Lemma 21.29.7. In Situation 21.29.1 we have $(V_ n) \Rightarrow (V_{n + 1})$.

**Proof.**
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 21.7.4. It follows from this that we may replace $U$ by the members of a $\tau '$-covering of $U$.

By locality of cohomology (Lemma 21.7.3) 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 (5). After replacing both $U$ by $V_ j$ and $\mathcal{F}$ by its restriction to $\mathcal{C}_\tau /V_ j$, which is allowed by (1), we reduce to the case discussed in the next paragraph.

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 $Y$.

By Lemma 21.29.5 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 21.29.6 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.

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 $Y$.

The assumptions tell us $\xi $ maps to zero under the map

Use Lemma 21.20.5. A simple argument using the distinguished triangle of truncations (Derived Categories, Remark 13.12.4) shows that $\xi $ maps to zero under the map

We will compare this with the map $\epsilon _{Y, *}\mathcal{F} \to K$ where

The equality $\epsilon _{X, *} f_\tau ^{-1} = f_{\tau '}^{-1} \epsilon _{Y, *}$ is a property of (21.29.0.1). Consider the map

used in the proof of Lemma 21.29.3 which induces by adjunction a map

Taking trunctions we find a map

which is an isomorphism by Lemma 21.29.3; the lemma applies because $f_\tau ^{-1}\mathcal{F}$ is in $\mathcal{A}_ X$ by Lemma 21.29.2. Choose a distinguished triangle

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 (2) and (4) we conclude that

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 (2) and (4). Pulling back the distinguished triangle above by $\epsilon _ Y$ we get the distinguished triangle

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 21.29.4 we have

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 21.7.3, we conclude. $\square$

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