**Proof.**
The equivalence of (1) and (2) follows immediately from Proposition 60.6.6.

Assume (3)(a) and (3)(b). Let $A \to B$ be faithfully flat and ind-Zariski. We will use without further mention the fact that a flat map $A \to B$ is faithfully flat if and only if every closed point of $\mathop{\mathrm{Spec}}(A)$ is in the image of $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ We will show that $A \to B$ has a section.

Let $I \subset A$ be an ideal such that $V(I) \subset \mathop{\mathrm{Spec}}(A)$ is the set of closed points of $\mathop{\mathrm{Spec}}(A)$. We may replace $B$ by the ring $C$ constructed in Lemma 60.5.8 for $A \to B$ and $I \subset A$. Thus we may assume $\mathop{\mathrm{Spec}}(B)$ is w-local such that the set of closed points of $\mathop{\mathrm{Spec}}(B)$ is $V(IB)$.

Assume $\mathop{\mathrm{Spec}}(B)$ is w-local and the set of closed points of $\mathop{\mathrm{Spec}}(B)$ is $V(IB)$. Choose a continuous section to the surjective continuous map $V(IB) \to V(I)$. This is possible as $V(I) \cong \pi _0(\mathop{\mathrm{Spec}}(A))$ is extremally disconnected, see Topology, Proposition 5.26.6. The image is a closed subspace $T \subset \pi _0(\mathop{\mathrm{Spec}}(B)) \cong V(JB)$ mapping homeomorphically onto $\pi _0(A)$. Replacing $B$ by the ind-Zariski quotient ring constructed in Lemma 60.6.1 we see that we may assume $\pi _0(\mathop{\mathrm{Spec}}(B)) \to \pi _0(\mathop{\mathrm{Spec}}(A))$ is bijective. At this point $A \to B$ is an isomorphism by Lemma 60.6.4.

Assume (1) or equivalently (2). Let $A \to A_ w$ be the ring map constructed in Lemma 60.5.3. By (1) there is a section $A_ w \to A$. Thus $\mathop{\mathrm{Spec}}(A)$ is homeomorphic to a closed subset of $\mathop{\mathrm{Spec}}(A_ w)$. By Lemma 60.2.4 we see (3)(a) holds. Finally, let $T \to \pi _0(A)$ be a surjective map with $T$ an extremally disconnected, quasi-compact, Hausdorff topological space (Topology, Lemma 5.26.9). Choose $A \to B$ as in Lemma 60.6.2 adapted to $T \to \pi _0(\mathop{\mathrm{Spec}}(A))$. By (1) there is a section $B \to A$. Thus we see that $T = \pi _0(\mathop{\mathrm{Spec}}(B)) \to \pi _0(\mathop{\mathrm{Spec}}(A))$ has a section. A formal categorical argument, using Topology, Proposition 5.26.6, implies that $\pi _0(\mathop{\mathrm{Spec}}(A))$ is extremally disconnected.
$\square$

## Comments (1)

Comment #5931 by Mingchen on