation to "sifted" topologies (those defined in terms of sieves). Aug 11, 2022 at 16:42
• In Vistoli's notes there is the notion of saturated topology (Definition 2.52). I believe this is more or less the same as what others call topology. Aug 11, 2022 at 16:45
• The condition of being subordinate implies containment in the saturation, saturation preserves containment, and saturation is an idempotent operation. At least, my tired brain suggested this line of attack. Aug 11, 2022 at 16:56

Let me break down the statement you are trying to prove into two independent facts. This answer is not really in the spirit of the question since I will make maximal use of sieves, but for such foundational matters I think it is much more efficient to embrace sieves rather than to avoid them (so my subjective answer to Question 1 is no: it is not easier without sieves). Let me also point out that the statement of your "main lemma" does not make sense since "the refinement" is not well-defined, and if you mean "any" or "some" refinement then the statement is false (it goes in the wrong direction!).

First, let $$\mathfrak U=\{U_i\to U\}_i$$ be a family of maps in a category $$C$$. It generates a sieve $$\langle\mathfrak U\rangle\subset \operatorname{Hom}(-,U)$$, consisting of all maps to $$U$$ that factor through one of the $$U_i$$.

Fact 1. For a presheaf $$F$$ on $$C$$, $$F$$ satisfies descent with respect to the family $$\mathfrak U$$ iff it satisfies descent with respect to the sieve $$\langle\mathfrak U\rangle$$.

Proof. Descent wrt the sieve means that $$F(U) \stackrel{\sim}{\to} \operatorname{Hom}(\langle\mathfrak U\rangle, F).$$ Descent wrt the family means that $$F(U) \stackrel{\sim}{\to} \operatorname{lim}_{n\in\Delta}\operatorname{Hom}(\check C_n(\mathfrak U),F),$$ where $$\check C_\bullet(\mathfrak U)\to \operatorname{Hom}(-,U)$$ is the Čech nerve of the morphism of presheaves $$\coprod_i \operatorname{Hom}(-,U_i) \to \operatorname{Hom}(-,U)$$. But the sieve $$\langle\mathfrak U\rangle \subset \operatorname{Hom}(-,U)$$ is by definition the image of this morphism, hence the colimit of its Čech nerve. So then $$\operatorname{Hom}(\langle\mathfrak U\rangle, F) = \operatorname{Hom}(\operatorname{colim}_{n\in\Delta^{\operatorname{op}}}\check C_n(\mathfrak U),F) = \operatorname{lim}_{n\in\Delta}\operatorname{Hom}(\check C_n(\mathfrak U),F),$$ which proves the claim.

Second, define a quasi-topology on a category $$C$$ to be a collection of sieves which is stable under pullbacks. For example, the sieves generated by the covering families of a pretopology form a quasi-topology.

Fact 2. Let $$J$$ be a quasi-topology on $$C$$ and $$F$$ a presheaf on $$C$$. Then $$F$$ is a sheaf for $$J$$ (i.e., satisfies descent wrt all sieves in $$J$$) iff it is a sheaf for the Grothendieck topology $$\bar J$$ generated by $$J$$.

Proof. Let $$K$$ be the finest quasi-topology on $$C$$ such that every $$J$$-sheaf is a $$K$$-sheaf: a sieve $$R\subset \operatorname{Hom}(-,U)$$ is in $$K$$ iff for every morphism $$f\colon V\to U$$ in $$C$$, every $$J$$-sheaf satisfies descent wrt $$f^*(R)$$. Tautologically $$J\subset K$$. It will suffice to show that $$K$$ is a topology, so that $$\bar J\subset K$$, so that every $$K$$-sheaf (in particular every $$J$$-sheaf) is a $$\bar J$$-sheaf. Only the "local character" of $$K$$ is not given: if $$R,S$$ are sieves on $$U$$ such that $$S$$ is in $$K$$ and $$f^*(R)$$ is in $$K$$ for every $$f$$ in $$S$$, then $$R$$ must be in $$K$$. Since $$K$$ is stable under pullbacks, it suffices to show that every $$J$$-sheaf $$F$$ satisfies descent wrt $$R$$. To see this consider the intersection $$R\cap S$$. Since $$S$$ is the colimit of the representable presheaves $$\operatorname{Hom}(-,V)$$ with $$f\colon V\to U$$ in $$S$$, $$R\cap S$$ is correspondingly the colimit of $$f^*(R)$$ for $$f$$ in $$S$$. By assumption each map $$f^*(R)\hookrightarrow \operatorname{Hom}(-,V)$$ becomes an isomorphism after applying $$\operatorname{Hom}(-,F)$$, hence also the map $$R\cap S\hookrightarrow S$$. Similarly, $$R$$ is the colimit of $$\operatorname{Hom}(-,V)$$ for $$f\colon V\to U$$ in $$R$$, so $$R\cap S$$ is the colimit of $$f^*(S)$$ for $$f$$ in $$R$$. Since $$K$$ is a quasi-topology, the sieve $$f^*(S)$$ on $$V$$ is in $$K$$, so we deduce as before that the inclusion $$R\cap S\hookrightarrow R$$ induces an isomorphism after applying $$\operatorname{Hom}(-,F)$$. Since both inclusions $$R\cap S\hookrightarrow S$$ and $$R\cap S\hookrightarrow R$$ as well as $$S\hookrightarrow \operatorname{Hom}(-,U)$$ become isomorphisms, so does the inclusion $$R\hookrightarrow \operatorname{Hom}(-,U)$$, as desired.

Remark. Fact 1 is Lemma C2.1.3 in Johnstone's Elephant and Fact 2 is a special case of Proposition C2.1.9. The statement of Fact 2 (and its proof) is from SGA4 Corollary II.2.3. I wrote both proofs in a way that makes sense for presheaves of animas on an ∞-category $$C$$. For presheaves of sets we can improve Fact 2 by weakening the notion of quasi-topology to Johnstone's notion of sifted coverage, which is a collection of sieves with the only requirement that the pullback of a covering sieve contains a covering sieve. However, I do not know if (and do not expect that) this refined statement holds for presheaves of animas.