Let $X$ in $\mathcal{C}$ and $\mathcal{F}$ in $\mathcal{A}_X$. Let $\xi \in H^{n + 1}_\tau(U, \mathcal{F})$ for some $U/X$. We have to show that $\xi$ restricts to zero on the members of a $\tau'$-covering of $U$. See Lemma \ref{lemma-higher-direct-images}. It follows from this that we may replace $U$ by the members of a $\tau'$-covering of $U$. \medskip\noindent By locality of cohomology (Lemma \ref{lemma-kill-cohomology-class-on-covering}) we can choose a $\tau$-covering $\{U_i \to U\}$ such that $\xi$ restricts to zero on $U_i$. Choose $\{V_j \to V\}$, $\{f_j : W_j \to V_j\}$, and $\{W_{jk} \to W_j\}$ as in (\ref{item-refine-tau-by-P}). After replacing both $U$ by $V_j$ and $\mathcal{F}$ by its restriction to $\mathcal{C}_\tau/V_j$, which is allowed by (\ref{item-base-change-P}), we reduce to the case discussed in the next paragraph. \medskip\noindent Here $f : X \to Y$ is an element of $\mathcal{P}$ such that $\{X \to Y\}$ is a $\tau$-covering, $\mathcal{F}$ is an object of $\mathcal{A}_Y$, and $\xi \in H^{n + 1}_\tau(Y, \mathcal{F})$ is such that there exists a $\tau'$-covering $\{X_i \to X\}_{i \in I}$ such that $\xi$ restricts to zero on $X_i$ for all $i \in I$. Problem: show that $\xi$ restricts to zero on a $\tau'$-covering of $Y$. \medskip\noindent By Lemma \ref{lemma-V-implies-cohomology-extra-general} there exists a unique $\tau'$-cohomology class $\theta \in H^{n + 1}_{\tau'}(X, \epsilon_{X, *}\mathcal{F})$ whose image is $\xi|_X$. Since $\xi|_X$ pulls back to the same class on $X \times_Y X$ via the two projections, we find that the same is true for $\theta$ (by uniqueness). By Lemma \ref{lemma-make-class-zero-general} we see that after replacing $Y$ by the members of a $\tau'$-covering, we may assume that $\theta = 0$. Consequently, we may assume that $\xi|_X$ is zero. \medskip\noindent Let $f : X \to Y$ be an element of $\mathcal{P}$ such that $\{X \to Y\}$ is a $\tau$-covering, $\mathcal{F}$ is an object of $\mathcal{A}_Y$, and $\xi \in H^{n + 1}_\tau(Y, \mathcal{F})$ maps to zero in $H^{n + 1}_\tau(X, \mathcal{F})$. Problem: show that $\xi$ restricts to zero on a $\tau'$-covering of $Y$. \medskip\noindent The assumptions tell us $\xi$ maps to zero under the map $$ \mathcal{F} \longrightarrow Rf_{\tau, *}f_\tau^{-1}\mathcal{F} $$ Use Lemma \ref{lemma-Leray-unbounded}. A simple argument using the distinguished triangle of truncations (Derived Categories, Remark \ref{derived-remark-truncation-distinguished-triangle}) shows that $\xi$ maps to zero under the map $$ \mathcal{F} \longrightarrow \tau_{\leq n}Rf_{\tau, *}f_\tau^{-1}\mathcal{F} $$ We will compare this with the map $\epsilon_{Y, *}\mathcal{F} \to K$ where $$ K = \tau_{\leq n}Rf_{\tau', *}f_{\tau'}^{-1}\epsilon_{Y, *}\mathcal{F} = \tau_{\leq n}Rf_{\tau', *}\epsilon_{X, *}f_{\tau}^{-1}\mathcal{F} $$ The equality $\epsilon_{X, *} f_\tau^{-1} = f_{\tau'}^{-1} \epsilon_{Y, *}$ is a property of (\ref{equation-commutative-epsilon}). Consider the map $$ Rf_{\tau', *}\epsilon_{X, *}f_{\tau}^{-1}\mathcal{F} \longrightarrow Rf_{\tau', *}R\epsilon_{X, *}f_{\tau}^{-1}\mathcal{F} = R\epsilon_{Y, *}Rf_{\tau, *}f_\tau^{-1}\mathcal{F} $$ used in the proof of Lemma \ref{lemma-V-implies-C-general} which induces by adjunction a map $$ \epsilon_Y^{-1} Rf_{\tau', *}\epsilon_{X, *}f_{\tau}^{-1}\mathcal{F} \to Rf_{\tau, *}f_\tau^{-1}\mathcal{F} $$ Taking trunctions we find a map $$ \epsilon_Y^{-1}K \longrightarrow \tau_{\leq n}Rf_{\tau, *}f_\tau^{-1}\mathcal{F} $$ which is an isomorphism by Lemma \ref{lemma-V-implies-C-general}; the lemma applies because $f_\tau^{-1}\mathcal{F}$ is in $\mathcal{A}_X$ by Lemma \ref{lemma-A}. Choose a distinguished triangle $$ \epsilon_{Y, *}\mathcal{F} \to K \to L \to \epsilon_{Y, *}\mathcal{F}[1] $$ The map $\mathcal{F} \to f_{\tau, *}f_\tau^{-1}\mathcal{F}$ is injective as $\{X \to Y\}$ is a $\tau$-covering. Thus $\epsilon_{Y, *}\mathcal{F} \to \epsilon_{Y, *}f_{\tau, *}f_\tau^{-1}\mathcal{F} = f_{\tau', *}f_{\tau'}^{-1}\epsilon_{Y, *}\mathcal{F}$ is injective too.  Hence $L$ only has nonzero cohomology sheaves in degrees $0, \ldots, n$. As $f_{\tau', *}f_{\tau'}^{-1}\epsilon_{Y, *}\mathcal{F}$ is in $\mathcal{A}'_Y$ by (\ref{item-restriction-A}) and (\ref{item-A-and-P}) we conclude that $$ H^0(L) = \Coker(\epsilon_{Y, *}\mathcal{F} \to f_{\tau', *}f_{\tau'}^{-1}\epsilon_{Y, *}\mathcal{F}) $$ is in the weak Serre subcategory $\mathcal{A}'_Y$. For $1 \leq i \leq n$ we see that $H^i(L) = R^if_{\tau', *}f_{\tau'}^{-1}\epsilon_{Y, *}\mathcal{F}$ is in $\mathcal{A}'_Y$ by (\ref{item-restriction-A}) and (\ref{item-A-and-P}). Pulling back the distinguished triangle above by $\epsilon_Y$ we get the distinguished triangle $$ \mathcal{F} \to \tau_{\leq n}Rf_{\tau, *}f_\tau^{-1}\mathcal{F} \to \epsilon_Y^{-1}L \to \mathcal{F}[1] $$ Since $\xi$ maps to zero in the middle term we find that $\xi$ is the image of an element $\xi' \in H^n_\tau(Y, \epsilon_Y^{-1}L)$. By Lemma \ref{lemma-V-implies-cohomology-general} we have $$ H^n_{\tau'}(Y, L) = H^n_\tau(Y, \epsilon_Y^{-1}L), $$ Thus we may lift $\xi'$ to an element of $H^n_{\tau'}(Y, L)$ and take the boundary into $H^{n + 1}_{\tau'}(Y, \epsilon_{Y, *}\mathcal{F})$ to see that $\xi$ is in the image of the canonical map $H^{n + 1}_{\tau'}(Y, \epsilon_{Y, *}\mathcal{F}) \to H^{n + 1}_\tau(Y, \mathcal{F})$. By locality of cohomology for $H^{n + 1}_{\tau'}(Y,\epsilon_{Y, *}\mathcal{F})$, see Lemma \ref{lemma-kill-cohomology-class-on-covering}, we conclude.