The Stacks project

Lemma 75.21.3. Let $S$ be a scheme. Let

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

be a cartesian diagram of algebraic spaces over $S$. 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 the equivalent conditions of Lemma 75.21.1 hold, then

  1. for $E \in D_\mathit{QCoh}(\mathcal{O}_ X)$ the equivalent conditions of Lemma 75.21.1 hold for $L(g')^*(E \otimes ^\mathbf {L} K) \to L(g')^*E \otimes ^\mathbf {L} K'$,

  2. if $E$ in $D(\mathcal{O}_ X)$ is perfect the equivalent conditions of Lemma 75.21.1 hold for $L(g')^*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (E, K) \to R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (L(g')^*E, K')$, and

  3. if $K$ is bounded below and $E$ in $D(\mathcal{O}_ X)$ pseudo-coherent the equivalent conditions of Lemma 75.21.1 hold for $L(g')^*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (E, K) \to R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (L(g')^*E, K')$.

Proof. The statement makes sense as the complexes involved have quasi-coherent cohomology sheaves by Lemmas 75.5.5, 75.5.6, and 75.13.10 and Cohomology on Sites, Lemmas 21.45.3 and 21.47.5. Having said this, we can check the maps ( are isomorphisms in case (1) by computing the source and target of ( using the transitive property of tensor product, see More on Algebra, Lemma 15.59.15. The map in (2) and (3) is the composition

\[ L(g')^*R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (E, K) \to R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (L(g')^*E, L(g')^*K) \to R\mathop{\mathcal{H}\! \mathit{om}}\nolimits (L(g')^*E, K') \]

where the first arrow is Cohomology on Sites, Remark 21.35.11 and the second arrow comes from the given map $L(g')^*K \to K'$. To prove the maps ( are isomorphisms one represents $E_ x$ by a bounded complex of finite projective $\mathcal{O}_{X. x}$-modules in case (2) or by a bounded above complex of finite free modules in case (3) and computes the source and target of the arrow. Some details omitted. $\square$

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