hcju/stacksqa huggingface dataset:
Assume (1). By definition this means there exists an open covering $X = \bigcup U_i$ such that $E|_{U_i}$ is represented by a strictly perfect complex. Thus $E$ is pseudo-coherent (i.e., $m$-pseudo-coherent for all $m$) by Lemma \ref{lemma-pseudo-coherent-independent-representative}. Moreover, a direct summand of a finite free module is flat, hence $E|_{U_i}$ has finite Tor dimension by Lemma \ref{lemma-tor-amplitude}. Thus (2) holds. \medskip\noindent Assume (2). After replacing $X$ by the members of an open covering we may assume there exist integers $a \leq b$ such that $E$ has tor amplitude in $[a, b]$. Since $E$ is $m$-pseudo-coherent for all $m$ we conclude using Lemma \ref{lemma-perfect-precise}.
First proof: Combine Lemmas \ref{lemma-perfect}, \ref{lemma-cone-pseudo-coherent}, and \ref{lemma-cone-tor-amplitude}. Second proof (sketch): Say $K$ and $L$ are perfect. After replacing $X$ by the members of an open covering we may assume that $K$ and $L$ are represented by strictly perfect complexes $\mathcal{K}^\bullet$ and $\mathcal{L}^\bullet$. After replacing $X$ by the members of an open covering we may assume the map $K \to L$ is given by a map of complexes $\alpha : \mathcal{K}^\bullet \to \mathcal{L}^\bullet$, see Lemma \ref{lemma-local-actual}. Then $M$ is isomorphic to the cone of $\alpha$ which is strictly perfect by Lemma \ref{lemma-cone}.
Omitted. Hints: as unit $\mathbf{1}$ we take the complex having $\mathcal{O}_X$ in degree $0$ and zero in other degrees with obvious isomorphisms $\text{Tot}(\mathbf{1} \otimes_{\mathcal{O}_X} \mathcal{G}^\bullet) = \mathcal{G}^\bullet$ and $\text{Tot}(\mathcal{F}^\bullet \otimes_{\mathcal{O}_X} \mathbf{1}) = \mathcal{F}^\bullet$. to prove the lemma you have to check the commutativity of various diagrams, see Categories, Definitions \ref{categories-definition-monoidal-category} and \ref{categories-definition-symmetric-monoidal-category}. The verifications are straightforward in each case.
By uniqueness of left duals (Categories, Remark \ref{categories-remark-left-dual-adjoint}) we get the final statement provided we show that $\mathcal{F}^\bullet$ is as stated. Let $\mathcal{G}^\bullet, \eta, \epsilon$ be a left dual. Write $\eta = \sum \eta_n$ and $\epsilon = \sum \epsilon_n$ where $\eta_n : \mathcal{O}_X \to \mathcal{F}^n \otimes_{\mathcal{O}_X} \mathcal{G}^{-n}$ and $\epsilon_n : \mathcal{G}^{-n} \otimes_{\mathcal{O}_X} \mathcal{F}^n \to \mathcal{O}_X$. Since $(1 \otimes \epsilon) \circ (\eta \otimes 1) = \text{id}_{\mathcal{F}^\bullet}$ and $(\epsilon \otimes 1) \circ (1 \otimes \eta) = \text{id}_{\mathcal{G}^\bullet}$ by Categories, Definition \ref{categories-definition-dual} we see immediately that we have $(1 \otimes \epsilon_n) \circ (\eta_n \otimes 1) = \text{id}_{\mathcal{F}^n}$ and $(\epsilon_n \otimes 1) \circ (1 \otimes \eta_n) = \text{id}_{\mathcal{G}^{-n}}$. Hence we see that $\mathcal{F}^n$ is locally a direct summand of a finite free $\mathcal{O}_X$-module by Modules, Lemma \ref{modules-lemma-left-dual-module}. Since the sum $\eta = \sum \eta_n$ is locally finite, we conclude that $\mathcal{F}^\bullet$ is locally bounded.
Since the map is globally defined and since formation of the right and left hand side commute with localization (see Lemma \ref{lemma-restriction-RHom-to-U}), to prove this we may work locally on $X$. Thus we may assume $K$ is represented by a strictly perfect complex $\mathcal{E}^\bullet$. \medskip\noindent If $K_1 \to K_2 \to K_3$ is a distinguished triangle in $D(\mathcal{O}_X)$, then we get distinguished triangles $$ R\SheafHom(L, M) \otimes_{\mathcal{O}_X}^\mathbf{L} K_1 \to R\SheafHom(L, M) \otimes_{\mathcal{O}_X}^\mathbf{L} K_2 \to R\SheafHom(L, M) \otimes_{\mathcal{O}_X}^\mathbf{L} K_3 $$ and $$ R\SheafHom(R\SheafHom(K_1, L), M) \to R\SheafHom(R\SheafHom(K_2, L), M) R\SheafHom(R\SheafHom(K_3, L), M) $$ See Section \ref{section-flat} and Lemma \ref{lemma-RHom-triangulated}. The arrow of Lemma \ref{lemma-internal-hom-evaluate} is functorial in $K$ hence we get a morphism between these distinguished triangles. Thus, if the result holds for $K_1$ and $K_3$, then the result holds for $K_2$ by Derived Categories, Lemma \ref{derived-lemma-third-isomorphism-triangle}. \medskip\noindent Combining the remarks above with the distinguished triangles $$ \sigma_{\geq n}\mathcal{E}^\bullet \to \mathcal{E}^\bullet \to \sigma_{\leq n - 1}\mathcal{E}^\bullet $$ of stupid trunctions, we reduce to the case where $K$ consists of a direct summand of a finite free $\mathcal{O}_X$-module placed in some degree. By an obvious compatibility of the problem with direct sums (similar to what was said above) and shifts this reduces us to the case where $K = \mathcal{O}_X^{\oplus n}$ for some integer $n$. This case is clear.
By Lemma \ref{lemma-internal-hom-evaluate} there is a canonical map $$ K = R\SheafHom(\mathcal{O}_X, \mathcal{O}_X) \otimes_{\mathcal{O}_X}^\mathbf{L} K \longrightarrow R\SheafHom(R\SheafHom(K, \mathcal{O}_X), \mathcal{O}_X) = (K^\vee)^\vee $$ which is an isomorphism by Lemma \ref{lemma-internal-hom-evaluate-isom}. To check the other statements we will use without further mention that formation of internal hom commutes with restriction to opens (Lemma \ref{lemma-restriction-RHom-to-U}). We may check $K^\vee$ is perfect locally on $X$. By Lemma \ref{lemma-dual} to see the final statement it suffices to check that the map (\ref{equation-eval}) $$ M \otimes^\mathbf{L}_{\mathcal{O}_X} K^\vee \longrightarrow R\SheafHom(K, M) $$ is an isomorphism. This is local on $X$ as well. Hence it suffices to prove these two statements $K$ is represented by a strictly perfect complex. \medskip\noindent Assume $K$ is represented by the strictly perfect complex $\mathcal{E}^\bullet$. Then it follows from Lemma \ref{lemma-Rhom-strictly-perfect} that $K^\vee$ is represented by the complex whose terms are $(\mathcal{E}^{-n})^\vee = \SheafHom_{\mathcal{O}_X}(\mathcal{E}^{-n}, \mathcal{O}_X)$ in degree $n$. Since $\mathcal{E}^{-n}$ is a direct summand of a finite free $\mathcal{O}_X$-module, so is $(\mathcal{E}^{-n})^\vee$. Hence $K^\vee$ is represented by a strictly perfect complex too and we see that $K^\vee$ is perfect. To see that (\ref{equation-eval}) is an isomorphism, represent $M$ by a complex $\mathcal{F}^\bullet$. By Lemma \ref{lemma-Rhom-strictly-perfect} the complex $R\SheafHom(K, M)$ is represented by the complex with terms $$ \bigoplus\nolimits_{n = p + q} \SheafHom_{\mathcal{O}_X}(\mathcal{E}^{-q}, \mathcal{F}^p) $$ On the other hand, the object $M \otimes^\mathbf{L}_{\mathcal{O}_X} K^\vee$ is represented by the complex with terms $$ \bigoplus\nolimits_{n = p + q} \mathcal{F}^p \otimes_{\mathcal{O}_X} (\mathcal{E}^{-q})^\vee $$ Thus the assertion that (\ref{equation-eval}) is an isomorphism reduces to the assertion that the canonical map $$ \mathcal{F} \otimes_{\mathcal{O}_X} \SheafHom_{\mathcal{O}_X}(\mathcal{E}, \mathcal{O}_X) \longrightarrow \SheafHom_{\mathcal{O}_X}(\mathcal{E}, \mathcal{F}) $$ is an isomorphism when $\mathcal{E}$ is a direct summand of a finite free $\mathcal{O}_X$-module and $\mathcal{F}$ is any $\mathcal{O}_X$-module. This follows immediately from the corresponding statement when $\mathcal{E}$ is finite free.
By Lemma \ref{lemma-dual-perfect-complex} we have $R\lim E \otimes^\mathbf{L}_{\mathcal{O}_X} K_n^\vee = R\lim R\SheafHom(K_n, E)$ which fits into the distinguished triangle $$ R\lim R\SheafHom(K_n, E) \to \prod R\SheafHom(K_n, E) \to \prod R\SheafHom(K_n, E) $$ Because $K$ similarly fits into the distinguished triangle $\bigoplus K_n \to \bigoplus K_n \to K$ it suffices to show that $\prod R\SheafHom(K_n, E) = R\SheafHom(\bigoplus K_n, E)$. This is a formal consequence of (\ref{equation-internal-hom}) and the fact that derived tensor product commutes with direct sums.
We will abbreviate $\otimes = \otimes_{\mathcal{O}_X}^\mathbf{L}$ and $\mathcal{O} = \mathcal{O}_X$. We will identify $E$ and $K$ with $R\SheafHom(\mathcal{O}, E)$ and $R\SheafHom(\mathcal{O}, K)$ and we will identify $E^\vee$ with $R\SheafHom(E, \mathcal{O})$. \medskip\noindent Let $\xi \in H^0(X, K \otimes E^\vee)$ and $\eta \in H^0(X, E)$. Denote $\tilde \xi : \mathcal{O} \to K \otimes E^\vee$ and $\tilde \eta : \mathcal{O} \to E$ the corresponding maps in $D(\mathcal{O})$. By Lemma \ref{lemma-second-cup-equals-first} the cup product $\xi \cup \eta$ corresponds to $\tilde \xi \otimes \tilde \eta : \mathcal{O} \to K \otimes E^\vee \otimes E$. \medskip\noindent We claim the map $\xi' : E \to K$ corresponding to $\xi$ by Lemma \ref{lemma-dual-perfect-complex} is the composition $$ E = \mathcal{O} \otimes E \xrightarrow{\tilde \xi \otimes 1_E} K \otimes E^\vee \otimes E \xrightarrow{1_K \otimes \epsilon} K $$ The construction in Lemma \ref{lemma-dual-perfect-complex} uses the evaluation map (\ref{equation-eval}) which in turn is constructed using the identification of $E$ with $R\SheafHom(\mathcal{O}, E)$ and the composition $\underline{\circ}$ constructed in Lemma \ref{lemma-internal-hom-composition}. Hence $\xi'$ is the composition \begin{align*} E = \mathcal{O} \otimes R\SheafHom(\mathcal{O}, E) & \xrightarrow{\tilde \xi \otimes 1} R\SheafHom(\mathcal{O}, K) \otimes R\SheafHom(E, \mathcal{O}) \otimes R\SheafHom(\mathcal{O}, E) \\ & \xrightarrow{\underline{\circ} \otimes 1} R\SheafHom(E, K) \otimes R\SheafHom(\mathcal{O}, E) \\ & \xrightarrow{\underline{\circ}} R\SheafHom(\mathcal{O}, K) = K \end{align*} The claim follows immediately from this and the fact that the composition $\underline{\circ}$ constructed in Lemma \ref{lemma-internal-hom-composition} is associative (insert future reference here) and the fact that $\epsilon$ is defined as the composition $\underline{\circ} : E^\vee \otimes E \to \mathcal{O}$ in Example \ref{example-dual-derived}. \medskip\noindent Using the results from the previous two paragraphs, we find the statement of the lemma is that $(1_K \otimes \epsilon) \circ (\tilde \xi \otimes \tilde \eta)$ is equal to $(1_K \otimes \epsilon) \circ (\tilde \xi \otimes 1_E) \circ (1_\mathcal{O} \otimes \tilde \eta)$ which is immediate.
Assume (1) and (2). Since $\Hom(j_!\mathcal{O}_U, \mathcal{F}) = \mathcal{F}(U)$ by Sheaves, Lemma \ref{sheaves-lemma-j-shriek-modules} we have $\Hom(j_!\mathcal{O}_U, K) = R\Gamma(U, K)$ for $K$ in $D(\mathcal{O}_X)$. Thus we have to show that $R\Gamma(U, -)$ commutes with direct sums. The first assumption means that the functor $F = H^0(U, -)$ has finite cohomological dimension. Moreover, the second assumption implies any direct sum of injective modules is acyclic for $F$. Let $K_i$ be a family of objects of $D(\mathcal{O}_X)$. Choose K-injective representatives $I_i^\bullet$ with injective terms representing $K_i$, see Injectives, Theorem \ref{injectives-theorem-K-injective-embedding-grothendieck}. Since we may compute $RF$ by applying $F$ to any complex of acyclics (Derived Categories, Lemma \ref{derived-lemma-unbounded-right-derived}) and since $\bigoplus K_i$ is represented by $\bigoplus I_i^\bullet$ (Injectives, Lemma \ref{injectives-lemma-derived-products}) we conclude that $R\Gamma(U, \bigoplus K_i)$ is represented by $\bigoplus H^0(U, I_i^\bullet)$. Hence $R\Gamma(U, -)$ commutes with direct sums as desired.
Let $K^\vee$ be the dual of $K$, see Lemma \ref{lemma-dual-perfect-complex}. Then we have $$ \Hom_{D(\mathcal{O}_X)}(K, M) = H^0(X, K^\vee \otimes_{\mathcal{O}_X}^\mathbf{L} M) $$ functorially in $M$ in $D(\mathcal{O}_X)$. Since $K^\vee \otimes_{\mathcal{O}_X}^\mathbf{L} -$ commutes with direct sums it suffices to show that $R\Gamma(X, -)$ commutes with the relevant direct sums. \medskip\noindent Proof of (b). Since $R\Gamma(X, K) = R\Hom(\mathcal{O}_X, K)$ and since $H^p(X, -)$ commutes with direct sums by Lemma \ref{lemma-quasi-separated-cohomology-colimit} this is a special case of Lemma \ref{lemma-when-jshriek-compact} \medskip\noindent Proof of (a). Let $\mathcal{I}_i$, $i \in I$ be a collection of injective $\mathcal{O}_X$-modules. By Lemma \ref{lemma-quasi-separated-cohomology-colimit} we see that $$ H^p(X, \bigoplus\nolimits_{i \in I} \mathcal{I}_i) = \bigoplus\nolimits_{i \in I} H^p(X, \mathcal{I}_i) = 0 $$ for all $p$. Now if $M = \bigoplus M_i$ is as in (a), then we see that there exists an $a \in \mathbf{Z}$ such that $H^n(M_i) = 0$ for $n < a$. Thus we can choose complexes of injective $\mathcal{O}_X$-modules $\mathcal{I}_i^\bullet$ representing $M_i$ with $\mathcal{I}_i^n = 0$ for $n < a$, see Derived Categories, Lemma \ref{derived-lemma-injective-resolutions-exist}. By Injectives, Lemma \ref{injectives-lemma-derived-products} we see that the direct sum complex $\bigoplus \mathcal{I}_i^\bullet$ represents $M$. By Leray acyclicity (Derived Categories, Lemma \ref{derived-lemma-leray-acyclicity}) we see that $$ R\Gamma(X, M) = \Gamma(X, \bigoplus \mathcal{I}_i^\bullet) = \bigoplus \Gamma(X, \bigoplus \mathcal{I}_i^\bullet) = \bigoplus R\Gamma(X, M_i) $$ as desired.
This is true because under the assumptions of the lemma we have $$ \Hom_{\mathcal{O}_X}(\mathcal{F}, \mathcal{E} \otimes_{\mathcal{O}_X} \mathcal{I}) = \Hom_{\mathcal{O}_X}( \mathcal{F} \otimes_{\mathcal{O}_X} \mathcal{E}^\vee, \mathcal{I}) $$ where $\mathcal{E}^\vee = \SheafHom_{\mathcal{O}_X}(\mathcal{E}, \mathcal{O}_X)$ is the dual of $\mathcal{E}$ which is finite locally free also. Since tensoring with a finite locally free sheaf is an exact functor we win by Homology, Lemma \ref{homology-lemma-characterize-injectives}.
Choose an injective resolution $\mathcal{F} \to \mathcal{I}^\bullet$ on $X$. Note that $f^*\mathcal{E}$ is finite locally free also, hence we get a resolution $$ f^*\mathcal{E} \otimes_{\mathcal{O}_X} \mathcal{F} \longrightarrow f^*\mathcal{E} \otimes_{\mathcal{O}_X} \mathcal{I}^\bullet $$ which is an injective resolution by Lemma \ref{lemma-injective-tensor-finite-locally-free}. Apply $f_*$ to see that $$ Rf_*(f^*\mathcal{E} \otimes_{\mathcal{O}_X} \mathcal{F}) = f_*(f^*\mathcal{E} \otimes_{\mathcal{O}_X} \mathcal{I}^\bullet). $$ Hence the lemma follows if we can show that $f_*(f^*\mathcal{E} \otimes_{\mathcal{O}_X} \mathcal{F}) = \mathcal{E} \otimes_{\mathcal{O}_Y} f_*(\mathcal{F})$ functorially in the $\mathcal{O}_X$-module $\mathcal{F}$. This is clear when $\mathcal{E} = \mathcal{O}_Y^{\oplus n}$, and follows in general by working locally on $Y$. Details omitted.
Since $f$ is a homeomorphism onto a closed subset, the functor $f_*$ is exact (Modules, Lemma \ref{modules-lemma-i-star-exact}). Hence $Rf_*$ is computed by applying $f_*$ to any representative complex. Choose a K-flat complex $\mathcal{K}^\bullet$ of $\mathcal{O}_Y$-modules representing $K$ and choose any complex $\mathcal{E}^\bullet$ of $\mathcal{O}_X$-modules representing $E$. Then $Lf^*K$ is represented by $f^*\mathcal{K}^\bullet$ which is a K-flat complex of $\mathcal{O}_X$-modules (Lemma \ref{lemma-pullback-K-flat}). Thus the right hand side of (\ref{equation-projection-formula-map}) is represented by $$ f_*\text{Tot}(\mathcal{E}^\bullet \otimes_{\mathcal{O}_X} f^*\mathcal{K}^\bullet) $$ By the same reasoning we see that the left hand side is represented by $$ \text{Tot}(f_*\mathcal{E}^\bullet \otimes_{\mathcal{O}_Y} \mathcal{K}^\bullet) $$ Since $f_*$ commutes with direct sums (Modules, Lemma \ref{modules-lemma-i-star-right-adjoint}) it suffices to show that $$ f_*(\mathcal{E} \otimes_{\mathcal{O}_X} f^*\mathcal{K}) = f_*\mathcal{E} \otimes_{\mathcal{O}_Y} \mathcal{K} $$ for any $\mathcal{O}_X$-module $\mathcal{E}$ and $\mathcal{O}_Y$-module $\mathcal{K}$. We will check this by checking on stalks. Let $y \in Y$. If $y \not \in f(X)$, then the stalks of both sides are zero. If $y = f(x)$, then we see that we have to show $$ \mathcal{E}_x \otimes_{\mathcal{O}_{X, x}} (\mathcal{O}_{X, x} \otimes_{\mathcal{O}_{Y, y}} \mathcal{F}_y) = \mathcal{E}_x \otimes_{\mathcal{O}_{Y, y}} \mathcal{F}_y $$ (using Sheaves, Lemma \ref{sheaves-lemma-stalks-closed-pushforward} and Lemma \ref{sheaves-lemma-stalk-pullback-modules}). This equality holds and therefore the lemma has been proved.
We prove this lemma by induction on $d$. So fix $d$ and assume the lemma holds for all Noetherian topological spaces of dimension $< d$. \medskip\noindent Let $\mathcal{F}$ be an abelian sheaf on $X$. Suppose $U \subset X$ is an open. Let $Z \subset X$ denote the closed complement. Denote $j : U \to X$ and $i : Z \to X$ the inclusion maps. Then there is a short exact sequence $$ 0 \to j_{!}j^*\mathcal{F} \to \mathcal{F} \to i_*i^*\mathcal{F} \to 0 $$ see Modules, Lemma \ref{modules-lemma-canonical-exact-sequence}. Note that $j_!j^*\mathcal{F}$ is supported on the topological closure $Z'$ of $U$, i.e., it is of the form $i'_*\mathcal{F}'$ for some abelian sheaf $\mathcal{F}'$ on $Z'$, where $i' : Z' \to X$ is the inclusion. \medskip\noindent We can use this to reduce to the case where $X$ is irreducible. Namely, according to Topology, Lemma \ref{topology-lemma-Noetherian} $X$ has finitely many irreducible components. If $X$ has more than one irreducible component, then let $Z \subset X$ be an irreducible component of $X$ and set $U = X \setminus Z$. By the above, and the long exact sequence of cohomology, it suffices to prove the vanishing of $H^p(X, i_*i^*\mathcal{F})$ and $H^p(X, i'_*\mathcal{F}')$ for $p > d$. By Lemma \ref{lemma-cohomology-and-closed-immersions} it suffices to prove $H^p(Z, i^*\mathcal{F})$ and $H^p(Z', \mathcal{F}')$ vanish for $p > d$. Since $Z'$ and $Z$ have fewer irreducible components we indeed reduce to the case of an irreducible $X$. \medskip\noindent If $d = 0$ and $X = \{*\}$, then every sheaf is constant and higher cohomology groups vanish (for example by Lemma \ref{lemma-irreducible-constant-cohomology-zero}). \medskip\noindent Suppose $X$ is irreducible of dimension $d$. By Lemma \ref{lemma-vanishing-generated-one-section} we reduce to the case where $\mathcal{F} = j_!\underline{\mathbf{Z}}_U$ for some open $U \subset X$. In this case we look at the short exact sequence $$ 0 \to j_!(\underline{\mathbf{Z}}_U) \to \underline{\mathbf{Z}}_X \to i_*\underline{\mathbf{Z}}_Z \to 0 $$ where $Z = X \setminus U$. By Lemma \ref{lemma-irreducible-constant-cohomology-zero} we have the vanishing of $H^p(X, \underline{\mathbf{Z}}_X)$ for all $p \geq 1$. By induction we have $H^p(X, i_*\underline{\mathbf{Z}}_Z) = H^p(Z, \underline{\mathbf{Z}}_Z) = 0$ for $p \geq d$. Hence we win by the long exact cohomology sequence.
The functor is fully faithful by Lemma \ref{lemma-completion-functor-fully-faithful}. Let $g : W \to X_{/T}$ be a morphism of formal algebraic spaces with $W$ locally Noetherian and $g$ rig-\'etale. We will prove $W$ is in the essential image to finish the proof. \medskip\noindent Choose an \'etale covering $\{X_i \to X\}$ with $X_i$ affine for all $i$. Denote $U_i \subset X_i$ the inverse image of $U$ and denote $T_i \subset X_i$ the inverse image of $T$. Recall that $(X_i)_{/T_i} = (X_i)_{/T} = (X_i \times_X X)_{/T}$ and $W_i = X_i \times_X W = (X_i)_{/T} \times_{X_{/T}} W$, see Lemma \ref{lemma-functoriality-completion-functor}. Observe that we obtain isomorphisms $$ \alpha_{ij} : W_i \times_{X_{/T}} (X_j)_{/T} \longrightarrow  (X_i)_{/T} \times_{X_{/T}} W_j $$ satisfying a suitable cocycle condition. By Lemma \ref{lemma-dilatations-affine} applied to $X_i, T_i, U_i, W_i \to (X_i)_{/T}$ there exists a morphism $X'_i \to X_i$ of algebraic spaces which is locally of finite type and an isomorphism over $U_i$ and an isomorphism $\beta_i : (X'_i)_{/T} \cong W_i$ over $(X_i)_{/T}$. By fully faithfullness we find an isomorphism $$ a_{ij} : X'_i \times_X X_j \longrightarrow X_i \times_X X'_j $$ over $X_i \times_X X_j$ such that $\alpha_{ij} = \beta_j|_{X_i \times_X X_j} \circ (a_{ij})_{/T} \circ \beta_i^{-1}|_{X_i \times_X X_j}$. By fully faithfulness again (this time over $X_i \times_X X_j \times_X X_k$) we see that these morphisms $a_{ij}$ satisfy the same cocycle condition as satisfied by the $\alpha_{ij}$. In other words, we obtain a descent datum (as in Descent on Spaces, Definition \ref{spaces-descent-definition-descent-datum-for-family-of-morphisms}) $(X'_i, a_{ij})$ relative to the family $\{X_i \to X\}$. By Bootstrap, Lemma \ref{bootstrap-lemma-descend-algebraic-space}, this descent datum is effective. Thus we find a morphism $f : X' \to X$ of algebraic spaces and isomorphisms $h_i : X' \times_X X_i \to X'_i$ over $X_i$ such that $a_{ij} = h_j|_{X_i \times_X X_j} \circ h_i^{-1}|_{X_i \times_X X_j}$. The reader can check that the ensuing isomorphisms $$ (X' \times_X X_i)_{/T} \xrightarrow{\beta_i \circ (h_i)_{/T}} W_i $$ over $X_i$ glue to an isomorphism $X'_{/T} \to W$ over $X_{/T}$; some details omitted.
Let $(B_n)$ be an object of $\mathcal{C}$. By Algebra, Lemma \ref{algebra-lemma-limit-complete} we see that $B = \lim B_n$ is $I$-adically complete and $B/I^nB = B_n$. Hence we see that $B$ is an object of $\mathcal{C}'$ and that we can recover the object $(B_n)$  by taking the quotients. Conversely, if $B$ is an object of $\mathcal{C}'$, then $B = \lim B/I^nB$ by assumption. Thus $B \mapsto (B/I^nB)$ is a quasi-inverse to the functor of the lemma. \medskip\noindent Since $A[x_1, \ldots, x_r]^\wedge = \lim A_n[x_1, \ldots, x_r]$ it is an object of $\mathcal{C}'$ by the first statement of the lemma. Finally, let $B$ be an object of $\mathcal{C}'$. Choose $b_1, \ldots, b_r \in B$ whose images in $B/IB$ generate $B/IB$ as an algebra over $A/I$. Since $B$ is $I$-adically complete, the $A$-algebra map $A[x_1, \ldots, x_r] \to B$, $x_i \mapsto b_i$ extends to an $A$-algebra map $A[x_1, \ldots, x_r]^\wedge \to B$. To finish the proof we have to show this map is surjective which follows from Algebra, Lemma \ref{algebra-lemma-completion-generalities} as our map $A[x_1, \ldots, x_r] \to B$ is surjective modulo $I$ and as $B = B^\wedge$.
Part (4) follows from Algebra, Lemma \ref{algebra-lemma-completion-Noetherian-Noetherian} as $A[x_1, \ldots, x_r]$ is Noetherian (Algebra, Lemma \ref{algebra-lemma-Noetherian-permanence}). To see (1) by Lemma \ref{lemma-topologically-finite-type} we reduce to the case of the completion of the polynomial ring which we just proved. Part (2) follows from Algebra, Lemma \ref{algebra-lemma-completion-tensor} which tells us that ever finite $B$-module is $IB$-adically complete. Part (3) follows in the same manner as part (4).
The lemma signifies that given a second presentation $B = A[y_1, \ldots, y_s]^\wedge / K$ the complexes of $B$-modules $$ (J/J^2 \to B\text{d}x_i) \quad\text{and}\quad (K/K^2 \to \bigoplus B\text{d}y_j) $$ are homotopy equivalent. To see this, we can argue exactly as in the proof of Algebra, Lemma \ref{algebra-lemma-NL-homotopy}. \medskip\noindent Step 1. If we choose $g_i(y_1, \ldots, y_s) \in A[y_1, \ldots, y_s]^\wedge$ mapping to the image of $x_i$ in $B$, then we obtain a (unique) continuous $A$-algebra homomorphism $$ A[x_1, \ldots, x_r]^\wedge \to A[y_1, \ldots, y_s]^\wedge,\quad x_i \mapsto g_i(y_1, \ldots, y_s) $$ compatible with the given surjections to $B$. Such a map is called a morphism of presentations. It induces a map from $J$ into $K$ and hence induces a $B$-module map $J/J^2 \to K/K^2$. Sending $\text{d}x_i$ to $\sum (\partial g_i/\partial y_j)\text{d}y_j$ we obtain a map of complexes $$ (J/J^2 \to \bigoplus B\text{d}x_i) \longrightarrow (K/K^2 \to \bigoplus B\text{d}y_j) $$ Of course we can do the same thing with the roles of the two presentations exchanged to get a map of complexes in the other direction. \medskip\noindent Step 2. The construction above is compatible with compositions of morphsms of presentations. Hence to finish the proof it suffices to show: given $g_i(x_1, \ldots, x_r) \in A[x_1, \ldots, x_n]^\wedge$ mapping to the image of $x_i$ in $B$, the induced map of complexes $$ (J/J^2 \to \bigoplus B\text{d}x_i) \longrightarrow (J/J^2 \to \bigoplus B\text{d}x_i) $$ is homotopic to the identity map. To see this consider the map $h : \bigoplus B \text{d}x_i \to J/J^2$ given by the rule $\text{d}x_i \mapsto g_i(x_1, \ldots, x_n) - x_i$ and compute.