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Let $X$ be a scheme and $\mathscr{F}$, $\mathscr{G}$ sheaves on $X$. Suppose we know that for every $x\in X$ the stalks are isomorphic: $$ \mathscr{F}_x \cong \mathscr{G}_x $$

Can we conclude immediately from here that $\mathscr{F}\cong \mathscr{G}$, or do we first need to find a morphism of sheaves $f:\mathscr{F}\to \mathscr{G}$ such that $f_x$ induces the above isomorphisms on the stalks for every $x$ ?

Remark: Proposition 1.1 (page 63) of Hartshorne says that:

A morphism of sheaves $f : \mathscr{F} \to \mathscr{G}$ on a topological space $X$ is an isomorphism iff the induced map on stalks $f_x : \mathscr{F}_x \to \mathscr{G}_x$ is an isomorphism for every $x\in X$.

Extra question: What can we say, instead, if we have isomorphisms between every fiber of $\mathscr{F}$ and $\mathscr{G}$ ?

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marked as duplicate by Zev Chonoles, Thomas Andrews, user1337, TMM, Joe Johnson 126 Dec 8 '13 at 22:27

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Is that true? For any sheaf of rank 1 $O_X$ modules, all there stacks are isomorphic. –  John Dec 7 '13 at 9:07
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I agree with John. Won't the twists $O(n)$ of the structure sheaf of the projective line provide many counterexamples? A sheaf is not just a collection of stalks. –  Jyrki Lahtonen Dec 7 '13 at 9:25
    
Please look at my edited question. There was a little mistake: I meant $f$ to be a morphism between the sheaves, not of the scheme $X$ to itself. –  Abramo Dec 7 '13 at 9:40
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The comments still stand, I think. –  Jyrki Lahtonen Dec 7 '13 at 9:44
    
Could you expand on your comments, please? It's not clear to me what is the fact you claim not to be true –  Abramo Dec 7 '13 at 10:07

2 Answers 2

up vote 7 down vote accepted

No, in the absence of a global morphism $f:\mathscr{F}\to \mathscr{G}$ inducing the isomorphisms on stalks, we cannot conclude immediately that the two sheaves are isomorphic.

As a counterexample consider two sheaves on the Riemann sphere (the same works with any projective line). Let $\mathscr{F}$ be the structure sheaf. Algebraic or holomorphic? Let's do holomorphic here, but it doesn't matter. The stalks consist of germs of holomorphic functions, i.e. power series $f(t)=\sum_{n\ge0}a_nt^n$ with a positive radius of convergence. Here $t$ denotes a local variable, so if we are interested in a stalk at a point $z_0\in\Bbb{C}$, then we can use $t=z-z_0$, and if we are working at the point of infinity, we can use $t=1/z$.

Then let $\mathscr{G}$ be the sheaf of holomorphic differential forms. Here the stalks look like $f(t)\,dt$, where again $t$ is a local variable, and $f(t)$ is the germ of a holomorphic function.

At the level of stalks we have an isomorphism $\mathscr{F}_{z_0}\cong\mathscr{G}_{z_0}$ given by $f(t)\mapsto f(t)\,dt$. Yet the sheaves $\mathscr{F}$ and $\mathscr{G}$ are not isomorphic. The sheaf $\mathscr{F}$ has the constant functions as global sections, but the sheaf $\mathscr{G}$ has no non-zero global sections. [Edit: Forgot to explain this] For if $\omega$ where such a non-zero differential form, then $F(z)=\int_{\gamma[0\to z]}\omega$ would define a non-constant holomorphic function on the Riemann sphere. But the image of that function would be compact, hence bounded, contradicting Liouville's theorem. [/Edit]

The same holds in an algebraic category. IIRC Hartshorne explains why the sheafs $\mathscr{O}[n],n<0,$ have no global sections on the projective line. The canonical sheaf $\mathscr{O}[-2]$ being the analogue of the above sheaf $\mathscr{G}$.

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Ok thanks! Just to make thinks clear, so your answer is "no, we cannot conclude immediately they are isomorphic". But still, if the isomorphisms on stalks are induced by a global morphism between the sheaves, then they are isomorphic. Is this right? Thanks a lot! –  Abramo Dec 7 '13 at 10:46
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That's correct. The point is (vaguely) that a local isomorphism cannot be lifted to a global one. But the fact that a global homomorphism is an isomorphism can be detected locally. –  Jyrki Lahtonen Dec 7 '13 at 10:50
    
Many thanks again Jirky! I edited your answer to make it more explicit. –  Abramo Dec 7 '13 at 10:55

If this were indeed true then all line bundles on any scheme would be isomorphic! However this is not true as for example take $\Bbb{P}^1$ that has an integers worth of non-isomorphic line bundles. The point is that an isomorphism $\varphi_x$ of $\mathcal{O}_{X,x}$-modules need not glue to give an isomorphism of $\mathcal{O}_X$-modules (for example how is it even clear to go from an isomorphism of modules at every point to an isomorphism of sheaves globally?).

However if say given a priori a morphism of sheaves $\varphi : \mathcal{F} \to \mathcal{G}$ that is an isomorphism on the stalks at every point, then it is indeed true that $\varphi$ will be an isomorphism. As for your last question, it is not clear to me what you mean by the fiber of a sheaf.

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