Lemma 21.31.9. Consider the comparison morphism $\epsilon : \textit{LC}_{qc} \to \textit{LC}_{Zar}$. Let $\mathcal{P}$ denote the class of proper maps of topological spaces. For $X$ in $\textit{LC}_{Zar}$ denote $\mathcal{A}'_ X \subset \textit{Ab}(\textit{LC}_{Zar}/X)$ the full subcategory consisting of sheaves of the form $\pi _ X^{-1}\mathcal{F}$ with $\mathcal{F}$ in $\textit{Ab}(X)$. Then (1), (2), (3), (4), and (5) of Situation 21.30.1 hold.

**Proof.**
We first show that $\mathcal{A}'_ X \subset \textit{Ab}(\textit{LC}_{Zar}/X)$ is a weak Serre subcategory by checking conditions (1), (2), (3), and (4) of Homology, Lemma 12.10.3. Parts (1), (2), (3) are immediate as $\pi _ X^{-1}$ is exact and fully faithful by Lemma 21.31.7 part (3). If $0 \to \pi _ X^{-1}\mathcal{F} \to \mathcal{G} \to \pi _ X^{-1}\mathcal{F}' \to 0$ is a short exact sequence in $\textit{Ab}(\textit{LC}_{Zar}/X)$ then $0 \to \mathcal{F} \to \pi _{X, *}\mathcal{G} \to \mathcal{F}' \to 0$ is exact by Lemma 21.31.7 part (2). Hence $\mathcal{G} = \pi _ X^{-1}\pi _{X, *}\mathcal{G}$ is in $\mathcal{A}'_ X$ which checks the final condition.

Property (1) holds by Lemma 21.31.1 and the fact that the base change of a proper map is a proper map (see Topology, Theorem 5.17.5 and Lemma 5.4.4).

Property (2) follows from the commutative diagram (5) in Lemma 21.31.7.

Property (3) is Lemma 21.31.6.

Property (4) is Lemma 21.31.7 part (7)(b).

Proof of (5). Suppose given a qc covering $\{ U_ i \to U\} $. For $u \in U$ pick $i_1, \ldots , i_ m \in I$ and quasi-compact subsets $E_ j \subset U_{i_ j}$ such that $\bigcup f_{i_ j}(E_ j)$ is a neighbourhood of $u$. Observe that $Y = \coprod _{j = 1, \ldots , m} E_ j \to U$ is proper as a continuous map between Hausdorff quasi-compact spaces (Topology, Lemma 5.17.7). Choose an open neighbourhood $u \in V$ contained in $\bigcup f_{i_ j}(E_ j)$. Then $Y \times _ U V \to V$ is a surjective proper morphism and hence a $qc$ covering by Lemma 21.31.4. Since we can do this for every $u \in U$ we see that (5) holds. $\square$

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