In Vakil's "the rising sea" book, it reads, and I quote:
"Suppose $\mathscr F$ is a sheaf (or indeed separated presheaf) of abelian groups on $X$, and $s$ is a global section of $\mathscr F$. Then let the support of $s$, denoted $Supp(s)$, be the points $p$ of $X$ where $s$ has a nonzero germ: $Supp(s)$ := {$p \in X : s_p \neq 0$ in $\mathscr F_p$}.
...Show that Supp(s) is a closed subset of $X$."
I just figure out a proof of this proposition without using the identity axiom... So does it require that $\mathscr F$ must be a sheaf (or indeed separated presheaf)?
Here is my proof:
Suppose $p \in X-Supp(s)$, then $[(s,X)] = 0$ in $\mathscr F_p$.([] marks the representative of the equivalence class), it implies that $\exists U$ such that $p \in U$ and $s \mid U$ is the $0$ element in $\mathscr F(U)$.
Now it suffices to prove that for any $p1 \in U$, we have $p1 \in X-Supp(s)$, that is, $[(s,X)] = 0$ in $\mathscr F_{p1}$.(Then $X-Supp(s)$ would be an open set of $X$.)
For any $[(t,V)] \in \mathscr F_{p1}$(which means $p1 \in V,t \in \mathscr F(V)$), $p1 \in U, V$, so $p1 \in U \cap V$. We know that $s \mid U$ is the $0$ element in $\mathscr F(U)$, so $s \mid U \cap V$ is the $0$ element in $\mathscr F(U \cap V)$ since a group homomorphism maps $0$ element to $0$ element.
$[(s,X)] + [(t,V)]=[(s \mid U,U)] + [(t,V)] = [(s \mid U \cap V,U \cap V)] + [(t \mid U \cap V,U \cap V)] = [(0,U \cap V)] + [(t \mid U \cap V,U \cap V)] = [(t \mid U \cap V,U \cap V)] = [(t,V)]$
End of proof.
I do not use the identity axiom in my proof! What is going wrong in the proof?