Lemma 6.29.4. In the situation described above, let $i \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{I})$ and let $U_ i \subset X_ i$ be a quasi-compact open. Then

$\mathop{\mathrm{colim}}\nolimits _{a : j \to i} \mathcal{F}_ j(f_ a^{-1}(U_ i)) = \mathcal{F}(p_ i^{-1}(U_ i))$

Proof. Recall that $p_ i^{-1}(U_ i)$ is a quasi-compact open of the spectral space $X$, see Topology, Lemma 5.24.5. Hence Lemma 6.29.1 applies and we have

$\mathcal{F}(p_ i^{-1}(U_ i)) = \mathop{\mathrm{colim}}\nolimits _{a : j \to i} p_ j^{-1}\mathcal{F}_ j(p_ i^{-1}(U_ i)).$

A formal argument shows that

$\mathop{\mathrm{colim}}\nolimits _{a : j \to i} \mathcal{F}_ j(f_ a^{-1}(U_ i)) = \mathop{\mathrm{colim}}\nolimits _{a : j \to i} \mathop{\mathrm{colim}}\nolimits _{b : k \to j} f_ b^{-1}\mathcal{F}_ j(f_{a \circ b}^{-1}(U_ i))$

Thus it suffices to show that

$p_ j^{-1}\mathcal{F}_ j(p_ i^{-1}(U_ i)) = \mathop{\mathrm{colim}}\nolimits _{b : k \to j} f_ b^{-1}\mathcal{F}_ j(f_{a \circ b}^{-1}(U_ i))$

This is Lemma 6.29.3 applied to $\mathcal{F}_ j$ and the quasi-compact open $f_ a^{-1}(U_ i)$. $\square$

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