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On nlab 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.

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Given that they defined a sheaf in terms of Grothendieck topologies, its not surprising that their other definitions involve them. If you're not dealing with such things I guess you could drop the Grothendieck bit. And the statement that presheaves are locally isomorphic to sheaves is almost tautological, given their definition of local isomorphism = isomorphism after passing to sheaves. – George Lowther Jan 11 '11 at 3:35
But, having said that, I don't think this question is really suitable here. Maybe math.stackexchange would be a better place? – George Lowther Jan 11 '11 at 3:37
Give a presheaf $F$, any the section $s\in Sh(F)(U)$ of the associated sheaf are locally sections of the presheaf i.e exist a covering $U=\cup_i U_i$ such that any restriction $s_{|U_i}$ come from (by the reflection map $r_{U_i}: F(U_i)\to Sh(F)(U_i)$) a section of the presheaf. – Buschi Sergio Jan 11 '11 at 14:13
The statement in the nLab entry about local isomorphism is correct, with the definition of local isomorphism as given at the page linked to: the canonical morphism from a presheaf to its sheaffification is a morphism that becomes an isomorphism under sheafification. Such morphisms are traditionally called local isomorphisms. And yes, this is a special case of the general theory of left exact reflective localizations. – Urs Schreiber Jan 11 '11 at 18:16
-1: Daniel, did you ever think to look up the page local isomorphism on the nLab: ? Also, the definition in terms of Grothendieck topologies is a necessary complication of the theory WRT local isomorphisms (or else we're left with only the trivial case). See my comment on Clark Barwick's answer. – Harry Gindi Jan 12 '11 at 13:24
up vote 13 down vote accepted

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 $\mathcal F'|U =0$

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An isomorphism on the level of stalks should probably be called an "infinitesimal isomorphism." – Kevin Ventullo Jan 11 '11 at 17:38
Also, Georges, I find it natural to require two objects to be globally morphic before we can talk about whether or not they're locally isomorphic. The OP's condition regarding restriction sheaves should instead be called "local satisfaction of the sheaf conditions". For instance, we say two manifolds are locally diffeomorphic if there exists a morphism admitting sections locally about each point. – Harry Gindi Jan 12 '11 at 13:36
Harry, all manifolds of the same dimension are locally diffeomorphic, independently of any morphisms betwen them. On the other hand a submersion of a manifold onto another one of lower dimension has local sections about each point but the manifolds are certainly not locally diffeomorphic. Finally I have never heard the notion of objects being globally morphic . – Georges Elencwajg Jan 12 '11 at 14:18
Dear Georges, two manifolds are locally diffeomorphic iff there exists a local diffeomorphism from one to the other (whence globally morphic). – Harry Gindi Jan 12 '11 at 15:21
Dear Harry, since our discussion hinges on terminology, I have nothing to add: we just seem to have different definitions. Thank you for your interest in this post and for sharing your point of view. – Georges Elencwajg Jan 12 '11 at 17:29

Here's one way to answer your question. Consider the category $\mathbf{PSh}(X)$ of presheaves (of sets) on a topological space $X$. A map $F\to G$ of $\mathbf{PSh}(X)$ is said to be a local isomorphism if for every point $x\in X$, the induced map $F_x\to G_x$ on stalks is a bijection. Denote by $W$ the class of local isomorphisms. Now the category $\mathbf{Sh}(X)$ of sheaves on $X$ is equivalent to the localization $W^{-1}\mathbf{PSh}(X)$. In particular, for any presheaf $F$, there is a local isomorphism $F\to F'$, where $F'$ is a sheaf.

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@Daniel: Clark's answer extends to the general case of an arbitrary base category with grothendieck topology $\tau$ by defining $W_\tau$ to be $Sh_\tau^{-1}(Iso(Psh(C))$. The rather more interesting part of this question, however, is that there exists a direct characterization of these "systems of local isomorphisms". It is then a theorem that Grothendieck topoogies on $C$ are in canonical bijection with systems of local isomorphisms on $Psh(C)$. – Harry Gindi Jan 11 '11 at 11:38

Arrgh, I wish to delete this, but do not know how. So i will make it into a question. Georges' nice answer is a presheaf that violates the existence axiom s2 on every nbhd of a point. Is there an example that violates the uniqueness axiom s1 on every nbhd of some point?

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Dear Roy, I have answered your interesting question in an addendum to my original answer. Please do not delete your question above, else readers will be disoriented by my orphaned answer! – Georges Elencwajg Jan 13 '11 at 21:19
Thank you Georges! If I understand correctly, in your example every element becomes zero near the point, but there is no set on which all do at once. It seems so clear after seeing it. I.e. this is all that is needed to make all the stalks zero, which makes the sheaf (but not the presheaf) zero. – roy smith Jan 14 '11 at 22:13

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