Let $X$ be a topological space, and let $\mathscr{F}, \mathscr{G}$ be sheaves of sets on $X$. It is well-known that a morphism $\varphi : \mathscr{F} \to \mathscr{G}$ is epic (in the category of sheaves on $X$) if and only if the induced map of stalks $\varphi_P : \mathscr{F}_P \to \mathscr{G}_P$ is surjective for every point $P$ in $X$, but the section maps $\varphi_U : \mathscr{F}(U) \to \mathscr{G}(U)$ need not be surjective. I know of a couple of examples from complex analysis:
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Let $X$ be the punctured complex plane, $\mathscr{F}$ the sheaf of meromorphic functions, $\mathscr{G}$ the sheaf of differential $1$-forms, and $\varphi$ the differential map; then $\varphi$ is epic and indeed the sequence $0 \to \mathscr{F} \to \mathscr{G} \to 0$ is even exact, but there are global sections of $\mathscr{G}$ which are not the image of a global section of $\mathscr{F}$, e.g. $z \mapsto \frac{1}{z} \, \mathrm{d}z$.
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Let $X$ be the punctured complex plane again, $\mathscr{F}$ the sheaf of meromorphic functions, $\mathscr{G}$ the sheaf of nowhere-zero meromorphic functions, and let $\varphi : \mathscr{F} \to \mathscr{G}$ be composition with $\exp : \mathbb{C} \to \mathbb{C}$; then $\varphi$ is epic but again fails to be surjective on (global) sections: after all, there is no holomorphic function $f : X \to \mathbb{C}$ such that $\exp f(z) = z$ for all non-zero $z$.
Question. Are there simpler examples which do not require much background knowledge beyond knowing the definition of sheaves and stalks?
Best Answer
Take $X=\mathbb R$ for your topological space, the constant sheaf $\underline {\mathbb Z} $ for $\mathcal F$ and for $\mathcal G$ the direct sum of two skyscraper sheaves with fibers $\mathbb Z$ at two distinct points $P,Q\in \mathbb R$, that is $\mathcal G=\mathbb Z^P \oplus \mathbb Z^Q$.
The natural restriction $\mathcal F=\underline {\mathbb Z} \to \mathcal G=\mathbb Z^P \oplus \mathbb Z^Q$ is a surjective sheaf morphism but the associated group morphism on global sections $\mathcal F(X)=\mathbb Z \to \mathcal G (X)=\mathbb Z \oplus \mathbb Z$ is not surjective [its image is the diagonal of $\mathbb Z \oplus \mathbb Z$, consisting of pairs $(z,w)$ with $z=w$].
Edit This example can easily be adapted to a three point space space: thanks to Pierre-Yves who, in a comment to Alex's answer, suggested that.
Take $X=\{P,Q, \eta\}$ with closed sets $X,\emptyset, \{P\}, \{Q\}, \{P,Q\}$ (this is the same space as Alex's).The rest is exactly the same as above. Namely $\mathcal F=\underline {\mathbb Z} $, $\mathcal G=\mathbb Z^P \oplus \mathbb Z^Q$, $\mathcal F=\underline {\mathbb Z} \to \mathcal G=\mathbb Z^P \oplus \mathbb Z^Q$ the restriction, which is again a surjective sheaf morphism, and $\mathcal F(X)=\mathbb Z \to \mathcal G (X)=\mathbb Z \oplus \mathbb Z \:$ not surjective (the image being again the diagonal of $\mathbb Z \oplus \mathbb Z$).
The main point is that the stalks of $\mathbb Z^P$ are:
$(\mathbb Z^P)_P=\mathbb Z,(\mathbb Z^P)_Q=0, (\mathbb Z^P)\eta=0$, because $P$ is a closed point. Ditto for $\mathbb Z^Q$.
Tangential remark It might be of some interest to notice that there is a scheme structure on $X$ which makes it the smallest possible non affine scheme. This is explained in the book The Geometry of Schemes by Eisenbud and Harris, on page 22.