Lemma 91.7.3. With notation as in (91.7.0.2) we have $Lg_{i, !} \circ g_ i^{-1} = \text{id}$ for $i = 1, 2, 3$ and hence also $L\pi _! \circ g_ i^{-1} = L\pi _{i, !}$ for $i = 1, 2, 3$.

Proof. Proof for $i = 1$. We claim the functor $\mathcal{C}_{C/B/A}$ is a fibred category over $\mathcal{C}_{B/A}$ Namely, suppose given $(P \to B, Q \to C)$ and a morphism $(P' \to B) \to (P \to B)$ of $\mathcal{C}_{B/A}$. Recall that this means we have an $A$-algebra homomorphism $P \to P'$ compatible with maps to $B$. Then we set $Q' = Q \otimes _ P P'$ with induced map to $C$ and the morphism

$(P' \to B, Q' \to C) \longrightarrow (P \to B, Q \to C)$

in $\mathcal{C}_{C/B/A}$ (note reversal arrows again) is strongly cartesian in $\mathcal{C}_{C/B/A}$ over $\mathcal{C}_{B/A}$. Moreover, observe that the fibre category of $u_1$ over $P \to B$ is the category $\mathcal{C}_{C/P}$. Let $\mathcal{F}$ be an abelian sheaf on $\mathcal{C}_{B/A}$. Since we have a fibred category we may apply Cohomology on Sites, Lemma 21.40.2. Thus $L_ ng_{1, !}g_1^{-1}\mathcal{F}$ is the (pre)sheaf which assigns to $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}_{B/A})$ the $n$th homology of $g_1^{-1}\mathcal{F}$ restricted to the fibre category over $U$. Since these restrictions are constant the desired result follows from Lemma 91.4.4 via our identifications of fibre categories above.

The case $i = 2$. We claim $\mathcal{C}_{C/B/A}$ is a fibred category over $\mathcal{C}_{C/A}$ is a fibred category. Namely, suppose given $(P \to B, Q \to C)$ and a morphism $(Q' \to C) \to (Q \to C)$ of $\mathcal{C}_{C/A}$. Recall that this means we have a $B$-algebra homomorphism $Q \to Q'$ compatible with maps to $C$. Then

$(P \to B, Q' \to C) \longrightarrow (P \to B, Q \to C)$

is strongly cartesian in $\mathcal{C}_{C/B/A}$ over $\mathcal{C}_{C/A}$. Note that the fibre category of $u_2$ over $Q \to C$ has an final (beware reversal arrows) object, namely, $(A \to B, Q \to C)$. Let $\mathcal{F}$ be an abelian sheaf on $\mathcal{C}_{C/A}$. Since we have a fibred category we may apply Cohomology on Sites, Lemma 21.40.2. Thus $L_ ng_{2, !}g_2^{-1}\mathcal{F}$ is the (pre)sheaf which assigns to $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}_{C/A})$ the $n$th homology of $g_1^{-1}\mathcal{F}$ restricted to the fibre category over $U$. Since these restrictions are constant the desired result follows from Cohomology on Sites, Lemma 21.39.5 because the fibre categories all have final objects.

The case $i = 3$. In this case we will apply Cohomology on Sites, Lemma 21.40.3 to $u = u_3 : \mathcal{C}_{C/B/A} \to \mathcal{C}_{C/B}$ and $\mathcal{F}' = g_3^{-1}\mathcal{F}$ for some abelian sheaf $\mathcal{F}$ on $\mathcal{C}_{C/B}$. Suppose $U = (\overline{Q} \to C)$ is an object of $\mathcal{C}_{C/B}$. Then $\mathcal{I}_ U = \mathcal{C}_{\overline{Q}/B/A}$ (again beware of reversal of arrows). The sheaf $\mathcal{F}'_ U$ is given by the rule $(P \to B, Q \to \overline{Q}) \mapsto \mathcal{F}(Q \otimes _ P B \to C)$. In other words, this sheaf is the pullback of a sheaf on $\mathcal{C}_{\overline{Q}/C}$ via the morphism $\mathop{\mathit{Sh}}\nolimits (\mathcal{C}_{\overline{Q}/B/A}) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{C}_{\overline{Q}/B})$. Thus Lemma 91.7.2 shows that $H_ n(\mathcal{I}_ U, \mathcal{F}'_ U) = 0$ for $n > 0$ and equal to $\mathcal{F}(\overline{Q} \to C)$ for $n = 0$. The aforementioned Cohomology on Sites, Lemma 21.40.3 implies that $Lg_{3, !}(g_3^{-1}\mathcal{F}) = \mathcal{F}$ and the proof is done. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 08QW. Beware of the difference between the letter 'O' and the digit '0'.