hcju/stacksqa huggingface dataset:
Let $U \to X$ be an \'etale morphism with $U$ affine. Say $U = \Spec(A)$ and let $\omega_A^\bullet$ be a dualizing complex for $A$ corresponding to $K|_U$ as in Lemma \ref{lemma-equivalent-definitions} and Duality for Schemes, Lemma \ref{duality-lemma-equivalent-definitions}. By Lemma \ref{lemma-affine-duality} the diagram $$ \xymatrix{ D_{\textit{Coh}}(A) \ar[r] \ar[d]_{R\Hom_A(-, \omega_A^\bullet)} & D_{\textit{Coh}}(\mathcal{O}_\etale) \ar[d]^{R\SheafHom_{\mathcal{O}_\etale}(-, K|_U)} \\ D_{\textit{Coh}}(A) \ar[r] & D(\mathcal{O}_\etale) } $$ commutes where $\mathcal{O}_\etale$ is the structure sheaf of the small \'etale site of $U$. Since formation of $R\SheafHom$ commutes with restriction, we conclude that $D$ sends $D_{\textit{Coh}}(\mathcal{O}_X)$ into $D_{\textit{Coh}}(\mathcal{O}_X)$. Moreover, the canonical map $$ L \longrightarrow R\SheafHom_{\mathcal{O}_X}(R\SheafHom_{\mathcal{O}_X}(L, K), K) $$ (Cohomology on Sites, Lemma \ref{sites-cohomology-lemma-internal-hom-evaluate}) is an isomorphism for all $L$ in $D_{\textit{Coh}}(\mathcal{O}_X)$ because this is true over all $U$ as above by Dualizing Complexes, Lemma \ref{dualizing-lemma-dualizing}. The statement on boundedness properties of the functor $D$ in the quasi-compact case also follows from the corresponding statements of Dualizing Complexes, Lemma \ref{dualizing-lemma-dualizing}.
Set $$ L = R\SheafHom_{\mathcal{O}_X}(K, K') $$ This is an invertible object of $D(\mathcal{O}_X)$, because affine locally this is true. Use Lemma \ref{lemma-affine-duality} and Dualizing Complexes, Lemma \ref{dualizing-lemma-dualizing-unique} and its proof. The evaluation map $L \otimes_{\mathcal{O}_X}^\mathbf{L} K \to K'$ is an isomorphism for the same reason.
Let $U$ be a scheme and let $U \to X$ be a surjective \'etale morphism. Let $\omega_U^\bullet$ be the dualizing complex on $U$ associated to $\omega_X^\bullet|_U$. If $u \in U$ maps to $x \in |X|$, then $\mathcal{O}_{X, \overline{x}}$ is the strict henselization of $\mathcal{O}_{U, u}$. By Dualizing Complexes, Lemma \ref{dualizing-lemma-flat-unramified} we see that if $\omega^\bullet$ is a normalized dualizing complex for $\mathcal{O}_{U, u}$, then $\omega^\bullet \otimes_{\mathcal{O}_{U, u}} \mathcal{O}_{X, \overline{x}}$ is a normalized dualizing complex for $\mathcal{O}_{X, \overline{x}}$. Hence we see that the dimension function $U \to \mathbf{Z}$ of Duality for Schemes, Lemma \ref{duality-lemma-dimension-function-scheme} for the scheme $U$ and the complex $\omega_U^\bullet$ is equal to the composition of $U \to |X|$ with $\delta$. Using the specializations in $|X|$ lift to specializations in $U$ and that nontrivial specializations in $U$ map to nontrivial specializations in $X$ (Decent Spaces, Lemmas \ref{decent-spaces-lemma-decent-specialization} and \ref{decent-spaces-lemma-decent-no-specializations-map-to-same-point}) an easy topological argument shows that $\delta$ is a dimension function on $|X|$.
We will prove a right adjoint exists by verifying the hypotheses of Derived Categories, Proposition \ref{derived-proposition-brown}. First off, the category $D_\QCoh(\mathcal{O}_X)$ has direct sums, see Derived Categories of Spaces, Lemma \ref{spaces-perfect-lemma-quasi-coherence-direct-sums}. The category $D_\QCoh(\mathcal{O}_X)$ is compactly generated by Derived Categories of Spaces, Theorem \ref{spaces-perfect-theorem-bondal-van-den-Bergh}. Since $X$ and $Y$ are quasi-compact and quasi-separated, so is $f$, see Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-compose-after-separated} and \ref{spaces-morphisms-lemma-quasi-compact-permanence}. Hence the functor $Rf_*$ commutes with direct sums, see Derived Categories of Spaces, Lemma \ref{spaces-perfect-lemma-quasi-coherence-pushforward-direct-sums}. This finishes the proof.
By Derived Categories of Spaces, Lemma \ref{spaces-perfect-lemma-quasi-coherence-direct-image} the functor $Rf_*$ has finite cohomological dimension. In other words, there exist an integer $N$ such that $H^i(Rf_*L) = 0$ for $i \geq N + c$ if $H^i(L) = 0$ for $i \geq c$. Say $K \in D^+_\QCoh(\mathcal{O}_Y)$ has $H^i(K) = 0$ for $i \leq c$. Then $$ \Hom_{D(\mathcal{O}_X)}(\tau_{\leq c - N}a(K), a(K)) = \Hom_{D(\mathcal{O}_Y)}(Rf_*\tau_{\leq c - N}a(K), K) = 0 $$ by what we said above. Clearly, this implies that $H^i(a(K)) = 0$ for $i \leq c - N$.
The statement makes sense as $DQ_Y$ exists by Derived Categories of Spaces, Lemma \ref{spaces-perfect-lemma-better-coherator}. Since $DQ_Y$ is the right adjoint to the inclusion functor $D_\QCoh(\mathcal{O}_Y) \to D(\mathcal{O}_Y)$ to prove the lemma we have to show that for any $M \in D_\QCoh(\mathcal{O}_Y)$ the map (\ref{equation-sheafy-trace}) induces an bijection $$ \Hom_Y(M, Rf_*R\SheafHom_{\mathcal{O}_X}(L, a(K))) \longrightarrow \Hom_Y(M, R\SheafHom_{\mathcal{O}_Y}(Rf_*L, K)) $$ To see this we use the following string of equalities \begin{align*} \Hom_Y(M, Rf_*R\SheafHom_{\mathcal{O}_X}(L, a(K))) & = \Hom_X(Lf^*M, R\SheafHom_{\mathcal{O}_X}(L, a(K))) \\ & = \Hom_X(Lf^*M \otimes_{\mathcal{O}_X}^\mathbf{L} L, a(K)) \\ & = \Hom_Y(Rf_*(Lf^*M \otimes_{\mathcal{O}_X}^\mathbf{L} L), K) \\ & = \Hom_Y(M \otimes_{\mathcal{O}_Y}^\mathbf{L} Rf_*L, K) \\ & = \Hom_Y(M, R\SheafHom_{\mathcal{O}_Y}(Rf_*L, K)) \end{align*} The first equality holds by Cohomology on Sites, Lemma \ref{sites-cohomology-lemma-adjoint}. The second equality by Cohomology on Sites, Lemma \ref{sites-cohomology-lemma-internal-hom}. The third equality by construction of $a$. The fourth equality by Derived Categories of Spaces, Lemma \ref{spaces-perfect-lemma-cohomology-base-change} (this is the important step). The fifth by Cohomology on Sites, Lemma \ref{sites-cohomology-lemma-internal-hom}.
By construction (Cohomology on Sites, Section \ref{sites-cohomology-section-global-RHom}) the complexes $$ R\Hom_X(L, a(K)) = R\Gamma(X, R\SheafHom_{\mathcal{O}_X}(L, a(K))) = R\Gamma(Y, Rf_*R\SheafHom_{\mathcal{O}_X}(L, a(K))) $$ and $$ R\Hom_Y(Rf_*L, K) = R\Gamma(Y, R\SheafHom_{\mathcal{O}_X}(Rf_*L, a(K))) $$ Thus the lemma is a consequence of Lemma \ref{lemma-iso-on-RSheafHom}. Namely, a map $E \to E'$ in $D(\mathcal{O}_Y)$ which induces an isomorphism $DQ_Y(E) \to DQ_Y(E')$ induces a quasi-isomorphism $R\Gamma(Y, E) \to R\Gamma(Y, E')$. Indeed we have $H^i(Y, E) = \Ext^i_Y(\mathcal{O}_Y, E) = \Hom(\mathcal{O}_Y[-i], E) = \Hom(\mathcal{O}_Y[-i], DQ_Y(E))$ because $\mathcal{O}_Y[-i]$ is in $D_\QCoh(\mathcal{O}_Y)$ and $DQ_Y$ is the right adjoint to the inclusion functor $D_\QCoh(\mathcal{O}_Y) \to D(\mathcal{O}_Y)$.
The base change map $Lg^* \circ Rf_* K \to Rf'_* \circ L(g')^*K$ is an isomorphism for every $K$ in $D_\QCoh(\mathcal{O}_X)$ by Derived Categories of Spaces, Lemma \ref{spaces-perfect-lemma-compare-base-change} (this uses the assumption of Tor independence). Thus the corresponding transformation between adjoint functors is an isomorphism as well.
It follows from the assumptions that $f$ and $h \circ h'$ are Tor independent (details omitted), hence the statement makes sense. In this proof we write $g^*$ in place of $Lg^*$ and $f_*$ instead of $Rf_*$. Let $a$, $a'$, and $a''$ be the right adjoints of Lemma \ref{lemma-twisted-inverse-image} for $f$, $f'$, and $f''$. The arrow corresponding to the right square is the composition $$ \gamma_{right} : g^* \circ a \to g^* \circ a \circ h_* \circ h^* \xleftarrow{\xi_{right}} g^* \circ g_* \circ a' \circ h^* \to a' \circ h^* $$ where $\xi_{right} : g_* \circ a' \to a \circ h_*$ is an isomorphism (hence can be inverted) and is the arrow ``dual'' to the base change map $h^* \circ f_* \to f'_* \circ g^*$. The outer arrows come from the canonical maps $1 \to h_* \circ h^*$ and $g^* \circ g_* \to 1$. Similarly for the left square we have $$ \gamma_{left} : (g')^* \circ a' \to (g')^* \circ a' \circ (h')_* \circ (h')^* \xleftarrow{\xi_{left}} (g')^* \circ (g')_* \circ a'' \circ (h')^* \to a'' \circ (h')^* $$ For the outer rectangle we get $$ \gamma_{rect} : k^* \circ a \to k^* \circ a \circ m_* \circ m^* \xleftarrow{\xi_{rect}} k^* \circ k_* \circ a'' \circ m^* \to a'' \circ m^* $$ where $k = g \circ g'$ and $m = h \circ h'$. We have $k^* = (g')^* \circ g^*$ and $m^* = (h')^* \circ h^*$. The statement of the lemma is that $\gamma_{rect}$ is equal to the composition $$ k^* \circ a = (g')^* \circ g^* \circ a \xrightarrow{\gamma_{right}} (g')^* \circ a' \circ h^* \xrightarrow{\gamma_{left}} a'' \circ (h')^* \circ h^* = a'' \circ m^* $$ To see this we contemplate the following diagram $$ \xymatrix{ & (g')^* \circ g^* \circ a \ar[d] \ar[ddl] \\ & (g')^* \circ g^* \circ a \circ h_* \circ h^* \ar[ld] \\ (g')^* \circ g^* \circ a \circ h_* \circ (h')_* \circ (h')^* \circ h^* & (g')^* \circ g^* \circ g_* \circ a' \circ h^* \ar[u]_{\xi_{right}} \ar[d] \ar[ld] \\ (g')^* \circ g^* \circ g_* \circ a' \circ (h')_* \circ (h')^* \circ h^* \ar[u]_{\xi_{right}} \ar[dr] & (g')^* \circ a' \circ h^* \ar[d] \\ (g')^* \circ g^* \circ g_* \circ (g')_* \circ a'' \circ (h')^* \circ h^* \ar[u]_{\xi_{left}} \ar[ddr] \ar[dr] & (g')^* \circ a' \circ (h')_* \circ (h')^* \circ h^* \\ & (g')^*\circ (g')_* \circ a'' \circ (h')^* \circ h^* \ar[u]_{\xi_{left}} \ar[d] \\ & a'' \circ (h')^* \circ h^* } $$ Going down the right hand side we have the composition and going down the left hand side we have $\gamma_{rect}$. All the quadrilaterals on the right hand side of this diagram commute by Categories, Lemma \ref{categories-lemma-properties-2-cat-cats} or more simply the discussion preceding Categories, Definition \ref{categories-definition-horizontal-composition}. Hence we see that it suffices to show that $$ g_* \circ (g')_* \circ a'' \xrightarrow{\xi_{left}} g_* \circ a' \circ (h')_* \xrightarrow{\xi_{right}} a \circ h_* \circ (h')_* $$ is equal to $\xi_{rect}$. This is the statement dual to Cohomology, Remark \ref{cohomology-remark-compose-base-change-horizontal} and the proof is complete.
In this proof we write $f_*$ for $Rf_*$ and $g^*$ for $Lg^*$ and we drop $\star$ products with identities as one can figure out which ones to add as long as the source and target of the transformation is known. Recall that $\beta : g^* \circ f_* \to f'_* \circ (g')^*$ is an isomorphism and that $\alpha$ is defined using the isomorphism $\beta^\vee : g'_* \circ a' \to a \circ g_*$ which is the adjoint of $\beta$, see Lemma \ref{lemma-flat-precompose-pus} and its proof. First we note that the top horizontal arrow of the diagram in the lemma is equal to the composition $$ g^* \circ f_* \circ a \to g^* \circ f_* \circ a \circ g_* \circ g^* \to g^* \circ g_* \circ g^* \to g^* $$ where the first arrow is the unit for $(g^*, g_*)$, the second arrow is $\text{Tr}_f$, and the third arrow is the counit for $(g^*, g_*)$. This is a simple consequence of the fact that the composition $g^* \to g^* \circ g_* \circ g^* \to g^*$ of unit and counit is the identity. Consider the diagram $$ \xymatrix{ & g^* \circ f_* \circ a \ar[ld]_\beta \ar[d] \ar[r]_{\text{Tr}_f} & g^* \\ f'_* \circ (g')^* \circ a \ar[dr] & g^* \circ f_* \circ a \circ g_* \circ g^* \ar[d]_\beta \ar[ru] & g^* \circ f_* \circ g'_* \circ a' \circ g^* \ar[l]_{\beta^\vee} \ar[d]_\beta & f'_* \circ a' \circ g^* \ar[lu]_{\text{Tr}_{f'}} \\ & f'_* \circ (g')^* \circ a \circ g_* \circ g^* & f'_* \circ (g')^* \circ g'_* \circ a' \circ g^* \ar[ru] \ar[l]_{\beta^\vee} } $$ In this diagram the two squares commute  Categories, Lemma \ref{categories-lemma-properties-2-cat-cats} or more simply the discussion preceding Categories, Definition \ref{categories-definition-horizontal-composition}. The triangle commutes by the discussion above. By Categories, Lemma \ref{categories-lemma-transformation-between-functors-and-adjoints} the square $$ \xymatrix{ g^* \circ f_* \circ g'_* \circ a' \ar[d]_{\beta^\vee} \ar[r]_-\beta & f'_* \circ (g')^* \circ g'_* \circ a' \ar[d] \\ g^* \circ f_* \circ a \circ g_* \ar[r] & \text{id} } $$ commutes which implies the pentagon in the big diagram commutes. Since $\beta$ and $\beta^\vee$ are isomorphisms, and since going on the outside of the big diagram equals $\text{Tr}_f \circ \alpha \circ \beta$ by definition this proves the lemma.
This proof is dual to the proof of Lemma \ref{lemma-trace-map-and-base-change}. In this proof we write $f_*$ for $Rf_*$ and $g^*$ for $Lg^*$ and we drop $\star$ products with identities as one can figure out which ones to add as long as the source and target of the transformation is known. Recall that $\beta : g^* \circ f_* \to f'_* \circ (g')^*$ is an isomorphism and that $\alpha$ is defined using the isomorphism $\beta^\vee : g'_* \circ a' \to a \circ g_*$ which is the adjoint of $\beta$, see Lemma \ref{lemma-flat-precompose-pus} and its proof. First we note that the left vertical arrow of the diagram in the lemma is equal to the composition $$ (g')^* \to (g')^* \circ g'_* \circ (g')^* \to (g')^* \circ g'_* \circ a' \circ f'_* \circ (g')^* \to a' \circ f'_* \circ (g')^* $$ where the first arrow is the unit for $((g')^*, g'_*)$, the second arrow is $\eta_{f'}$, and the third arrow is the counit for $((g')^*, g'_*)$. This is a simple consequence of the fact that the composition $(g')^* \to (g')^* \circ (g')_* \circ (g')^* \to (g')^*$ of unit and counit is the identity. Consider the diagram $$ \xymatrix{ & (g')^* \circ a \circ f_* \ar[r] & (g')^* \circ a \circ g_* \circ g^* \circ f_* \ar[ld]_\beta \\ (g')^* \ar[ru]^{\eta_f} \ar[dd]_{\eta_{f'}} \ar[rd] & (g')^* \circ a \circ g_* \circ f'_* \circ (g')^* & (g')^* \circ g'_* \circ a' \circ g^* \circ f_* \ar[u]_{\beta^\vee} \ar[ld]_\beta \ar[d] \\ & (g')^* \circ g'_* \circ a' \circ f'_* \circ (g')^* \ar[ld] \ar[u]_{\beta^\vee} & a' \circ g^* \circ f_* \ar[lld]^\beta \\ a' \circ f'_* \circ (g')^* } $$ In this diagram the two squares commute  Categories, Lemma \ref{categories-lemma-properties-2-cat-cats} or more simply the discussion preceding Categories, Definition \ref{categories-definition-horizontal-composition}. The triangle commutes by the discussion above. By the dual of Categories, Lemma \ref{categories-lemma-transformation-between-functors-and-adjoints} the square $$ \xymatrix{ \text{id} \ar[r] \ar[d] & g'_* \circ a' \circ g^* \circ f_* \ar[d]^\beta \\ g'_* \circ a' \circ g^* \circ f_* \ar[r]^{\beta^\vee} & a \circ g_* \circ f'_* \circ (g')^* } $$ commutes which implies the pentagon in the big diagram commutes. Since $\beta$ and $\beta^\vee$ are isomorphisms, and since going on the outside of the big diagram equals $\beta \circ \alpha \circ \eta_f$ by definition this proves the lemma.
Let $K^\vee$ be the ``dual'' to $K$, see Cohomology on Sites, Lemma \ref{sites-cohomology-lemma-dual-perfect-complex}. For $M \in D_\QCoh(\mathcal{O}_X)$ we have \begin{align*} \Hom_{D(\mathcal{O}_Y)}(Rf_*M, K \otimes^\mathbf{L}_{\mathcal{O}_Y} L) & = \Hom_{D(\mathcal{O}_Y)}( Rf_*M \otimes^\mathbf{L}_{\mathcal{O}_Y} K^\vee, L) \\ & = \Hom_{D(\mathcal{O}_X)}( M \otimes^\mathbf{L}_{\mathcal{O}_X} Lf^*K^\vee, a(L)) \\ & = \Hom_{D(\mathcal{O}_X)}(M, Lf^*K \otimes^\mathbf{L}_{\mathcal{O}_X} a(L)) \end{align*} Second equality by the definition of $a$ and the projection formula (Cohomology on Sites, Lemma \ref{sites-cohomology-lemma-projection-formula}) or the more general Derived Categories of Spaces, Lemma \ref{spaces-perfect-lemma-cohomology-base-change}. Hence the result by the Yoneda lemma.
In this proof we will write $f_*$ for $Rf_*$ and $f^*$ for $Lf^*$, etc, and we will write $\otimes$ for $\otimes^\mathbf{L}_{\mathcal{O}_X}$, etc. Let us write (\ref{equation-compare-with-pullback}) as the composition \begin{align*} f^*K \otimes a(\mathcal{O}_Y) & \to a(f_*(f^*K \otimes a(\mathcal{O}_Y))) \\ & \leftarrow a(K \otimes f_*a(\mathcal{O}_K)) \\ & \to a(K \otimes \mathcal{O}_Y) \\ & \to a(K) \end{align*} Here the first arrow is the unit $\eta_f$, the second arrow is $a$ applied to Cohomology on Sites, Equation (\ref{sites-cohomology-equation-projection-formula-map}) which is an isomorphism by Derived Categories of Spaces, Lemma \ref{spaces-perfect-lemma-cohomology-base-change}, the third arrow is $a$ applied to $\text{id}_K \otimes \text{Tr}_f$, and the fourth arrow is $a$ applied to the isomorphism $K \otimes \mathcal{O}_Y = K$. The proof of the lemma consists in showing that each of these maps gives rise to a commutative square as in the statement of the lemma. For $\eta_f$ and $\text{Tr}_f$ this is Lemmas \ref{lemma-unit-and-base-change} and \ref{lemma-trace-map-and-base-change}. For the arrow using Cohomology on Sites, Equation (\ref{sites-cohomology-equation-projection-formula-map}) this is Cohomology on Sites, Remark \ref{sites-cohomology-remark-compatible-with-diagram}. For the multiplication map it is clear. This finishes the proof.
Let $P$ be a perfect object of $D(\mathcal{O}_X)$. By Derived Categories of Spaces, Lemma \ref{spaces-perfect-lemma-flat-proper-perfect-direct-image-general} the complex $Rf_*P$ is perfect on $Y$. Let $K_i$ be a family of objects of $D_\QCoh(\mathcal{O}_Y)$. Then \begin{align*} \Hom_{D(\mathcal{O}_X)}(P, a(\bigoplus K_i)) & = \Hom_{D(\mathcal{O}_Y)}(Rf_*P, \bigoplus K_i) \\ & = \bigoplus \Hom_{D(\mathcal{O}_Y)}(Rf_*P, K_i) \\ & = \bigoplus \Hom_{D(\mathcal{O}_X)}(P, a(K_i)) \end{align*} because a perfect object is compact (Derived Categories of Spaces, Proposition \ref{spaces-perfect-proposition-compact-is-perfect}). Since $D_\QCoh(\mathcal{O}_X)$ has a perfect generator (Derived Categories of Spaces, Theorem \ref{spaces-perfect-theorem-bondal-van-den-Bergh}) we conclude that the map $\bigoplus a(K_i) \to a(\bigoplus K_i)$ is an isomorphism, i.e., $a$ commutes with direct sums.
By Lemma \ref{lemma-proper-flat} we know that $a$ commutes with direct sums. Hence the collection of objects of $D_\QCoh(\mathcal{O}_Y)$ for which (\ref{equation-compare-with-pullback}) is an isomorphism is a strictly full, saturated, triangulated subcategory of $D_\QCoh(\mathcal{O}_Y)$ which is moreover preserved under taking direct sums. Since $D_\QCoh(\mathcal{O}_Y)$ is a module category (Derived Categories of Spaces, Theorem \ref{spaces-perfect-theorem-DQCoh-is-Ddga}) generated by a single perfect object (Derived Categories of Spaces, Theorem \ref{spaces-perfect-theorem-bondal-van-den-Bergh}) we can argue as in More on Algebra, Remark \ref{more-algebra-remark-P-resolution} to see that it suffices to prove (\ref{equation-compare-with-pullback}) is an isomorphism for a single perfect object. However, the result holds for perfect objects, see Lemma \ref{lemma-compare-with-pullback-perfect}.
Consider $g : Y' \to Y$ \'etale with $Y'$ an affine scheme and denote $X' = Y' \times_Y X$ the base change. By Definition \ref{definition-relative-dualizing-proper-flat} and the discussion following, there is a unique isomorphism $\iota : (\omega_1^\bullet|_{X'}, \tau_1|_{Y'}) \to (\omega_2^\bullet|_{X'}, \tau_2|_{Y'})$. If $Y'' \to Y'$ is a further \'etale morphism of affines and $X'' = Y'' \times_Y X$, then $\iota|_{X''}$ is the unique isomorphism $(\omega_1^\bullet|_{X''}, \tau_1|_{Y''}) \to (\omega_2^\bullet|_{X''}, \tau_2|_{Y''})$ (by uniqueness). Also we have $$ \text{Ext}^p_{X'}(\omega_1^\bullet|_{X'}, \omega_2^\bullet|_{X'}) = 0, \quad p < 0 $$ because $\mathcal{O}_{X'} \cong R\SheafHom_{\mathcal{O}_{X'}}(\omega_1^\bullet|_{X'}, \omega_1^\bullet|_{X'}) \cong R\SheafHom_{\mathcal{O}_{X'}}(\omega_1^\bullet|_{X'}, \omega_2^\bullet|_{X'})$ by Lemma \ref{lemma-relative-dualizing-RHom}. \medskip\noindent Choose a \'etale hypercovering $b : V \to Y$ such that each $V_n = \coprod_{i \in I_n} Y_{n, i}$ with $Y_{n, i}$ affine. This is possible by Hypercoverings, Lemma \ref{hypercovering-lemma-hypercovering-object} and Remark \ref{hypercovering-remark-take-unions-hypercovering-X} (to replace the hypercovering produced in the lemma by the one having disjoint unions in each degree). Denote $X_{n, i} = Y_{n, i} \times_Y X$ and $U_n = V_n \times_Y X$ so that we obtain an \'etale hypercovering $a : U \to X$ (Hypercoverings, Lemma \ref{hypercovering-lemma-hypercovering-morphism-sites}) with $U_n = \coprod X_{n, i}$. The assumptions of Simplicial Spaces, Lemma \ref{spaces-simplicial-lemma-fppf-neg-ext-zero-hom} are satisfied for $a : U \to X$ and the complexes $\omega_1^\bullet$ and $\omega_2^\bullet$. Hence we obtain a unique morphism $\iota : \omega_1^\bullet \to \omega_2^\bullet$ whose restriction to $X_{0, i}$ is the unique isomorphism $(\omega_1^\bullet|_{X_{0, i}}, \tau_1|_{Y_{0, i}}) \to (\omega_2^\bullet|_{X_{0, i}}, \tau_2|_{Y_{0, i}})$ We still have to see that the diagram $$ \xymatrix{ Rf_*\omega_1^\bullet \ar[rd]_{\tau_1} \ar[rr]_{Rf_*\iota} & & Rf_*\omega_1^\bullet \ar[ld]^{\tau_2} \\ & \mathcal{O}_Y } $$ is commutative. However, we know that $Rf_*\omega_1^\bullet$ and $Rf_*\omega_2^\bullet$ have vanishing cohomology sheaves in positive degrees (Lemma \ref{lemma-relative-dualizing-RHom}) thus this commutativity may be proved after restricting to the affines $Y_{0, i}$ where it holds by construction.
Let $g : Y' \to Y$ and $X', f', g', a'$ be as in Definition \ref{definition-relative-dualizing-proper-flat}. Set $((\omega')^\bullet, \tau') = (L(g')^*\omega^\bullet, Lg^*\tau)$. We can find a finite \'etale covering $\{Y'_j \to Y'\}$ by affines which refines $\{Y_i \times_Y Y' \to Y'\}$ (Topologies, Lemma \ref{topologies-lemma-etale-affine}). Thus for each $j$ there is an $i_j$ and a morphism $k_j : Y'_j \to Y_{i_j}$ over $Y$. Consider the fibre products $$ \xymatrix{ X'_j \ar[r]_{h_j'} \ar[d]_{f'_j} & X' \ar[d]^{f'} \\ Y'_j \ar[r]^{h_j} & Y' } $$ Denote $k'_j : X'_j \to X_{i_j}$ the induced morphism (base change of $k_j$ by $f_{i_j}$). Restricting the given isomorphisms to $Y'_j$ via the morphism $k'_j$ we get isomorphisms of pairs $((\omega')^\bullet|_{X'_j}, \tau'|_{Y'_j}) \to (a_j(\mathcal{O}_{Y'_j}), \text{Tr}_{f'_j, \mathcal{O}_{Y'_j}})$. After replacing $f : X \to Y$ by $f' : X' \to Y'$ we reduce to the problem solved in the next paragraph. \medskip\noindent Assume $Y$ is affine. Problem: show $(\omega^\bullet, \tau)$ is isomorphic to $(\omega_{X/Y}^\bullet, \text{Tr}) = (a(\mathcal{O}_Y), \text{Tr}_{f, \mathcal{O}_Y})$. We may assume our covering $\{Y_i \to Y\}$ is given by a single surjective \'etale morphism $\{g : Y' \to Y\}$ of affines. Namely, we can first replace $\{g_i: Y_i \to Y\}$ by a finite subcovering, and then we can set $g = \coprod g_i :  Y' = \coprod Y_i \to Y$; some details omitted. Set $X' = Y' \times_Y X$ with maps $f', g'$ as in Definition \ref{definition-relative-dualizing-proper-flat}. Then all we're given is that we have an isomorphism $$ (\omega^\bullet|_{X'}, \tau|_{Y'}) \to (a'(\mathcal{O}_{Y'}), \text{Tr}_{f', \mathcal{O}_{Y'}}) $$ Since $(\omega_{X/Y}^\bullet, \text{Tr})$ is a relative dualizing complex (see discussion following Definition \ref{definition-relative-dualizing-proper-flat}) there is a unique isomorphism $$ (\omega_{X/Y}^\bullet|_{X'}, \text{Tr}|_{Y'}) \to (a'(\mathcal{O}_{Y'}), \text{Tr}_{f', \mathcal{O}_{Y'}}) $$ Uniqueness by Lemma \ref{lemma-uniqueness-relative-dualizing} for example. Combining the displayed isomorphisms we find an isomorphism $$ \alpha : (\omega^\bullet|_{X'}, \tau|_{Y'}) \to (\omega_{X/Y}^\bullet|_{X'}, \text{Tr}|_{Y'}) $$ Set $Y'' = Y' \times_Y Y'$ and $X'' = Y'' \times_Y X$ the two pullbacks of $\alpha$ to $X''$ have to be the same by uniqueness again. Since we have vanishing negative self exts for $\omega_{X'/Y'}^\bullet$ over $X'$ (Lemma \ref{lemma-relative-dualizing-RHom}) and since this remains true after pulling back by any projection $Y' \times_Y \ldots \times_Y Y' \to Y'$ (small detail omitted -- compare with the proof of Lemma \ref{lemma-uniqueness-relative-dualizing}), we find that $\alpha$ descends to an isomorphism $\omega^\bullet \to \omega_{X/Y}^\bullet$ over $X$ by Simplicial Spaces, Lemma \ref{spaces-simplicial-lemma-fppf-neg-ext-zero-hom}.