Prove that the associated sheaf $\mathcal{F}^+$ of a presheaf $\mathcal{F}$ is indeed a sheaf I was reading Perrin's Algebraic Geometry and i wonder how to prove the gluing axiom for the following construction.
Given a presheaf $\mathcal{F}$ of functions we set:
$\mathcal{F}^+(U)$={$f$: $U$$\rightarrow$$K$ $\vert$ $\forall x\in$$U$,$\exists$$V$ such that $x\in$$V$$\subset U$, and $g \in \mathcal{F}(V)$ such that $f\vert_V=g$}
 A: Let $U = \bigcup_i U_i$ be an open cover and assume there are functions $f_i \in \mathcal F^+(U_i)$, that agree on intersections.
Let $x \in U$. Define $f(x) := f_i(x)$ for some $i$ with $x \in U_i$. Since the $f_i$ agree on intersections, this is independent of the choice of $i$, thus this gives us a well defined function $f:U \to K$.
It remains to show $f \in \mathcal F^+(U)$. Therefore you have to use, that $f_{|U_i}=f_i$ holds by construction and that $f_i \in \mathcal F^+(U_i)$.
You should be aware of the following: A presheaf of function always satisfies the first sheaf axiom, which states, that we can test the vanishing of sections locally. This is because a function is zero iff its evaluation is zero at any point. Furthermore local functions, that agree on intersections, always glue to a unique global function. But this function might not be contained in $\mathcal F(U)$. The intuitive picture of the associated sheaf is, that we just add this unique function to $\mathcal F(U)$, if it is not contained in $\mathcal F(U)$.
Let us consider an example:
Let $X = \mathbb R$ and $\mathcal F$ the presheaf of continuous bounded functions. This is not a sheaf, because boundedness cannot be tested locally. The identity on any open bounded set is a local section. Those local sections glue uniquely to the identity on $\mathbb R$, which is not contained in $\mathcal F(X)$.
The associated sheaf thus does contain the identity. Since every continuous function is locally bounded, we can do this argument with any such function and obtain that the associated sheaf is the sheaf of continuous functions.
You should also be aware of the fact, that in general sheafification is a little bit more involved, since you have to take care of the first sheaf axiom, too. Of course it can happen, that we have a non-zero presheaf, whose associated sheaf is the zero sheaf: For instance take a two-pointed discrete space and define the global sections to be $\mathbb Z$ and the sections on the two open points to be trivial. The stalks at both points are zero, hence the associated sheaf is trivial, because it has the same stalks.
In particular the presheaf is not always contained in the associated sheaf (which is true with your presheaves of functions).
