Proof.
Let $Y' \subset X'$ be the exceptional divisor. Let $Z' \subset Y'$ be the inverse image of $Z \subset Y$. Then $U' = X' \setminus Z'$ is the inverse image of $U$. With $\delta ^{Y'}_{Z'}$ as in (52.18.0.1) we set
\[ T' = \{ y' \in Y' \mid \delta ^{Y'}_{Z'}(y') = 1\text{ or }2\} \subset T = \{ y' \in Y' \mid \delta ^{Y'}_{Z'}(y') = 1\} \]
These are specialization stable subsets of $U' \cap Y' = Y' \setminus Z'$. Consider the object $(b|_{U'}^*\mathcal{F}_ n)$ of $\textit{Coh}(U', I\mathcal{O}_{U'})$, see Cohomology of Schemes, Lemma 30.23.9. For $y' \in U' \cap Y'$ let us denote
\[ \mathcal{F}_{y'}^\wedge = \mathop{\mathrm{lim}}\nolimits (b|_{U'}^*\mathcal{F}_ n)_{y'} \]
the “stalk” of this pullback at $y'$. We claim that conditions (a), (b), (c), (d), and (e) of Lemma 52.21.2 hold for the object $(b|_{U'}^*\mathcal{F}_ n)$ on $U'$ with $d$ replaced by $1$ and the subsets $T' \subset T \subset U' \cap Y'$. Condition (a) holds because $Y'$ is an effective Cartier divisor and hence locally cut out by $1$ equation. Condition (e) holds by Lemma 52.18.1 parts (1) and (2). To prove (b), (c), and (d) we need some preparation.
Let $y' \in U' \cap Y'$ and let $\mathfrak p' \subset \mathcal{O}_{X', y'}^\wedge $ be a prime ideal not contained in $V(I\mathcal{O}_{X', y'}^\wedge )$. Denote $y = b(y') \in U \cap Y$. Choose $f \in I$ such that $y'$ is contained in the spectrum of the affine blowup algebra $A[\frac{I}{f}]$, see Divisors, Lemma 31.32.2. For any $A$-algebra $B$ denote $B' = B[\frac{IB}{f}]$ the corresponding affine blowup algebra. Denote $I$-adic completion by ${\ }^\wedge $. By our choice of $f$ we get a ring map $(\mathcal{O}_{X, y}^\wedge )' \to \mathcal{O}_{X', y'}^\wedge $. If we let $\mathfrak q' \subset (\mathcal{O}_{X, y}^\wedge )'$ be the inverse image of $\mathfrak m_{y'}^\wedge $, then we see that $((\mathcal{O}_{X, y}^\wedge )'_{\mathfrak q'})^\wedge = \mathcal{O}_{X', y'}^\wedge $. Let $\mathfrak p \subset \mathcal{O}_{X, y}^\wedge $ be the corresponding prime. At this point we have a commutative diagram
\[ \xymatrix{ \mathcal{O}_{X, y}^\wedge \ar[d] \ar[r] & (\mathcal{O}_{X, y}^\wedge )' \ar[d]_\alpha \ar[r] & (\mathcal{O}_{X, y}^\wedge )'_{\mathfrak q'} \ar[d] \ar[r]_\beta & \mathcal{O}_{X', y'}^\wedge \ar[d] \\ \mathcal{O}_{X, y}^\wedge /\mathfrak p \ar[r] & (\mathcal{O}_{X, y}^\wedge /\mathfrak p)' \ar[r] & (\mathcal{O}_{X, y}^\wedge /\mathfrak p)'_{\mathfrak q'} \ar[r]^\gamma & ((\mathcal{O}_{X, y}^\wedge /\mathfrak p)'_{\mathfrak q'})^\wedge \ar[d] \\ & & & \mathcal{O}_{X', y'}^\wedge /\mathfrak p' } \]
whose vertical arrows are surjective. By More on Algebra, Lemma 15.43.1 and the dimension formula (Algebra, Lemma 10.113.1) we have
\[ \dim (((\mathcal{O}_{X, y}^\wedge /\mathfrak p)'_{\mathfrak q'})^\wedge ) = \dim ((\mathcal{O}_{X, y}^\wedge /\mathfrak p)'_{\mathfrak q'}) = \dim (\mathcal{O}_{X, y}^\wedge /\mathfrak p) - \text{trdeg}(\kappa (y')/\kappa (y)) \]
Tracing through the definitions of pullbacks, stalks, localizations, and completions we find
\[ (\mathcal{F}_ y^\wedge )_{\mathfrak p} \otimes _{(\mathcal{O}_{X, y}^\wedge )_\mathfrak p} (\mathcal{O}_{X', y'}^\wedge )_{\mathfrak p'} = (\mathcal{F}_{y'}^\wedge )_{\mathfrak p'} \]
Details omitted. The ring maps $\beta $ and $\gamma $ in the diagram are flat with Gorenstein (hence Cohen-Macaulay) fibres, as these are completions of rings having a dualizing complex. See Dualizing Complexes, Lemmas 47.23.1 and 47.23.2 and the discussion in More on Algebra, Section 15.51. Observe that $(\mathcal{O}_{X, y}^\wedge )_\mathfrak p = (\mathcal{O}_{X, y}^\wedge )'_{\tilde{\mathfrak p}}$ where $\tilde{\mathfrak p}$ is the kernel of $\alpha $ in the diagram. On the other hand, $(\mathcal{O}_{X, y}^\wedge )'_{\tilde{\mathfrak p}} \to (\mathcal{O}_{X', y'}^\wedge )_{\mathfrak p'}$ is flat with CM fibres by the above. Whence $(\mathcal{O}_{X, y}^\wedge )_\mathfrak p \to (\mathcal{O}_{X', y'}^\wedge )_{\mathfrak p'}$ is flat with CM fibres. Using Algebra, Lemma 10.163.1 we see that
\[ \text{depth}((\mathcal{F}_{y'}^\wedge )_{\mathfrak p'}) = \text{depth}((\mathcal{F}_ y^\wedge )_{\mathfrak p}) + \dim (F_\mathfrak r) \]
where $F$ is the generic formal fibre of $(\mathcal{O}_{X, y}^\wedge /\mathfrak p)'_{\mathfrak q'}$ and $\mathfrak r$ is the prime corresponding to $\mathfrak p'$. Since $(\mathcal{O}_{X, y}^\wedge /\mathfrak p)'_{\mathfrak q'}$ is a universally catenary local domain, its $I$-adic completion is equidimensional and (universally) catenary by Ratliff's theorem (More on Algebra, Proposition 15.109.5). It then follows that
\[ \dim (((\mathcal{O}_{X, y}^\wedge /\mathfrak p)'_{\mathfrak q'})^\wedge ) = \dim (F_\mathfrak r) + \dim (\mathcal{O}_{X', y'}^\wedge /\mathfrak p') \]
Combined with Lemma 52.18.2 we get
52.22.3.1
\begin{equation} \label{algebraization-equation-one} \begin{aligned} & \text{depth}((\mathcal{F}_{y'}^\wedge )_{\mathfrak p'}) + \delta ^{Y'}_{Z'}(y') \\ & = \text{depth}((\mathcal{F}_ y^\wedge )_{\mathfrak p}) + \dim (F_\mathfrak r) + \delta ^{Y'}_{Z'}(y') \\ & \geq \text{depth}((\mathcal{F}_ y^\wedge )_{\mathfrak p}) + \delta ^ Y_ Z(y) + \dim (F_\mathfrak r) + \text{trdeg}(\kappa (y')/\kappa (y)) - (d - 1) \\ & = \text{depth}((\mathcal{F}_ y^\wedge )_{\mathfrak p}) + \delta ^ Y_ Z(y) - (d - 1) + \dim (\mathcal{O}_{X, y}^\wedge /\mathfrak p) - \dim (\mathcal{O}_{X', y'}^\wedge /\mathfrak p') \end{aligned} \end{equation}
Please keep in mind that $\dim (\mathcal{O}_{X, y}^\wedge /\mathfrak p) \geq \dim (\mathcal{O}_{X', y'}^\wedge /\mathfrak p')$. Rewriting this we get
52.22.3.6
\begin{equation} \label{algebraization-equation-two} \begin{aligned} & \text{depth}((\mathcal{F}_{y'}^\wedge )_{\mathfrak p'}) + \dim (\mathcal{O}_{X', y'}^\wedge /\mathfrak p') + \delta ^{Y'}_{Z'}(y') \\ & \geq \text{depth}((\mathcal{F}_ y^\wedge )_{\mathfrak p}) + \dim (\mathcal{O}_{X, y}^\wedge /\mathfrak p) + \delta ^ Y_ Z(y) - (d - 1) \end{aligned} \end{equation}
This inequality will allow us to check the remaining conditions.
Conditions (b) and (d) of Lemma 52.21.2. Assume $V(\mathfrak p') \cap V(I\mathcal{O}_{X', y'}^\wedge ) = \{ \mathfrak m_{y'}^\wedge \} $. This implies that $\dim (\mathcal{O}_{X', y'}^\wedge /\mathfrak p') = 1$ because $Z'$ is an effective Cartier divisor. The combination of (b) and (d) is equivalent with
\[ \text{depth}((\mathcal{F}_{y'}^\wedge )_{\mathfrak p'}) + \delta ^{Y'}_{Z'}(y') > 2 \]
If $(\mathcal{F}_ n)$ satisfies the inequalities in (3)(b) then we immediately conclude this is true by applying (52.22.3.6). If $(\mathcal{F}_ n)$ satisfies (3)(a), i.e., the $(d + 1, d + 2)$-inequalities, then we see that in any case
\[ \text{depth}((\mathcal{F}_ y^\wedge )_{\mathfrak p}) + \delta ^ Y_ Z(y) \geq d + 1 \quad \text{or}\quad \text{depth}((\mathcal{F}_ y^\wedge )_{\mathfrak p}) + \dim (\mathcal{O}_{X, y}^\wedge /\mathfrak p) + \delta ^ Y_ Z(y) > d + 2 \]
Looking at (52.22.3.1) and (52.22.3.6) above this gives what we want except possibly if $\dim (\mathcal{O}_{X, y}^\wedge /\mathfrak p) = 1$. However, if $\dim (\mathcal{O}_{X, y}^\wedge /\mathfrak p) = 1$, then we have $V(\mathfrak p) \cap V(I\mathcal{O}_{X, y}^\wedge ) = \{ \mathfrak m_ y^\wedge \} $ and we see that actually
\[ \text{depth}((\mathcal{F}_ y^\wedge )_{\mathfrak p}) + \delta ^ Y_ Z(y) > d + 1 \]
as $(\mathcal{F}_ n)$ satisfies the $(d + 1, d + 2)$-inequalities and we conclude again.
Condition (c) of Lemma 52.21.2. Assume $V(\mathfrak p') \cap V(I\mathcal{O}_{X', y'}^\wedge ) \not= \{ \mathfrak m_{y'}^\wedge \} $. Then condition (c) is equivalent to
\[ \text{depth}((\mathcal{F}_{y'}^\wedge )_{\mathfrak p'}) + \delta ^{Y'}_{Z'}(y') \geq 2 \quad \text{or}\quad \text{depth}((\mathcal{F}_{y'}^\wedge )_{\mathfrak p'}) + \dim (\mathcal{O}_{X', y'}^\wedge /\mathfrak p') + \delta ^{Y'}_{Z'}(y') > 3 \]
If $(\mathcal{F}_ n)$ satisfies the inequalities in (3)(b) then we see the second of the two displayed inequalities holds true by applying (52.22.3.6). If $(\mathcal{F}_ n)$ satisfies (3)(a), i.e., the $(d + 1, d + 2)$-inequalities, then this follows immediately from (52.22.3.1) and (52.22.3.6). This finishes the proof of our claim.
Choose $(b|_{U'}^*\mathcal{F}_ n) \to (\mathcal{F}_ n'')$ and $(\mathcal{H}_ n)$ in $\textit{Coh}(U', I\mathcal{O}_{U'})$ as in Lemma 52.21.2. For any affine open $W \subset X'$ observe that $\delta ^{W \cap Y'}_{W \cap Z'}(y') \geq \delta ^{Y'}_{Z'}(y')$ by Lemma 52.18.1 part (7). Hence we see that $(\mathcal{H}_ n|_ W)$ satisfies the assumptions of Lemma 52.22.1. Thus $(\mathcal{H}_ n|_ W)$ extends canonically to $W$. Let $(\mathcal{G}_{W, n})$ in $\textit{Coh}(W, I\mathcal{O}_ W)$ be the canonical extension as in Lemma 52.16.8. By Lemma 52.16.9 we see that for $W' \subset W$ there is a unique isomorphism
\[ (\mathcal{G}_{W, n}|_{W'}) \longrightarrow (\mathcal{G}_{W', n}) \]
compatible with the given isomorphisms $(\mathcal{G}_{W, n}|_{W \cap U}) \cong (\mathcal{H}_ n|_{W \cap U})$. We conclude that there exists an object $(\mathcal{G}_ n)$ of $\textit{Coh}(X', I\mathcal{O}_{X'})$ whose restriction to $U$ is isomorphic to $(\mathcal{H}_ n)$.
If $A$ is $I$-adically complete we can finish the proof as follows. By Grothedieck's existence theorem (Cohomology of Schemes, Lemma 30.24.3) we see that $(\mathcal{G}_ n)$ is the completion of a coherent $\mathcal{O}_{X'}$-module. Then by Cohomology of Schemes, Lemma 30.23.6 we see that $(b|_{U'}^*\mathcal{F}_ n)$ is the completion of a coherent $\mathcal{O}_{U'}$-module $\mathcal{F}'$. By Cohomology of Schemes, Lemma 30.25.3 we see that there is a map
\[ (\mathcal{F}_ n) \longrightarrow ((b|_{U'})_*\mathcal{F}')^\wedge \]
whose kernel and cokernel is annihilated by a power of $I$. Then finally, we win by applying Lemma 52.17.1.
If $A$ is not complete, then, before starting the proof, we may replace $A$ by its completion, see Lemma 52.16.6. After completion the assumptions still hold: this is immediate for condition (3), follows from Dualizing Complexes, Lemma 47.22.4 for condition (1), and from Divisors, Lemma 31.32.3 for condition (2). Thus the complete case implies the general case.
$\square$
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