Given a presheaf, in Angelo Vistoli's 2007 [Notes on Grothendieck topologies,
fibered categories
and descent theory](http://homepage.sns.it/vistoli/descent.pdf) there is a construction of the sheafification (Proof for theorem 2.64).



> 
> **Note:** In this context, a Grothendieck topology is a Singleton Grothendieck topology, such that every covering of an object $U$ in $C$ is a single map $\phi:T\rightarrow U$.
> 
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> 


The second part of the proof is the construction of a sheaf. We have a given separated presheaf $F$. Then we define for an object $U\in C$ the set
$$P\_U:=\{(\phi,f)\,|\,\phi:T\rightarrow U \text{ is a covering, }f\in F(T)\text{ such that }pr\_1^\*f=pr\_2^\*\text{ in }F(T\times\_UT)\}.$$ On this set we impose an equivalence relation, by declaring $(\phi,f)\sim(\phi',f')$ if $pr\_1^\*f=pr\_2^\*f'$ in $F'(T\times\_UT')$. Now we define $F^+(U):=P\_U/\sim$. Given an arrow $X\rightarrow U$ in $C$, we define the a function $F^+(U)\rightarrow F^+(X)$ by $[\phi:T\rightarrow U,f]\mapsto [pr\_1:X\times\_UT\rightarrow X,pr\_2^\*f]$. This is from the fiber product
$\require{AMScd}$
\begin{CD}
X\times\_UT @>{pr\_2}>> T\\
@V{pr\_1}VV @VV{\phi}V\\
X @>>> U.
\end{CD}
That defines our functor $F^+$. There is also a natural transformation $(+):F\Rightarrow F^+$, obtained by sending an element $f\in F(U)$ into $[U\rightarrow U, f]\in F^+(U)$.


Now let $U\in C$ be arbitrary. Then we have the component $(+)\_U:F(U)\rightarrow F^+(U)$ with $f\mapsto [id\_U,f]$.


Now I want prove, that $(+)\_U$ is an isomorphism?


I got the injectivity. But I'm struggling with surjectivity. So let $[\phi:T\rightarrow U,g]\in F^+(U)$ be arbitrary. Then we need a $f\in F(U)$ with $(+)\_U(f)=[id\_U,f]=[\phi:T\rightarrow U,g]$.


So my **question** is: How do I get that $f$?



Does the (Vistoli-)sheafification induce isomorphism?
You cannot prove this since it is not true. The map $F \to F^+$ is an isomorphism if and only if $F$ is already a sheaf. (Also notice that Vistoli, of course, does not claim that $F \to F^+$ is an isomorphism. In that case there would also be no need to construct $F^+$ in the first place.)


