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If i have a morpism of Sheaves, there is only one morphism of the stalks. Because i want to know, if i have two morphisms say $\phi, \psi: F \rightarrow G$ and i know for some reason that $\psi_x = \phi_x$ can i deduce from that $\phi = \psi$ already?

Thanks

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    Passing to the stalk is a functor, i.e. if we have $\phi: F \to G$, then $\phi_x: F_x \to G_x$ arises in a functorial way. Furthermore your latest claim (we can test equality of maps on stalks) is true, but this is an whole other story... – 2017-02-03
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    ok i will change that one but how can i show the equality of the maps on stalks? – 2017-02-03
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    Two morphisms are equal if all sections of $F$ are mapped to the same sections of $G$. You can test equality of section on stalks (this is basically the first sheaf axiom), hence you are done. – 2017-02-03

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