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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