MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ be a topological space and let $\mathcal{F}$ and $\mathcal{G}$ be two sheaves over $X$.

Of course, if one has a morphism $f : \mathcal{F} \to \mathcal{G}$ such that for all $x\in X$, $f_x : \mathcal{F}_x \to \mathcal{G}_x$ is an isomorphism, then it is known that $f$ itself is an isomorphism.

My question is the following: if we don't have such a morphism $f$, but if we know that for all $x\in X$, $\mathcal{F}_x$ and $\mathcal{G}_x$ are isomorphic, is it true that $\mathcal{F}$ and $\mathcal{G}$ are isomorphic ?

share|cite|improve this question
No: think of non-isomorphic complex vector bundles over $X$. – Qfwfq May 12 '10 at 18:20
up vote 11 down vote accepted

Definitely not. If $X$ is a ringed space, then a $\mathcal{O}_X$-module $F$ is called locally free of rank $1$, if $X$ is covered by open subsets $U_i$ such that $F|_{U_i}$ is free of rank $1$ over $\mathcal{O}_{U_i}$. The correspond to line bundles on $X$. Line bundles form a group, called the Picard group of $X$ and denoted by $\text{Pic}(X)$. This group does not have to vanish: If $X$ is a CW complex and we take the sheaf of continuous functions, then $\text{Pic}(X)$ is isomorphic to $H^1(X,\mathbb{Z}/2)$. For $X=S^1$, the moebius strip is the nontrivial element here. The corresponding example in algebraic geometry is $\text{Pic}(\mathbb{P}^n_k)=\mathbb{Z}$, the generator given by the Serre twist $\mathcal{O}(1)$.

Of course, there are more easy counterexamples, but I wanted to indicate that there is a rich theory coming from the observation that two locally isomorphic sheaves are not isomorphic.

share|cite|improve this answer

Since there are locally free sheaves that are not globally free, the answer is clearly no. Indeed the existence of the term "locally free" is a meta-proof of this. For a concrete example, consider the circle embedded as the central circle in a smooth Mobius strip. Its normal bundle is locally isomorphic to the constant sheaf with fibres $\mathbb{R}$ but it is not a constant sheaf as it has no global sections which are everywhere nonzero.

share|cite|improve this answer

A conceptual explanation (since the other answers have given counterexamples) for why this is untrue is as follows: A sheaf consists of local data and global data specifying how those local data fit together. Even if all of the local data of two sheaves are isomorphic, there is no reason to believe that those isomorphisms can be fit together in a compatible way. This is why we require that the isomorphisms on stalks arise from a map that is already a morphism of sheaves, since this exactly says that the data fit together in the proper way.

We even encounter the same problems when we work with presheaves, for instance. Since presheaves are functors, a morphism of presheaves must be a natural transformation of functors. However, simply having isomorphisms "pointwise", as it were, is not enough. The isomorphisms must also commute with the restriction maps.

share|cite|improve this answer

No, indeed there exist sheaves which are locally isomorphic to locally constant sheaves, without being locally constant. These are usually called local coefficient systems.

It is not hard to see that, for a nice spaces $X$, e.g. a manifold, to give a sheaf locally isomorphic to the locally constant sheaf $\underline{G}$ associated to the group $G$ is the same as to give a homomorphism $\pi(X) \to \mathop{Aut}(G)$. In particular whenever there is a nontrivial homomorphism, there exist nontrivial local coefficient systems.

share|cite|improve this answer

Actually, the examples given in the answers so far are even counter-examples to the weaker statement that two sheaves $F,G$ for which there exists a covering $U_i$ such that $F|U_i\cong G|U_i$ be isomorphic.

For the original question regarding stalks there is a simpler example: Let $X=\{\eta,s\}$ be the topological space having $\{\eta\}$ as the only non-trivial open set. To give an abelian sheaf $F$ on $X$ is equivalent to give two groups $F(X)=F_s$ and $F(\{\eta\})=F_\eta$ and a restriction homomorphism $F_s\to F_\eta$. Taking $F_s=F_\eta=A$ for some abelian group $A\ne0$ and choosing either $\mathrm{id}_A$ or $0$ as restriction defines two non-isomorphic sheaves.

share|cite|improve this answer
A very elegant minimalist example – Georges Elencwajg Nov 19 '10 at 12:02

How about a positive answer for a twist:-))? Pick an open cover $X=\cup_i U_i$ and try to reglue $F|_{U_i}$ into a brand new sheaf. To do that you need to pick automorphism $\sigma_{i,j}\in Aut(F|_{U_i \cap U_j})$ that agree on triple intersection. This is some Chech cocycle $Z^1_{U_i} (X, Aut (F))$. Two Chech cocycle will give you the same sheaf if they are different by a coboundary. Hence if $H^1_{U_i} (X, Aut (F))=1$ then any regluing will give the same sheaf.

Now you have to take care of all possible covers by going to the limit. Here is your positive answer then: true if and only if $H^1 (X, Aut (F))=1$.

share|cite|improve this answer
Do you mean dim H¹=1? – Ketil Tveiten May 12 '10 at 13:45
No, I do not. $H^1$ with coefficients in a sheaf of groups (not necessarily abelian) is a group. I mean that this group is trivial. – Bugs Bunny May 12 '10 at 17:28
@BugsBunny: H^1(X,F) is NOT a group for a non-abelian sheaf of groups F, it's just a pointed set. – Qfwfq May 12 '10 at 18:25
Thanks, unknown google! You are right, it is just a set if $Aut(F)$ is non-abelian. But it still parametrizes isomorphisms classes of sheaves, locally isomorphic to $F$. – Bugs Bunny May 12 '10 at 22:08

No: think of non-isomorphic vector bundles over $X$. They are stalkwise isomorphic even as modules (over $\mathcal{O}_{X,x}$ at various $x \in X$, where $\mathcal{O}_X$ is the sheaf of continuous functions on $X$), hence as abelian groups.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.