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

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

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

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 (75.21.0.1) are isomorphisms in case (1) by computing the source and target of (75.21.0.1) 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 (75.21.0.1) 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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