Lemma 61.6.2. Let A be a ring and let X = \mathop{\mathrm{Spec}}(A). Let T be a profinite space and let T \to \pi _0(X) be a continuous map. There exists an ind-Zariski ring map A \to B such that with Y = \mathop{\mathrm{Spec}}(B) the diagram
\xymatrix{ Y \ar[r] \ar[d] & \pi _0(Y) \ar[d] \\ X \ar[r] & \pi _0(X) }
is cartesian in the category of topological spaces and such that \pi _0(Y) = T as spaces over \pi _0(X).
Proof.
Namely, write T = \mathop{\mathrm{lim}}\nolimits T_ i as the limit of an inverse system finite discrete spaces over a directed set (see Topology, Lemma 5.22.2). For each i let Z_ i = \mathop{\mathrm{Im}}(T \to \pi _0(X) \times T_ i). This is a closed subset. Observe that X \times T_ i is the spectrum of A_ i = \prod _{t \in T_ i} A and that A \to A_ i is a local isomorphism. By Lemma 61.6.1 we see that Z_ i \subset \pi _0(X \times T_ i) = \pi _0(X) \times T_ i corresponds to a surjection A_ i \to B_ i which is ind-Zariski such that \mathop{\mathrm{Spec}}(B_ i) = X \times _{\pi _0(X)} Z_ i as subsets of X \times T_ i. The transition maps T_ i \to T_{i'} induce maps Z_ i \to Z_{i'} and X \times _{\pi _0(X)} Z_ i \to X \times _{\pi _0(X)} Z_{i'}. Hence ring maps B_{i'} \to B_ i (Lemmas 61.3.8 and 61.4.6). Set B = \mathop{\mathrm{colim}}\nolimits B_ i. Because T = \mathop{\mathrm{lim}}\nolimits Z_ i we have X \times _{\pi _0(X)} T = \mathop{\mathrm{lim}}\nolimits X \times _{\pi _0(X)} Z_ i and hence Y = \mathop{\mathrm{Spec}}(B) = \mathop{\mathrm{lim}}\nolimits \mathop{\mathrm{Spec}}(B_ i) fits into the cartesian diagram
\xymatrix{ Y \ar[r] \ar[d] & T \ar[d] \\ X \ar[r] & \pi _0(X) }
of topological spaces. By Lemma 61.2.5 we conclude that T = \pi _0(Y).
\square
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