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A Sheaf is a Presheaf with locality and gluing axiom (https://en.wikipedia.org/wiki/Sheaf_(mathematics)). Now suppose we assume a Presheaf satisfies a sort of Unique Gluing axiom i.e. let $F$ be a Presheaf from a topological space $X$ to a category of sets or abelian groups such that If $\{U_i\}$ is an open covering of an open set $U$ , and if for each $i$ a section $s_i \in F(U_i)$ is given such that for each pair $U_i,U_j$ of the covering sets the restrictions of $s_i$ and $s_j$ agree on the overlaps: $s_i|U_i\cap U_j = s_j|U_i \cap U_j$, then there is a $\mathbf {unique}$ section $s \in F(U)$ such that $s|U_i = s_i, \forall i$ . Then is it true that $F$ is a Sheaf ?

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  • $\begingroup$ Yes, your definition is equivalent to $F$ being a sheaf. $\endgroup$
    – Watson
    Feb 17, 2018 at 19:36
  • $\begingroup$ @Watson : Could you please provide a proof of the locality condition ? $\endgroup$
    – user
    Feb 17, 2018 at 19:37
  • $\begingroup$ What did you try? What direction (either your definition $\implies$ sheaf, or sheaf $\implies$ your definition) is more difficult for you? $\endgroup$
    – Watson
    Feb 17, 2018 at 19:38
  • $\begingroup$ @Watson : of course I only want a proof of my definition to the locality axiom of sheaf ... $\endgroup$
    – user
    Feb 17, 2018 at 19:41
  • $\begingroup$ This definition can be found in the book of Robin Hartshorne,Algebraic geometry. A presheaf is just a contravariant functor, a sheaf is a presheaf with extra glueing conditions as you described above. $\endgroup$
    – Jiabin Du
    Mar 15, 2018 at 5:23

1 Answer 1

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As for the "locality" condition, let $(U_i)_{i \in I}$ be an open covering of an open set $U$, and let $s,t \in F(U)$ be such that $s\vert_{U_i} = t\vert_{U_i}$ for each $i \in I$. We show that $s=t$.

Let $s_i := s\vert_{U_i} \in F(U_i)$. We have $s_i\vert_{U_i \cap U_j} = s_j\vert_{U_i \cap U_j}$ for any $i,j \in I$, by compatibility of the restriction morphisms. By your definition, the element $s$ is the unique element of $F(U)$ such that $s\vert_{U_i} = s_i$. But $t \in F(U)$ also satisfies $s_i = t\vert_{U_i}$, by hypothesis. Uniqueness ensures that $s=t$, as desired.

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