The constant sheaf of integers on the line is not flasque

$A=(-2,-1)$ and $B=(1,2)$ are open interval forms of the line, so by [F4] each is an open subset of $\mathbb{R}$; by the union axiom for open sets [F5] their union $U=A\cup B$ is open as well, and $U\mathrm{\ne }\mathbb{R}$ since $0\in \mathbb{R}\setminus U$. The two sets are nonempty, disjoint, and $U=A\cup B$; both are open in the subspace $U$ as well, being traces of open sets of $\mathbb{R}$.

Because $A\cap B=\mathrm{\varnothing }$ and $U=A\cup B$, the rule that assigns $0$ to every point of $A$ and $1$ to every point of $B$ defines a function $s:U\to \mathbb{Z}$. It is locally constant in the sense of [F2]: a point of $A$ has the open neighbourhood $A$ inside $U$, on which $s$ is constantly $0$, and a point of $B$ has the open neighbourhood $B$, on which $s$ is constantly $1$. By [F3] the locally constant $\mathbb{Z}$-valued functions on $U$ are the sections of $\underset{‾}{\mathbb{Z}}$ over $U$, so $s\in \underset{‾}{\mathbb{Z}}(U)$.

Suppose that $t\in \underset{‾}{\mathbb{Z}}(\mathbb{R})$ restricts to $s$, that is, $t{\mathrm{\mid }}_U=s$. By [F3] the element $t$ is a locally constant $\mathbb{Z}$-valued function on $\mathbb{R}$. Every such function is constant: its fibres $t^{-1}(n)$, $n\in \mathbb{Z}$, are open by local constancy [F2], pairwise disjoint, and cover $\mathbb{R}$, so if two distinct fibres were nonempty, one of them and the union of all the other fibres would be nonempty disjoint open sets covering $\mathbb{R}$, hence would form a separation [F7], which is impossible because $\mathbb{R}=(-\mathrm{\infty },\mathrm{\infty })$ is connected [F6]. Hence $t\equiv n$ for some $n\in \mathbb{Z}$. Restricting to the nonempty sets $A$ and $B$ of [step 1.1] gives $n=t{\mathrm{\mid }}_A=s{\mathrm{\mid }}_A=0$ and $n=t{\mathrm{\mid }}_B=s{\mathrm{\mid }}_B=1$, so $0=1$, a contradiction. Therefore no global section restricts to $s$.

By [step 2.1] the element $s$ lies in $\underset{‾}{\mathbb{Z}}(U)$, and by [step 3.1] it has no preimage under the restriction map $\underset{‾}{\mathbb{Z}}(\mathbb{R})\to \underset{‾}{\mathbb{Z}}(U)$. That map is therefore not surjective, and since a sheaf is flasque exactly when all of its restriction maps are surjective [F1], the constant sheaf $\underset{‾}{\mathbb{Z}}$ on $\mathbb{R}$ is not flasque. The section $s$ itself is the witness: it takes the two distinct values $0$ and $1$ on the two components $A$ and $B$ of $U$, and a global locally constant function on the connected line has only one value. Note that $s$ does extend to the global function equal to $0$ on $A$, $1$ on $B$, and, say, $0$ on $\mathbb{R}\setminus U$; that function is not locally constant, and correspondingly the sheaf of all functions on $\mathbb{R}$ is flasque (The sheaf of all functions to an abelian group is flasque), so the failure is exactly the locally constant requirement and not the extension of functions. No choice principle is used: the sets and the section are given by explicit formulas. ∎