$\xymatrix{ X' \ar[r]_{g'} \ar[d]_{f'} & X \ar[d]^ f \\ S' \ar[r]^ g & S }$

be a cartesian diagram of schemes. Let $K \in D_\mathit{QCoh}(\mathcal{O}_ X)$ and let $L(g')^*K \to K'$ be a map in $D_\mathit{QCoh}(\mathcal{O}_{X'})$. If

1. the equivalent conditions of Lemma 36.26.1 hold, and

2. $f$ is quasi-compact and quasi-separated,

then the composition $Lg^*Rf_*K \to Rf'_*L(g')^*K \to Rf'_*K'$ is an isomorphism.

Proof. We could prove this using the same method as in the proof of Lemma 36.22.5 but instead we will prove it using the induction principle and relative Mayer-Vietoris.

To check the map is an isomorphism we may work locally on $S'$. Hence we may assume $g : S' \to S$ is a morphism of affine schemes. In particular $X$ is a quasi-compact and quasi-separated scheme. We will use the induction principle of Cohomology of Schemes, Lemma 30.4.1 to prove that for any quasi-compact open $U \subset X$ the similarly constructed map $Lg^*R(U \to S)_*K|_ U \to R(U' \to S')_*K'|_{U'}$ is an isomorphism. Here $U' = (g')^{-1}(U)$.

If $U \subset X$ is an affine open, then we find that the result is true by assumption, see Lemma 36.26.1 part (2) and the translation into algebra afforded to us by Lemmas 36.3.5 and 36.3.8.

The induction step. Suppose that $X = U \cup V$ is an open covering with $U$, $V$, $U \cap V$ quasi-compact such that the result holds for $U$, $V$, and $U \cap V$. Denote $a = f|_ U$, $b = f|_ V$ and $c = f|_{U \cap V}$. Let $a' : U' \to S'$, $b' : V' \to S'$ and $c' : U' \cap V' \to S'$ be the base changes of $a$, $b$, and $c$. Using the distinguished triangles from relative Mayer-Vietoris (Cohomology, Lemma 20.33.5) we obtain a commutative diagram

$\xymatrix{ Lg^*Rf_*K \ar[r] \ar[d] & Rf'_* K' \ar[d] \\ Lg^*Ra_* K|_ U \oplus Lg^*Rb_* K|_ V \ar[r] \ar[d] & Ra'_* K'|_{U'} \oplus Rb'_* K'|_{V'} \ar[d] \\ Lg^*Rc_* K|_{U \cap V} \ar[r] \ar[d] & Rc'_* K'|_{U' \cap V'} \ar[d] \\ Lg^*Rf_* K[1] \ar[r] & Rf'_* K'[1] }$

Since the 2nd and 3rd horizontal arrows are isomorphisms so is the first (Derived Categories, Lemma 13.4.3) and the proof of the lemma is finished. $\square$

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