On [nlab](http://ncatlab.org/nlab/show/sheafification) it says that a presheaf is locally isomorphic to a sheaf. What do they mean by locally isomorphic? Their definition of locally isomorphic is given in terms of Grothendieck topologies which i think is overkill. 


When I first read the nlab page, I thought that it might mean that every presheaf, when restricted to a small enough open set is a sheaf, but I have doubts now because I can't find a proof in the literature and I can't prove it myself.



Presheaves are locally sheaves?
Dear Daniel, the reason you couldn't find a proof of your statement nor locate one in the literature is that it is false ; so you were quite right to "have doubts now" ! Here are two (essentially equivalent) statements that hopefully clarify the situation.


**I) Given a presheaf $\mathcal F$ on a topological space, it is not true that there exists a non-empty open subset $U\subset X$ such that the restriction $\mathcal F |U$ is a sheaf.**


For example take $X=\mathbb R$ and define the presheaf $\mathcal F$ by $\mathcal F(V)= \mathbb Z$ for all open 
$V\subset \mathbb R$ (constant presheaf on $\mathbb R$ with values in $\mathbb Z$). Since every open $U$ contains disjoint open subsets,
the restriction $\mathcal F |U$ is never a sheaf.


**II) Given a presheaf $\mathcal F$ on a topological space and its sheafification  

$\mathcal F \to \mathcal F'$ it is not true that there exists a non-empty open subset $U\subset X$ such that the restricted morphism $\mathcal F |U \to \mathcal F'|U$ is an isomorphism of presheaves.**


In the preceding example the sheafification $\mathcal F'$ is the sheaf of *locally* constant $\mathbb Z $-valued functions and again for every $U\subset \mathbb R$ you will find *disjoint* open intervals $I\_1,I\_2 \subset U$ for which
$\mathcal F(I\_1\sqcup I\_2)= \mathbb Z \neq \mathcal F'(I\_1\sqcup I\_2)= \mathbb Z^2$ . So 
the restricted morphism $\mathcal F |U \to \mathcal F'|U$ is not an isomorphism of presheaves. 


**Conclusion** I find it ambiguous, as proved by this very question, to call a morphism of sheaves a "local isomorphism" if it is an isomorphism on the stalks. I don't know how widespread this usage is but in my opinion people using it should warn their readers if they decide to adopt it. On the other hand, I must concede that everybody (myself included) calls $\mathcal F'$ a *constant* sheaf. This terminology also seems a little misleading but it is firmly entranched now and is here to stay.


**An answer to Roy's question** He asks (in his answer below) for an example of a presheaf all of whose restrictions to open subsets are non-separated. [Recall that a presheaf 
 $\mathcal F$ is said to be separated if given a covering $U=\cup U\_i $ of an open set $U$ by open subsets $U\_i$, you can deduce for two sections $f,g\in \mathcal F (U)$ that $f=g$ as soon as you know that $f| U\_i=g| U\_i$ for all $i$ . This is equivalent to saying that, if $\mathcal F'$ denotes the sheafification of $\mathcal F$, all morphisms 
$\mathcal F (U) \to \mathcal F'(U)$ are injective.]


Here is the example. On the topological space $\mathbb R$ consider the sheaf of continuous functions $\mathcal C$, its subpresheaf $\mathcal C\_b $ of continuous *bounded* functions ( Caution: this is *not* a sheaf !) and the quotient presheaf 
$\mathcal F= \mathcal C / \mathcal C\_b $ i.e. for $V$ open in $\mathbb R$, $\mathcal F(V)=\mathcal C (V)/ \mathcal C\_b (V) $. It is then clear that for all non-empty open $V\subset \mathbb R$ we have $\mathcal F(V) \neq 0$ but for the sheafification $\mathcal F'$ we have $\mathcal F'(V)= 0$ (because every continuous function is *locally* bounded !). And this is the example requested by Roy: for every non-empty $U$ the restriction $\mathcal F |U $ is a non-separated presheaf on $U$ : $\mathcal F |U \neq 0$ certainly does not inject into $\mathcal F'|U =0$


