Failure of isomorphisms on stalks to arise from an isomorphism of sheaves It is well-known (Hartshorne 2.1.1) that if $F$ and $G$ are sheaves on a space $X$, then $\phi:F\rightarrow G$ is an isomorphism if and only if the induced stalk map $\phi_p:F_p\rightarrow G_p$ is an isomorphism for all $p\in X$. However, if we have a collection of isomorphisms $\{\psi_p:F_p\rightarrow G_p\}_{p\in X}$, this does not guarantee that $F$ and $G$ are isomorphic, because the $\psi_p$ might not be related to each other, i.e. there might not be a sheaf map $\psi:F\rightarrow G$ such that $\psi_p$ is the induced stalk map for all $p\in X$.
However, I was recently making this point to someone and was unable to think of a good example of non-isomorphic $F$ and $G$ having isomorphisms $\psi_p:F_p\rightarrow G_p$. I'm sure I knew one at some point, but I'm blanking on it now. Can someone provide an illustrative example, e.g. an example that occurs in some natural or basic problem, or one that captures the essential pattern of any example where this issue arises, or one where it is clear that $F$ and $G$ could not be isomorphic?
 A: Let $X$ be a topological space, and the open sets are:
$X=\{a,b\}$, $U=\{a\}$ and empty set.
$F$  and $G$ are two sheaves on $X$ of Abelian groups. Set
$F(X)=F(U)=G(X)=G(U)=\mathbb{Z}/2$, $\phi:F(X)\to F(U)$ is identity. But $\psi:G(X)\to G(U)$ is zero map. Verify $F$ and $G$ are sheaves on $X$. But they cannot be isomorphic.
A: On a locally ringed space $(X,\mathcal O_X)$, a locally free sheaf $\mathcal E$ of rank $r$ has (as the name indicates) your property $\mathcal E_p \simeq \mathcal O^{\oplus r}_{X,p}$ (isomorphism of $O_{X,p}$-modules). Hence you get any number of natural examples by considering non-trivial   locally free sheaves $\mathcal E \not\simeq \mathcal O^{\oplus r}_X $  in your favourite category: schemes, topological manifolds, differentiable manifolds, analytic spaces, ...
A: Let $X$ be a Hausdorff space. Consider:


*

*The constant sheaf $\mathbb{Z}$.

*The sheaf $\bigoplus_{x \in X} i_*(\mathbb{Z})$ where $i_*$ means the inclusion of the point $x \in X$. Note that the stalks are $\mathbb{Z}$ at each $x \in X$ as taking stalks commutes with colimits (it's left adjoint to the skyscraper sheaf functor, and is in fact a special case of the inverse image functor).


But these will almost never be isomorphic. If $X$ is $\mathbb{R}$, for instance, then the global sections of the first example will just be $\mathbb{Z}$. There will be a lot more global sections in the second example.
A: A bit late to the party, but my favourite examples are the twisting sheaves. In the proof of Proposition 5.12(a) of Hartshorne, the isomorphism of restricted sheaves does not lift to an isomorphism of the entire sheaf.
A: Question: "However, I was recently making this point to someone and was unable to think of a good example of non-isomorphic F and G having isomorphisms ψp:Fp→Gp. I'm sure I knew one at some point, but I'm blanking on it now. Can someone provide an illustrative example, e.g. an example that occurs in some natural or basic problem, or one that captures the essential pattern of any example where this issue arises, or one where it is clear that F and G could not be isomorphic?"
Answer: Assume $A$ is a commutative ring with $1$ and $E$ a rank $r$ locally trivial $A$-module in the sense that there is an open cover $D(f_i)$ of $X:=Spec(A)$ with $E_{f_i} \cong A_{f_i}^r$. Let $F:=A^r$. It follows $E_x \cong F_x$ for any point $x\in X$, but $E$ and $F$ are not neccessarily isomorphic.
Ex. An explicit example is the "tangent bundle" on the real 2-sphere $S:=Spec(A)$ with $A:=\mathbb{R}[x,y,z]/(x^2+y^2+z^2-1)$. If $E:=Der_{\mathbb{R}}(A)$ it follows $E$ is a non trivial rank 2 locally trivial $A$-module. Hence $E_x \cong A^2_x$ for any $x\in S$ but $E \neq A^2$.
"Homological algebra" proves the following: Let $f,g:E\rightarrow F$ be two maps of $A$-modules and let $\phi:=f-g$. It follows $\phi$ is a map of $A$-modules. Let $K:=ker(\phi), C:=E/K$. We get an exact sequence
$$0 \rightarrow K \rightarrow E \rightarrow C \rightarrow 0$$
and it follows for any $x\in Spec(A):=X$ that $C_x=0$ iff $f_x=g_x$ are equal on the stalk at $x$.
Moreover $C=0$ iff $C_x=0$ for all $x\in X$. It follows $C=0$ iff $f=g$ iff $C_x=0$ for all $x\in X$ iff $f_x=g_x$. Hence the two maps $f,g$ are equal iff they are equal on the  stalks.
Note moreover: It is enough to check this at maximal ideals: $C=0$ iff $C_x=0$ for all closed points $x\in X$. Hence the prime ideals (or "non closed points") are not needed to distinguish two maps of vector bundles (or quasi coherent sheaves).
