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Let $\mathscr{Z}$ be an arbitrary sheaf on $\mathbb{R}/\mathbb{Z}=X$ (with the quotient topology). Let $\mathscr{F}$ and $\mathscr{G}$ denote the sheaves of continuous functions on $X$ with values in $\mathbb{R}$ and $X$, respectively.
Given a morphism $f\colon \mathscr{G}\to \mathscr{Z}$ such that the composition with the canonical map $j\colon \mathscr{F}\to \mathscr{G}$ is the zero morphism, how can I see that $f$ has to be zero?

It suffices to prove that for each $x\in X$, the map on the stalks $f_x$ is zero. Can we say that for each open $x\in U$ in $X$ and each $g\colon U\to X$, there exists some $x\in V \subset U$ open such that $g\vert V$ factors through the projection $\mathbb{R}\to X$, i.e. there exists some $h\colon V\to \mathbb{R}$ in $\mathscr{F}(V)$ with $j_V(h)=g_V$?

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Here is a sketch without many details, just intuition (but it can be made rigourous).

Since $X$ locally looks like $\mathbb R$, the stalk $\mathscr G_x$ can be thought of as functions $\mathbb R \to X$. These are periodic functions. But the stalk $\mathscr F_x$ can be thought of as arbitrary continous functions $\mathbb R \to \mathbb R$.

Hence we have an injection $k:\mathscr G \to \mathscr F$ such that $jk=id_{\mathscr G}$ (because the set of periodic functions inject into the set of all functions).

Hence $f=f \circ id_{\mathscr G}=f \circ j \circ k= 0 \circ k = 0$.

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