Lemma 20.20.6. Let $X$ be an irreducible topological space. Let $\mathcal{H} \subset \underline{\mathbf{Z}}$ be an abelian subsheaf of the constant sheaf. Then there exists a nonempty open $U \subset X$ such that $\mathcal{H}|_ U = \underline{d\mathbf{Z}}_ U$ for some $d \in \mathbf{Z}$.

**Proof.**
Recall that $\underline{\mathbf{Z}}(V) = \mathbf{Z}$ for any nonempty open $V$ of $X$ (see proof of Lemma 20.20.2). If $\mathcal{H} = 0$, then the lemma holds with $d = 0$. If $\mathcal{H} \not= 0$, then there exists a nonempty open $U \subset X$ such that $\mathcal{H}(U) \not= 0$. Say $\mathcal{H}(U) = n\mathbf{Z}$ for some $n \geq 1$. Hence we see that $\underline{n\mathbf{Z}}_ U \subset \mathcal{H}|_ U \subset \underline{\mathbf{Z}}_ U$. If the first inclusion is strict we can find a nonempty $U' \subset U$ and an integer $1 \leq n' < n$ such that $\underline{n'\mathbf{Z}}_{U'} \subset \mathcal{H}|_{U'} \subset \underline{\mathbf{Z}}_{U'}$. This process has to stop after a finite number of steps, and hence we get the lemma.
$\square$

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