Lemma 91.7.2. With notation as in (91.7.0.2) suppose that $C$ is a polynomial algebra over $B$. Then $L\pi _!(g_3^{-1}\mathcal{F}) = L\pi _{3, !}\mathcal{F} = \pi _{3, !}\mathcal{F}$ for any abelian sheaf $\mathcal{F}$ on $\mathcal{C}_{C/B}$

**Proof.**
Write $C = B[E]$ for some set $E$. Choose a resolution $P_\bullet \to B$ of $B$ over $A$. For every $n$ consider the object $U_ n = (P_ n \to B, P_ n[E] \to C)$ of $\mathcal{C}_{C/B/A}$. Then $U_\bullet $ is a cosimplicial object of $\mathcal{C}_{C/B/A}$. Note that $u_3(U_\bullet )$ is the constant cosimplicial object of $\mathcal{C}_{C/B}$ with value $(C \to C)$. We will prove that the object $U_\bullet $ of $\mathcal{C}_{C/B/A}$ satisfies the hypotheses of Cohomology on Sites, Lemma 21.39.7. This implies the lemma as it shows that $L\pi _!(g_3^{-1}\mathcal{F})$ is computed by the constant simplicial abelian group $\mathcal{F}(C \to C)$ which is the value of $L\pi _{3, !}\mathcal{F} = \pi _{3, !}\mathcal{F}$ by Lemma 91.4.6.

Let $U = (\beta : P \to B, \gamma : Q \to C)$ be an object of $\mathcal{C}_{C/B/A}$. We may write $P = A[S]$ and $Q = A[S \amalg T]$ by the definition of our category $\mathcal{C}_{C/B/A}$. We have to show that

is homotopy equivalent to a singleton simplicial set $*$. Observe that this simplicial set is the product

where $F_ s$ is the corresponding simplicial set for $U_ s = (A[\{ s\} ] \to B, A[\{ s\} ] \to C)$ and $F'_ t$ is the corresponding simplicial set for $U_ t = (A \to B, A[\{ t\} ] \to C)$. Namely, the object $U$ is the product $\prod U_ s \times \prod U_ t$ in $\mathcal{C}_{C/B/A}$. It suffices each $F_ s$ and $F'_ t$ is homotopy equivalent to $*$, see Simplicial, Lemma 14.26.10. The case of $F_ s$ follows as $P_\bullet \to B$ is a trivial Kan fibration (as a resolution) and $F_ s$ is the fibre of this map over $\beta (s)$. (Use Simplicial, Lemmas 14.30.3 and 14.30.8). The case of $F'_ t$ is more interesting. Here we are saying that the fibre of

over $\gamma (t) \in C$ is homotopy equivalent to a point. In fact we will show this map is a trivial Kan fibration. Namely, $P_\bullet \to B$ is a trivial can fibration. For any ring $R$ we have

(filtered colimit). Thus the displayed map of simplicial sets is a filtered colimit of trivial Kan fibrations, whence a trivial Kan fibration by Simplicial, Lemma 14.30.7. $\square$

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