hcju/stacksqa huggingface dataset:
Let $Y = V(I)$ and $Y' = V(I_\mathfrak p) \subset \Spec(A_\mathfrak p)$. Recall that $R\Gamma_Y(A) \otimes_A A_\mathfrak p = R\Gamma_{Y'}(A_\mathfrak p)$ by Dualizing Complexes, Lemma \ref{dualizing-lemma-torsion-change-rings}. Thus we conclude by Algebra, Lemma \ref{algebra-lemma-characterize-zero-local}.
We first prove the second statement. Recall that $\dim(A)$ denotes the Krull dimension. By Lemma \ref{lemma-cd-local} we may assume $A$ is local. If $V(I) = \emptyset$, then the result is true. If $V(I) \not = \emptyset$, then $\dim(\Spec(A) \setminus V(I)) < \dim(A)$ because the closed point is missing. Observe that $U = \Spec(A) \setminus V(I)$ is a quasi-compact open of the spectral space $\Spec(A)$, hence a spectral space itself. See Algebra, Lemma \ref{algebra-lemma-spec-spectral} and Topology, Lemma \ref{topology-lemma-spectral-sub}. Thus Cohomology, Proposition \ref{cohomology-proposition-cohomological-dimension-spectral} implies $H^i(U, \mathcal{F}) = 0$ for $i \geq \dim(A)$ which implies what we want by Lemma \ref{lemma-cd}. In the Noetherian case the reader may use Grothendieck's Cohomology, Proposition \ref{cohomology-proposition-vanishing-Noetherian}. \medskip\noindent We will deduce the first statement from the second. Let $\mathfrak a$ be the annihilator of the finite $A$-module $M$. Set $B = A/\mathfrak a$. Recall that $\Spec(B) = \text{Supp}(M)$, see Algebra, Lemma \ref{algebra-lemma-support-closed}. Set $J = IB$. Then $M$ is a $B$-module and $H^i_{V(I)}(M) = H^i_{V(J)}(M)$, see Dualizing Complexes, Lemma \ref{dualizing-lemma-local-cohomology-and-restriction}. Since $\text{cd}(B, J) \leq \dim(B) = \dim(\text{Supp}(M))$ by the first part we conclude.
By one of the characterizations of dimension, there exists an ideal of definition for $A$ generated by $d$ elements, see Algebra, Proposition \ref{algebra-proposition-dimension}. Hence $\text{cd}(A, \mathfrak m) \leq d$ by Lemma \ref{lemma-bound-cd}. Thus $H^d_\mathfrak m(A)$ is nonzero if and only if $\text{cd}(A, \mathfrak m) = d$ if and only if $\text{cd}(A, \mathfrak m) \geq d$. \medskip\noindent Let $A \to A^\wedge$ be the map from $A$ to its completion. Observe that $A^\wedge$ is a Noetherian local ring of the same dimension as $A$ with maximal ideal $\mathfrak m A^\wedge$. See Algebra, Lemmas \ref{algebra-lemma-completion-Noetherian-Noetherian}, \ref{algebra-lemma-completion-complete}, and \ref{algebra-lemma-completion-faithfully-flat} and More on Algebra, Lemma \ref{more-algebra-lemma-completion-dimension}. By Lemma \ref{lemma-cd-change-rings} it suffices to prove the lemma for $A^\wedge$. \medskip\noindent By the previous paragraph we may assume that $A$ is a complete local ring. Then $A$ has a normalized dualizing complex $\omega_A^\bullet$ (Dualizing Complexes, Lemma \ref{dualizing-lemma-ubiquity-dualizing}). The local duality theorem (in the form of Dualizing Complexes, Lemma \ref{dualizing-lemma-special-case-local-duality}) tells us $H^d_\mathfrak m(A)$ is Matlis dual to $\text{Ext}^{-d}(A, \omega_A^\bullet) = H^{-d}(\omega_A^\bullet)$ which is nonzero for example by Dualizing Complexes, Lemma \ref{dualizing-lemma-nonvanishing-generically-local}.
By Lemma \ref{lemma-cd-change-rings} we have $\text{cd}(A, I) \geq \text{cd}(A/\mathfrak p, I(A/\mathfrak p))$. Since $V(I) \cap V(\mathfrak p) = \{\mathfrak m\}$ we have $\text{cd}(A/\mathfrak p, I(A/\mathfrak p)) = \text{cd}(A/\mathfrak p, \mathfrak m/\mathfrak p)$. By Lemma \ref{lemma-cd-maximal} this is equal to $\dim(A/\mathfrak p)$.
Set $U = X \setminus V(I)$. Denote $j : U \to X'$ the canonical open immersion, see Divisors, Section \ref{divisors-section-blowing-up}. Since the exceptional divisor is an effective Cartier divisor (Divisors, Lemma \ref{divisors-lemma-blowing-up-gives-effective-Cartier-divisor}) we see that $j$ is affine, see Divisors, Lemma \ref{divisors-lemma-complement-locally-principal-closed-subscheme}. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_U$-module. Then $R^pj_*\mathcal{F} = 0$ for $p > 0$, see Cohomology of Schemes, Lemma \ref{coherent-lemma-relative-affine-vanishing}. On the other hand, we have $R^qb_*(j_*\mathcal{F}) = 0$ for $q \geq d$ by Limits, Lemma \ref{limits-lemma-higher-direct-images-zero-above-dimension-fibre}. Thus by the Leray spectral sequence (Cohomology, Lemma \ref{cohomology-lemma-relative-Leray}) we conclude that $R^n(b \circ j)_*\mathcal{F} = 0$ for $n \geq d$. Thus $H^n(U, \mathcal{F}) = 0$ for $n \geq d$ (by Cohomology, Lemma \ref{cohomology-lemma-apply-Leray}). This means that $\text{cd}(A, I) \leq d$ by definition.
The equivalence holds because the support of an element of $M$ is contained in the support of $M$ and conversely the support of $M$ is the union of the supports of its elements. The category of these modules is a Serre subcategory (Homology, Definition \ref{homology-definition-serre-subcategory}) of $\text{Mod}_A$ by Algebra, Lemma \ref{algebra-lemma-support-quotient}. We omit the proof of the statement on direct sums.
Let $M$ be an object of $\text{Mod}_{A, T}$. Choose an embedding $M \to J$ into an injective $A$-module. By Dualizing Complexes, Proposition \ref{dualizing-proposition-structure-injectives-noetherian} the module $J$ is a direct sum of injective hulls of residue fields. Let $E$ be an injective hull of the residue field of $\mathfrak p$. Since $E$ is $\mathfrak p$-power torsion we see that $H^0_T(E) = 0$ if $\mathfrak p \not \in T$ and $H^0_T(E) = E$ if $\mathfrak p \in T$. Thus $H^0_T(J)$ is injective as a direct sum of injective hulls (by the proposition) and we have an embedding $M \to H^0_T(J)$. Thus every object $M$ of $\text{Mod}_{A, T}$ has an injective resolution $M \to J^\bullet$ with $J^n$ also in $\text{Mod}_{A, T}$. It follows that $RH^0_T(M) = M$. \medskip\noindent Next, suppose that $K \in D_T^+(A)$. Then the spectral sequence $$ R^qH^0_T(H^p(K)) \Rightarrow R^{p + q}H^0_T(K) $$ (Derived Categories, Lemma \ref{derived-lemma-two-ss-complex-functor}) converges and above we have seen that only the terms with $q = 0$ are nonzero. Thus we see that $RH^0_T(K) \to K$ is an isomorphism. Thus the functor $D^+(\text{Mod}_{A, T}) \to D^+_T(A)$ is an equivalence with quasi-inverse given by $RH^0_T$.
Say $\dim(A) = d$. Then we see that $H^i_Z(M) = 0$ for $i > d$ for every closed subset $Z$ of $\Spec(A)$, see Lemma \ref{lemma-cd-dimension}. By Lemma \ref{lemma-adjoint-ext} we find that $H^0_T$ has bounded cohomological dimension. \medskip\noindent Let $K \in D_T(A)$. We claim that $RH^0_T(K) \to K$ is an isomorphism. We know this is true when $K$ is bounded below, see Lemma \ref{lemma-equal-plus}. However, since $H^0_T$ has bounded cohomological dimension, we see that the $i$th cohomology of $RH_T^0(K)$ only depends on $\tau_{\geq -d + i}K$ and we conclude. Thus $D(\text{Mod}_{A, T}) \to D_T(A)$ is an equivalence with quasi-inverse $RH^0_T$.
From the map $R\Gamma_T(K) \to K$ we get a map $R\Gamma_T(K) \otimes_A^\mathbf{L} B \to K \otimes_A^\mathbf{L} B$. The cohomology modules of $R\Gamma_T(K) \otimes_A^\mathbf{L} B$ are supported on $T'$ and hence we get the arrow of the lemma. This arrow is an isomorphism if $T$ is a closed subset of $\Spec(A)$ by Dualizing Complexes, Lemma \ref{dualizing-lemma-torsion-change-rings}. Recall that $H^i_T(K)$ is the colimit of $H^i_Z(K)$ where $Z$ runs over the (directed set of) closed subsets of $T$, see Lemma \ref{lemma-adjoint-ext}. Correspondingly $H^i_{T'}(K \otimes_A^\mathbf{L} B) = \colim H^i_{Z'}(K \otimes_A^\mathbf{L} B)$ where $Z'$ is the inverse image of $Z$. Thus the result because $\otimes_A B$ commutes with filtered colimits and there are no higher Tors.
By construction we may represent $R\Gamma_T(A)$ by a complex $J^\bullet$ in $\text{Mod}_{A, T}$. Thus if we represent $K$ by a K-flat complex $K^\bullet$ then we see that $R\Gamma_T(A) \otimes_A^\mathbf{L} K$ is represented by the complex $\text{Tot}(J^\bullet \otimes_A K^\bullet)$ in $\text{Mod}_{A, T}$. Using the map $R\Gamma_T(A) \to A$ we obtain a map $R\Gamma_T(A) \otimes_A^\mathbf{L} K\to K$. Thus by the adjointness property of $R\Gamma_T$ we obtain a canonical map $$ R\Gamma_T(A) \otimes_A^\mathbf{L} K \longrightarrow R\Gamma_T(K) $$ factoring the just constructed map. Observe that $R\Gamma_T$ commutes with direct sums in $D(A)$ for example by Lemma \ref{lemma-adjoint-ext}, the fact that directed colimits commute with direct sums, and the fact that usual local cohomology commutes with direct sums (for example by Dualizing Complexes, Lemma \ref{dualizing-lemma-local-cohomology-adjoint}). Thus by More on Algebra, Remark \ref{more-algebra-remark-P-resolution} it suffices to check the map is an isomorphism for $K = A[k]$ where $k \in \mathbf{Z}$. This is clear. \medskip\noindent The final statements follow from the result we've just shown by transitivity of derived tensor products.
For every $Z$ and $f$ the spectral sequence of Dualizing Complexes, Lemma \ref{dualizing-lemma-local-cohomology-ss} degenerates to give short exact sequences $$ 0 \to H^1_f(H^{i - 1}_Z(M)) \to H^i_{Z \cap V(f)}(M) \to H^0_f(H^i_Z(M)) \to 0 $$ We will use this without further mention below. \medskip\noindent Let $\xi \in H^i_T(M)$ map to zero in the direct sum. Then we first write $\xi$ as the image of some $\xi' \in H^i_Z(M)$ for some closed subset $Z \subset T$, see Lemma \ref{lemma-adjoint-ext}. Then $\xi'$ maps to zero in $H^i_{\mathfrak p A_\mathfrak p}(M_\mathfrak p)$ for every $\mathfrak p \in Z$, $\mathfrak p \not \in T'$. Since there are finitely many of these primes, we may choose $f \in A$ not contained in any of these such that $f$ annihilates $\xi'$. Then $\xi'$ is the image of some $\xi'' \in H^i_{Z'}(M)$ where $Z' = Z \cap V(f)$. By our choice of $f$ we have $Z' \subset T'$ and we get exactness at the penultimate spot. \medskip\noindent Let $\xi \in H^i_{T'}(M)$ map to zero in $H^i_T(M)$. Choose closed subsets $Z' \subset Z$ with $Z' \subset T'$ and $Z \subset T$ such that $\xi$ comes from $\xi' \in H^i_{Z'}(M)$ and maps to zero in $H^i_Z(M)$. Then we can find $f \in A$ with $V(f) \cap Z = Z'$ and we conclude.
We first make a general remark: suppose we have an exact sequence $$ (A_n) \to (B_n) \to (C_n) $$ of inverse systems of abelian groups. Suppose that for every $m$ there exists an integer $m'(m) \geq m$ such that $$ A_k \to A_m \quad\text{and}\quad C_k \to C_m $$ are zero for $k \geq m'(m)$. Then for $k \geq m'(m'(m))$ the map $B_k \to B_m$ is zero. \medskip\noindent We will prove the lemma by induction on $\dim(T)$ which is finite because $\dim(A)$ is finite. Let $T' \subset T$ be the set of nonminimal primes in $T$. Then $T'$ is a subset of $\Spec(A)$ stable under specialization and the hypotheses of the lemma apply to $T'$. Since $\dim(T') < \dim(T)$ we know the lemma holds for $T'$. For every $A$-module $M$ there is an exact sequence $$ H^i_{T'}(M) \to H^i_T(M) \to \bigoplus\nolimits_{\mathfrak p \in T \setminus T'} H^i_{\mathfrak p A_\mathfrak p}(M_\mathfrak p) $$ by Lemma \ref{lemma-filter-local-cohomology}. Thus we conclude by the initial remark of the proof.
We first make a general remark: suppose we have an exact sequence $$ (A_n) \to (B_n) \to (C_n) \to (D_n) $$ of inverse systems of abelian groups. Suppose that there exists an integer $m_0$ such that for every $m \geq m_0$ there exists an integer $m'(m) \geq m$ such that the maps $$ \Im(B_k \to B_m) \longrightarrow B_{m_0} \quad\text{and}\quad \Im(D_k \to D_m) \longrightarrow D_{m_0} $$ are injective for $k \geq m'(m)$ and $A_k \to A_m$ is zero for $k \geq m'(m)$. Then for $m \geq m'(m_0)$ and $k \geq m'(m'(m))$ the map $$ \Im(C_k \to C_m) \to C_{m'(m_0)} $$ is injective. Namely, let $c_0 \in C_m$ be the image of $c_3 \in C_k$ and say $c_0$ maps to zero in $C_{m'(m_0)}$. Picture $$ C_k \to C_{m'(m'(m))} \to C_{m'(m)} \to C_m \to C_{m'(m_0)},\quad c_3 \mapsto c_2 \mapsto c_1 \mapsto c_0 \mapsto 0 $$ We have to show $c_0 = 0$. The image $d_3$ of $c_3$ maps to zero in $C_{m_0}$ and hence we see that the image $d_1 \in D_{m'(m)}$ is zero. Thus we can choose $b_1 \in B_{m'(m)}$ mapping to the image $c_1$. Since $c_3$ maps to zero in $C_{m'(m_0)}$ we find an element $a_{-1} \in A_{m'(m_0)}$ which maps to the image $b_{-1} \in B_{m'(m_0)}$ of $b_1$. Since $a_{-1}$ maps to zero in $A_{m_0}$ we conclude that $b_1$ maps to zero in $B_{m_0}$. Thus the image $b_0 \in B_m$ is zero which of course implies $c_0 = 0$ as desired. \medskip\noindent We will prove the lemma by induction on $\dim(T)$ which is finite because $\dim(A)$ is finite. Let $T' \subset T$ be the set of nonminimal primes in $T$. Then $T'$ is a subset of $\Spec(A)$ stable under specialization and the hypotheses of the lemma apply to $T'$. Since $\dim(T') < \dim(T)$ we know the lemma holds for $T'$. For every $A$-module $M$ there is an exact sequence $$ 0 \to \colim_{Z, f} H^1_f(H^{i - 1}_Z(M)) \to H^i_{T'}(M) \to H^i_T(M) \to \bigoplus\nolimits_{\mathfrak p \in T \setminus T'} H^i_{\mathfrak p A_\mathfrak p}(M_\mathfrak p) $$ by Lemma \ref{lemma-filter-local-cohomology}. Thus we conclude by the initial remark of the proof and the fact that we've seen the system of groups $$ \left\{\colim_{Z, f} H^1_f(H^{i - 1}_Z(M_n))\right\}_{n \geq 0} $$ is pro-zero in Lemma \ref{lemma-zero}; this uses that the function $m''(m)$ in that lemma for $H^{i - 1}_Z(M)$ is independent of $Z$.
We prove the equivalence of (1) and (2) by induction on $n$. For $n = 0$ we have $H^0_T(M) \subset M$ is finite. Hence (1) is true. Since $H^0_T(M) = \colim H^0_{V(J)}(M)$ with $J$ as in (2) we see that (2) is true. Assume that $n > 0$. \medskip\noindent Assume (1) is true. Recall that $H^i_J(M) = H^i_{V(J)}(M)$, see Dualizing Complexes, Lemma \ref{dualizing-lemma-local-cohomology-noetherian}. Thus $H^i_T(M) = \colim H^i_J(M)$ where the colimit is over ideals $J \subset A$ with $V(J) \subset T$, see Lemma \ref{lemma-adjoint-ext}. Since $H^i_T(M)$ is finitely generated for $i \leq n$ we can find a $J \subset A$ as in (2) such that $H^i_J(M) \to H^i_T(M)$ is surjective for $i \leq n$. Thus the finite list of generators are $J$-power torsion elements and we see that (2) holds with $J$ replaced by some power. \medskip\noindent Assume we have $J$ as in (2). Let $N = H^0_T(M)$ and $M' = M/N$. By construction of $R\Gamma_T$ we find that $H^i_T(N) = 0$ for $i > 0$ and $H^0_T(N) = N$, see Remark \ref{remark-upshot}. Thus we find that $H^0_T(M') = 0$ and $H^i_T(M') = H^i_T(M)$ for $i > 0$. We conclude that we may replace $M$ by $M'$. Thus we may assume that $H^0_T(M) = 0$. This means that the finite set of associated primes of $M$ are not in $T$. By prime avoidance (Algebra, Lemma \ref{algebra-lemma-silly}) we can find $f \in J$ not contained in any of the associated primes of $M$. Then the long exact local cohomology sequence associated to the short exact sequence $$ 0 \to M \to M \to M/fM \to 0 $$ turns into short exact sequences $$ 0 \to H^i_T(M) \to H^i_T(M/fM) \to H^{i + 1}_T(M) \to 0 $$ for $i < n$. We conclude that $J^2$ annihilates $H^i_T(M/fM)$ for $i < n$. By induction hypothesis we see that $H^i_T(M/fM)$ is finite for $i < n$. Using the short exact sequence once more we see that $H^{i + 1}_T(M)$ is finite for $i < n$ as desired. \medskip\noindent We omit the proof of the equivalence of (2) and (3) in case $T = V(I)$.
The implication (1) $\Rightarrow$ (2) is immediate. We prove the converse by induction on $n$. The case $n = 0$ is clear because both (1) and (2) are always true in that case. \medskip\noindent Assume $n > 0$ and that (2) is true. Let $N = H^0_Z(M)$ and $M' = M/N$. By Dualizing Complexes, Lemma \ref{dualizing-lemma-divide-by-torsion} we may replace $M$ by $M'$. Thus we may assume that $H^0_Z(M) = 0$. This means that $\text{depth}_I(M) > 0$ (Dualizing Complexes, Lemma \ref{dualizing-lemma-depth}). Pick $f \in I$ a nonzerodivisor on $M$ and consider the short exact sequence $$ 0 \to M \to M \to M/fM \to 0 $$ which produces a long exact sequence $$ 0 \to H^0_Z(M/fM) \to H^1_Z(M) \to H^1_Z(M) \to H^1_Z(M/fM) \to H^2_Z(M) \to \ldots $$ and similarly after localization. Thus assumption (2) implies that the modules $H^i_Z(M/fM)_\mathfrak p$ are finite for $i < n$. Hence by induction assumption $H^i_Z(M/fM)$ are finite for $i < n$. \medskip\noindent Let $\mathfrak p$ be a prime of $A$ which is associated to $H^i_Z(M)$ for some $i \leq n$. Say $\mathfrak p$ is the annihilator of the element $x \in H^i_Z(M)$. Then $\mathfrak p \in Z$, hence $f \in \mathfrak p$. Thus $fx = 0$ and hence $x$ comes from an element of $H^{i - 1}_Z(M/fM)$ by the boundary map $\delta$ in the long exact sequence above. It follows that $\mathfrak p$ is an associated prime of the finite module $\Im(\delta)$. We conclude that $\text{Ass}(H^i_Z(M))$ is finite for $i \leq n$, see Algebra, Lemma \ref{algebra-lemma-finite-ass}. \medskip\noindent Recall that $$ H^i_Z(M) \subset \prod\nolimits_{\mathfrak p \in \text{Ass}(H^i_Z(M))} H^i_Z(M)_\mathfrak p $$ by Algebra, Lemma \ref{algebra-lemma-zero-at-ass-zero}. Since by assumption the modules on the right hand side are finite and $I$-power torsion, we can find integers $e_{\mathfrak p, i} \geq 0$, $i \leq n$, $\mathfrak p \in \text{Ass}(H^i_Z(M))$ such that $I^{e_{\mathfrak p, i}}$ annihilates $H^i_Z(M)_\mathfrak p$. We conclude that $I^e$ with $e = \max\{e_{\mathfrak p, i}\}$ annihilates $H^i_Z(M)$ for $i \leq n$. By Lemma \ref{lemma-check-finiteness-local-cohomology-by-annihilator} we see that $H^i_Z(M)$ is finite for $i \leq n$.
The assumption that $H^i_Z(A)$ is finite for $0 \leq i \leq n$ implies there exists an $e \geq 0$ such that $I^e$ annihilates $H^i_Z(A)$ for $0 \leq i \leq n$, see Lemma \ref{lemma-check-finiteness-local-cohomology-by-annihilator}. Then Lemma \ref{lemma-annihilate-local-cohomology} implies that $H^i_Z(M)$, $0 \leq i \leq n$ is annihilated by $I^m$ for some $m = m(M, i)$. We may take the same $m$ for all $0 \leq i \leq n$. Then Lemma \ref{lemma-check-finiteness-local-cohomology-by-annihilator} implies that $H^i_Z(M)$ is finite for $0 \leq i \leq n$ as desired.
By Properties, Lemma \ref{properties-lemma-lift-finite-presentation} there exists a coherent $\mathcal{O}_X$-module $\mathcal{F}'$ whose restriction to $U$ is isomorphic to $\mathcal{F}$. Say $\mathcal{F}'$ corresponds to the finite $A$-module $M$ as in (1). Note that $R^pj_*\mathcal{F}$ is quasi-coherent (Cohomology of Schemes, Lemma \ref{coherent-lemma-quasi-coherence-higher-direct-images}) and corresponds to the $A$-module $H^p(U, \mathcal{F})$. By Lemma \ref{lemma-local-cohomology-is-local-cohomology} and the discussion in Cohomology, Sections \ref{cohomology-section-cohomology-support} and \ref{cohomology-section-cohomology-support-bis} we obtain an exact sequence $$ 0 \to H^0_Z(M) \to M \to H^0(U, \mathcal{F}) \to H^1_Z(M) \to 0 $$ and isomorphisms $H^p(U, \mathcal{F}) = H^{p + 1}_Z(M)$ for $p \geq 1$. Here we use that $H^j(X, \mathcal{F}') = 0$ for $j > 0$ as $X$ is affine and $\mathcal{F}'$ is quasi-coherent (Cohomology of Schemes, Lemma \ref{coherent-lemma-quasi-coherent-affine-cohomology-zero}). This proves (2). Parts (3) and (4) are straightforward from (2); see also Lemma \ref{lemma-local-cohomology}.
By Lemma \ref{lemma-check-finiteness-pushforward-on-associated-points} it suffices to prove $j_*i_{x, *}\mathcal{O}_{W_x}$ is coherent for $x \in \text{Ass}(\mathcal{F})$. Let $\pi : Y \to X$ be the normalization of $X$ in $\Spec(\kappa(x))$, see Morphisms, Section \ref{morphisms-section-normalization}. By Morphisms, Lemma \ref{morphisms-lemma-nagata-normalization-finite-general} the morphism $\pi$ is finite. Since $\pi$ is finite $\mathcal{G} = \pi_*\mathcal{O}_Y$ is a coherent $\mathcal{O}_X$-module by Cohomology of Schemes, Lemma \ref{coherent-lemma-finite-pushforward-coherent}. Observe that $W_x = U \cap \pi(Y)$. Thus $\pi|_{\pi^{-1}(U)} : \pi^{-1}(U) \to U$ factors through $i_x : W_x \to U$ and we obtain a canonical map $$ i_{x, *}\mathcal{O}_{W_x} \longrightarrow (\pi|_{\pi^{-1}(U)})_*(\mathcal{O}_{\pi^{-1}(U)}) = (\pi_*\mathcal{O}_Y)|_U = \mathcal{G}|_U $$ This map is injective (for example by Divisors, Lemma \ref{divisors-lemma-check-injective-on-ass}). Hence $j_*i_{x, *}\mathcal{O}_{W_x} \subset j_*\mathcal{G}|_U$ and it suffices to show that $j_*\mathcal{G}|_U$ is coherent. \medskip\noindent It remains to prove that $j_*(\mathcal{G}|_U)$ is coherent. We claim Divisors, Lemma \ref{divisors-lemma-depth-2-hartog} applies to $$ \mathcal{G} \longrightarrow j_*(\mathcal{G}|_U) $$ which finishes the proof. It suffices to show that $\text{depth}(\mathcal{G}_z) \geq 2$ for $z \in Z$. Let $y_1, \ldots, y_n \in Y$ be the points mapping to $z$. By Algebra, Lemma \ref{algebra-lemma-depth-goes-down-finite} it suffices to show that $\text{depth}(\mathcal{O}_{Y, y_i}) \geq 2$ for $i = 1, \ldots, n$. If not, then by Properties, Lemma \ref{properties-lemma-criterion-normal} we see that $\dim(\mathcal{O}_{Y, y_i}) = 1$ for some $i$. This is impossible by the dimension formula (Morphisms, Lemma \ref{morphisms-lemma-dimension-formula}) for $\pi : Y \to \overline{\{x\}}$ and assumption (3).
We may assume $X$ is affine. Using Lemmas \ref{lemma-check-finiteness-local-cohomology-locally} and \ref{lemma-finiteness-pushforwards-and-H1-local} we reduce to $X = \Spec(A)$ where $(A, \mathfrak m)$ is a Noetherian local domain and $\mathfrak m \in Z$. Then we can use induction on $d = \dim(A)$. (The base case is $d = 0, 1$ which do not happen by our assumption on the local rings.) Set $V = \Spec(A) \setminus \{\mathfrak m\}$. Observe that the local rings of $V$ have dimension strictly smaller than $d$. Repeating the arguments for $j' : U \to V$ we and using induction we conclude that $j'_*\mathcal{O}_U$ is a coherent $\mathcal{O}_V$-module. Pick a nonzero $f \in A$ which vanishes on $Z$. Since $D(f) \cap V \subset U$ we find an $n$ such that multiplication by $f^n$ on $U$ extends to a map $f^n : j'_*\mathcal{O}_U \to \mathcal{O}_V$ over $V$ (for example by Cohomology of Schemes, Lemma \ref{coherent-lemma-homs-over-open}). This map is injective hence there is an injective map $$ j_*\mathcal{O}_U = j''_* j'_* \mathcal{O}_U \to j''_*\mathcal{O}_V $$ on $X$ where $j'' : V \to X$ is the inclusion morphism. Hence it suffices to show that $j''_*\mathcal{O}_V$ is coherent. In other words, we may assume that $X$ is the spectrum of a local Noetherian domain and that $Z$ consists of the closed point. \medskip\noindent Assume $X = \Spec(A)$ with $(A, \mathfrak m)$ local and $Z = \{\mathfrak m\}$. Let $A^\wedge$ be the completion of $A$. Set $X^\wedge = \Spec(A^\wedge)$, $Z^\wedge = \{\mathfrak m^\wedge\}$, $U^\wedge = X^\wedge \setminus Z^\wedge$, and $\mathcal{F}^\wedge = \mathcal{O}_{U^\wedge}$. The ring $A^\wedge$ is universally catenary and Nagata (Algebra, Remark \ref{algebra-remark-Noetherian-complete-local-ring-universally-catenary} and Lemma \ref{algebra-lemma-Noetherian-complete-local-Nagata}). Moreover, condition (3) of Lemma \ref{lemma-finiteness-pushforward} for $X^\wedge, Z^\wedge, U^\wedge, \mathcal{F}^\wedge$ holds by assumption! Thus we see that $(U^\wedge \to X^\wedge)_*\mathcal{O}_{U^\wedge}$ is coherent. Since the morphism $c : X^\wedge \to X$ is flat we conclude that the pullback of $j_*\mathcal{O}_U$ is $(U^\wedge \to X^\wedge)_*\mathcal{O}_{U^\wedge}$ (Cohomology of Schemes, Lemma \ref{coherent-lemma-flat-base-change-cohomology}). Finally, since $c$ is faithfully flat we conclude that $j_*\mathcal{O}_U$ is coherent by Descent, Lemma \ref{descent-lemma-finite-type-descends}.