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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.

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)?

Question 2: How do I prove the main-lemma?

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.

I'm currently learning about sheaf theory with 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.

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)?

Question 2: How do I prove the main-lemma?

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.

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)?

Question 2: How do I prove the main-lemma?

I'm currently learning about sheaf theory with 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. My first idea was using the the 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 lemma:

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'?

Question 2: How do I prove the main-lemma?

I'm currently learning about sheaf theory with 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.

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)?

Question 2: How do I prove the main-lemma?