I know that this question might be too easy for you, but I have to study on my own, so please explain for me.
In the page 64, Hartshone defined the sheafification of a presheaf $\mathcal{F}$ by assign for each open set $U$ a set $\mathcal{F}^{+}(U)$ containing all the function $s$ from $U$ to $\cup_{P\in U}\mathcal{F}_{P}$, satisfying some specific conditions.
I can see that $\mathcal{F}^{+}$ is a presheaf.
My question is: How can I prove that $\mathcal{F}^{+}$ is a sheaf by proving it satisfies two gluing condition (on page 61). I really want to see that proof rigorously.
Please help me. Thanks.
Best Answer
As stated in the comments, the axioms of sheaf can be checked directly using the conditions imposed on the functions in $\mathcal{F}^+ (U)$. Let's just write them down as they are presented in Hartshorne's book:
Using these we need to prove two conditions for $\mathcal{F}^+$ being a sheaf.
First: let $U$ be an open set, $ \lbrace V_i \rbrace $ an open covering of $U$, and $s\in \mathcal{F}^+(U) $ such that $s_{|V_i}=0$ for all $i$. Then $s$ must be equal to $0$.
This is true: $s=0$ means that $s(P)=0$ for each $P$ in $U$. Take any $P$ in $U$, then there exists an open set $V_i$ that contains $P$, hence $0=s_{|V_i}(P)=s(P)$, and this verifies the axiom.
Second: suppose we have, for each $i$, an element $s_i\in \mathcal{F}^+(V_i)$, so that $s_{i|V_i \cap V_j}=s_{j|V_i\cap V_j}$ whenever this makes sense. Then we have to find $s\in \mathcal{F}^+(U)$ such that $s_i=s_{|V_i}$ for each $i$.
Such $s$ can be constructed directly: take any point $P$ in $U$ and the open set $V_i$ which contains $P$, and define $s(P)=s_i(P)$. This is a good definition because the $s_i$'s agree on the intersections, all is left to check is that $s$ satisfies the conditions of the sheafification. The first one is ok, because it's verified by the $s_i$'s, for the second one, just consider that you can apply the condition to the $s_i$: this will give you an open neighborhood $W_i$ contained in $V_i$ and containing $P$, with an element $t_i\in \mathcal{F}(W_i)$ as above. Since $W_i$ is open in $V_i$, which is open in $U$, the set $W_i$ is suitable also for the function $s$ we have just defined.
This proves that $\mathcal{F}^+$ is a sheaf, and a bit more of similar manipulation yields that all of its stalks are isomorphic to the stalks of $\mathcal{F}$.
I hope this can work.
P.S. There is a very compact way to describe the sheafification (easy to remember at least): one can write $$ \mathcal{F}^+(U)=\varprojlim_{p\in U}\varinjlim_{p\in V} \mathcal{F}(V) $$