Failure of isomorphisms on stalks to arise from an isomorphism of sheaves
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, ...
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.
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.