Lemma 69.23.9. In Situation 69.23.1 suppose that we have a cartesian diagram

$\xymatrix{ X^1 \ar[r]_ p \ar[d]_ q & X^3 \ar[d]^ a \\ X^2 \ar[r]^ b & X^4 }$

of algebraic spaces quasi-separated and of finite type over $B$. For each $j = 1, 2, 3, 4$ choose $i_ j \in I$ and a diagram

$\xymatrix{ X^ j \ar[r] \ar[d] & W^ j \ar[d] \\ B \ar[r] & B_{i_ j} }$

as in (69.23.2.1). Let $X^ j = \mathop{\mathrm{lim}}\nolimits _{i \geq i_ j} X^ j_ i$ be the corresponding limit descriptions as in Lemma 69.23.4. Let $(a_ i)_{i \geq i_5}$, $(b_ i)_{i \geq i_6}$, $(p_ i)_{i \geq i_7}$, and $(q_ i)_{i \geq i_8}$ be the corresponding morphisms of inverse systems contructed in Lemma 69.23.4. Then there exists an $i_9 \geq \max (i_5, i_6, i_7, i_8)$ such that for $i \geq i_9$ we have $a_ i \circ p_ i = b_ i \circ q_ i$ and such that

$(q_ i, p_ i) : X^1_ i \longrightarrow X^2_ i \times _{b_ i, X^4_ i, a_ i} X^3_ i$

is a closed immersion. If $a$ and $b$ are flat and of finite presentation, then there exists an $i_{10} \geq \max (i_5, i_6, i_7, i_8, i_9)$ such that for $i \geq i_{10}$ the last displayed morphism is an isomorphism.

Proof. According to the discussion in Remark 69.23.5 the choice of $W^1$ fitting into a diagram as in (69.23.2.1) is immaterial for the truth of the lemma. Thus we may choose $W^1 = W^2 \times _{W^4} W^3$. Then it is immediate from the construction of $X^1_ i$ that $a_ i \circ p_ i = b_ i \circ q_ i$ and that

$(q_ i, p_ i) : X^1_ i \longrightarrow X^2_ i \times _{b_ i, X^4_ i, a_ i} X^3_ i$

is a closed immersion.

If $a$ and $b$ are flat and of finite presentation, then so are $p$ and $q$ as base changes of $a$ and $b$. Thus we can apply Lemma 69.23.6 to each of $a$, $b$, $p$, $q$, and $a \circ p = b \circ q$. It follows that there exists an $i_9 \in I$ such that

$(q_ i, p_ i) : X^1_ i \to X^2_ i \times _{X^4_ i} X^3_ i$

is the base change of $(q_{i_9}, p_{i_9})$ by the morphism by the morphism $X^4_ i \to X^4_{i_9}$ for all $i \geq i_9$. We conclude that $(q_ i, p_ i)$ is an isomorphism for all sufficiently large $i$ by Lemma 69.6.10. $\square$

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