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 followingmain-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.

Sketches of an Elephanthas 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). $\endgroup$saturatedtopology (Definition 2.52). I believe this is more or less the same as what others call topology. $\endgroup$