Lemma 21.29.3. In Situation 21.29.1 assume $(V_ n)$ holds. For $f : X \to Y$ in $\mathcal{P}$ and $\mathcal{F}$ in $\mathcal{A}_ X$ we have $R^ if_{\tau ', *}\epsilon _{X, *}\mathcal{F} = \epsilon _{Y, *}R^ if_{\tau , *}\mathcal{F}$ for $i \leq n$.

Proof. We will use the commutative diagram (21.29.0.1) without further mention. In particular have

$Rf_{\tau ', *}R\epsilon _{X, *}\mathcal{F} = R\epsilon _{Y, *}Rf_{\tau , *}\mathcal{F}$

Assumption $(V_ n)$ tells us that $\epsilon _{X, *}\mathcal{F} \to R\epsilon _{X, *}\mathcal{F}$ is an isomorphism in degrees $\leq n$. Hence $Rf_{\tau ', *}\epsilon _{X, *}\mathcal{F} \to Rf_{\tau ', *}R\epsilon _{X, *}\mathcal{F}$ is an isomorphism in degrees $\leq n$. We conclude that

$R^ if_{\tau ', *}\epsilon _{X, *}\mathcal{F} \to H^ i(R\epsilon _{Y, *}Rf_{\tau , *}\mathcal{F})$

is an isomorphism for $i \leq n$. We will prove the lemma by looking at the second page of the spectral sequence of Lemma 21.14.7 for $R\epsilon _{Y, *}Rf_{\tau , *}\mathcal{F}$. Here is a picture:

$\begin{matrix} \ldots & \ldots & \ldots & \ldots \\ \epsilon _{Y, *}R^2f_{\tau , *}\mathcal{F} & R^1\epsilon _{Y, *}R^2f_{\tau , *}\mathcal{F} & R^2\epsilon _{Y, *}R^2f_{\tau , *}\mathcal{F} & \ldots \\ \epsilon _{Y, *}R^1f_{\tau , *}\mathcal{F} & R^1\epsilon _{Y, *}R^1f_{\tau , *}\mathcal{F} & R^2\epsilon _{Y, *}R^1f_{\tau , *}\mathcal{F} & \ldots \\ \epsilon _{Y, *}f_{\tau , *}\mathcal{F} & R^1\epsilon _{Y, *}f_{\tau , *}\mathcal{F} & R^2\epsilon _{Y, *}f_{\tau , *}\mathcal{F} & \ldots \end{matrix}$

Let $(C_ m)$ be the hypothesis: $R^ if_{\tau ', *}\epsilon _{X, *}\mathcal{F} = \epsilon _{Y, *}R^ if_{\tau , *}\mathcal{F}$ for $i \leq m$. Observe that $(C_0)$ holds. We will show that $(C_{m - 1}) \Rightarrow (C_ m)$ for $m < n$. Namely, if $(C_{m - 1})$ holds, then for $n \geq p > 0$ and $q \leq m - 1$ we have

\begin{align*} R^ p\epsilon _{Y, *}R^ qf_{\tau , *}\mathcal{F} & = R^ p\epsilon _{Y, *} \epsilon _ Y^{-1} \epsilon _{Y, *} R^ qf_{\tau , *}\mathcal{F} \\ & = R^ p\epsilon _{Y, *} \epsilon _ Y^{-1}R^ qf_{\tau ', *}\epsilon _{X, *}\mathcal{F} = 0 \end{align*}

First equality as $\epsilon _ Y^{-1}\epsilon _{Y, *} = \text{id}$, the second by $(C_{m - 1})$, and the final by by $(V_ n)$ because $\epsilon _ Y^{-1}R^ qf_{\tau ', *}\epsilon _{X, *}\mathcal{F}$ is in $\mathcal{A}_ Y$ by (4). Looking at the spectral sequence we see that $E_2^{0, m} = \epsilon _{Y, *}R^ mf_{\tau , *}\mathcal{F}$ is the only nonzero term $E_2^{p, q}$ with $p + q = m$. Recall that $\text{d}_ r^{p, q} : E_ r^{p, q} \to E_ r^{p + r, q - r + 1}$. Hence there are no nonzero differentials $\text{d}_ r^{p, q}$, $r \geq 2$ either emanating or entering this spot. We conclude that $H^ m(R\epsilon _{Y, *}Rf_{\tau , *}\mathcal{F}) = \epsilon _{Y, *}R^ mf_{\tau , *}\mathcal{F}$ which implies $(C_ m)$ by the discussion above.

Finally, assume $(C_{n - 1})$. The same analysis shows that $E_2^{0, n} = \epsilon _{Y, *}R^ nf_{\tau , *}\mathcal{F}$ is the only nonzero term $E_2^{p, q}$ with $p + q = n$. We do still have no nonzero differentials entering this spot, but there can be a nonzero differential emanating it. Namely, the map $d_{n + 1}^{0, n} : \epsilon _{Y, *}R^ nf_{\tau , *}\mathcal{F} \to R^{n + 1}\epsilon _{Y, *}f_{\tau , *}\mathcal{F}$. We conclude that there is an exact sequence

$0 \to R^ nf_{\tau ', *}\epsilon _{X, *}\mathcal{F} \to \epsilon _{Y, *}R^ nf_{\tau , *}\mathcal{F} \to R^{n + 1}\epsilon _{Y, *}f_{\tau , *}\mathcal{F}$

By (4) and (3) the sheaf $R^ nf_{\tau ', *}\epsilon _{X, *}\mathcal{F}$ satisfies the sheaf property for $\tau$-coverings as does $\epsilon _{Y, *}R^ nf_{\tau , *}\mathcal{F}$ (use the description of $\epsilon _*$ in Section 21.26). However, the $\tau$-sheafification of the $\tau '$-sheaf $R^{n + 1}\epsilon _{Y, *}f_{\tau , *}\mathcal{F}$ is zero (by locality of cohomology; use Lemmas 21.7.3 and 21.7.4). Thus $R^ nf_{\tau ', *}\epsilon _{X, *}\mathcal{F} \to \epsilon _{Y, *}R^ nf_{\tau , *}\mathcal{F}$ has to be an isomorphism and the proof is complete. $\square$

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