Lemma 61.6.5. Let $A \to B$ be ring map such that

1. $A \to B$ identifies local rings,

2. the topological spaces $\mathop{\mathrm{Spec}}(B)$, $\mathop{\mathrm{Spec}}(A)$ are w-local, and

3. $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is w-local.

Then $A \to B$ is ind-Zariski.

Proof. Set $X = \mathop{\mathrm{Spec}}(A)$ and $Y = \mathop{\mathrm{Spec}}(B)$. Let $X_0 \subset X$ and $Y_0 \subset Y$ be the set of closed points. Let $A \to A'$ be the ind-Zariski morphism of affine schemes such that with $X' = \mathop{\mathrm{Spec}}(A')$ the diagram

$\xymatrix{ X' \ar[r] \ar[d] & \pi _0(X') \ar[d] \\ X \ar[r] & \pi _0(X) }$

is cartesian in the category of topological spaces and such that $\pi _0(X') = \pi _0(Y)$ as spaces over $\pi _0(X)$, see Lemma 61.6.2. By Lemma 61.2.5 we see that $X'$ is w-local and the set of closed points $X'_0 \subset X'$ is the inverse image of $X_0$.

We obtain a continuous map $Y \to X'$ of underlying topological spaces over $X$ identifying $\pi _0(Y)$ with $\pi _0(X')$. By Lemma 61.3.8 (and Lemma 61.4.6) this corresponds to a morphism of affine schemes $Y \to X'$ over $X$. Since $Y \to X$ maps $Y_0$ into $X_0$ we see that $Y \to X'$ maps $Y_0$ into $X'_0$, i.e., $Y \to X'$ is w-local. By Lemma 61.6.4 we see that $Y \cong X'$ and we win. $\square$

Comment #5930 by Mingchen on

There is a typo in the proof: "this is corresponds", remove "is".

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