Example of non-isomorphic sheaves, with isomorphic stalks at every point?

Question

Basically what the title asks. I'd like to see an example of two sheaves on a topological space which have the same (isomorphic) stalks at every point, but are not isomorphic as sheaves.

Answer 1

For a very simple example, let $X=\{0,1\}$ with the topology in which $\{0\}$ is open and $\{1\}$ is not. Consider the following two sheaves of abelian groups $F$ and $G$ on $X$. We have $F(U)=G(U)=\mathbb{Z}$ for each nonempty open set $U$. The restriction map $F(X)\to F(\{0\})$ is the identity map and the restriction map $G(X)\to G(\{0\})$ is multiplication by $2$. All the stalks of $F$ and $G$ are isomorphic to $\mathbb{Z}$, but $F$ and $G$ are not isomorphic since the restriction map $F(X)\to F(\{0\})$ is an isomorphism and the restriction map $G(X)\to G(\{0\})$ is not.

Answer 2

Take as the topological space $S^1$ the circle. Take as stalks, $\mathbb{Z}_3$. There is an automorphism switching the two non identity elements. One sheaf is the trivial sheaf, the other is twisted like a Möbius strip.