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
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
of topological spaces. By Lemma 61.2.5 we conclude that $T = \pi _0(Y)$. $\square$
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