Isomorphisms of sheaves, are they equal in this situation?

Question

Suppose I have a topological space $X$ and sheaves of rings $F$ and $G$ on $X$. Suppose I also have two isomorphisms $\phi, \psi : F \rightarrow G$ and that $\phi(X) = \psi(X)$ as maps from $F(X)$ to $G(X)$. Does it then follow that $\phi = \psi$? Thanks!

Answer 1

Not really, no.

It is true for instance if for every open $U \subset X$ the "restriction" map $F(X) \to F(U)$ is surjective, that is, $F$ is a flabby sheaf: http://en.wikipedia.org/wiki/Injective_sheaf#Flasque_or_flabby_sheaves.

Another case where the morphism between global sections determines the morphism of sheaves. Assume for instance that there exists a basis $U_i$ of $X$ so that the restriction $F(X) \to F(U_i)$ and $G(X)\to G(U_i)$ "are" localizations with respect to a multiplicative set, or even that for every $x \in X$ the restriction $F(X) \to F_x$, $G(X)\to G_x$ are morphisms to a ring of fractions. This happens for instance if $F$ is the structure sheaf of an affine scheme. Also works if for instance $F(X) \to F_x$ surjective for every $x$. For instance, $F$ is the sheaf of smooth functions on a $C^{\infty}$ manifold.

But not in general. Here is a minimal example:

Consider the topological space $X=\{1,2\}$ with topology $\tau = \{ \emptyset, X, \{1\} \}$.

Define the sheaves of rings $F=G$ on $X$ as by $F(X)=G(X) = \mathbb{Z}$ and $F(\{1\}) = G(\{1\})= \mathbb{Z}[t]$. Consider the morphism of sheaves $\phi$, $\psi$, $\ \ \phi= \mathbb{1}_F$, while $\psi(X)= \mathbb{1}_{\mathbb{Z}}$ and $\psi(\{1\})$ acts on $\mathbb{Z}[t]$ by $t \mapsto -t$.

Obs: one can assign to the empty set the ring $0$ : $F(\emptyset) = 0$.