The Stacks project

Proof. Let $(M, K', \alpha )$ be an object of the target of the functor of the lemma. Here $\alpha : Lf^*M \to Li^*K'$ is an isomorphism which is adjoint to a map $\beta : M \to Rf_*Li^*K'$. Thus we obtain maps

where the arrow pointing left comes from $K' \to Ri_*Li^*K'$. Choose a distinguished triangle

in $D(\mathcal{O}_{Y'})$. The first arrow defines canonical maps $Lj^*M' \to M$ and $L(f')^*M' \to K'$ compatible with $\alpha$. Thus it suffices to show that the maps $Lj^*M' \to M$ and $L(f')^*M' \to K$ are isomorphisms. This we may check étale locally on $Y'$, hence we may assume $Y'$ is étale.

Assume $Y'$ affine and $M \in D(\mathcal{O}_ Y)$ and $K' \in D(\mathcal{O}_{X'})$ are pseudo-coherent. Say our pushout corresponds to the fibre product

of rings where $B' \to B$ is surjective with locally nilpotent kernel $I$ (and hence $A' \to A$ is surjective with locally nilpotent kernel $I$ as well). The assumption on $M$ and $K'$ imply that $M$ comes from a pseudo-coherent object of $D(A)$ and $K'$ comes from a pseudo-coherent object of $D(B')$, see Derived Categories of Spaces, Lemmas 75.13.6, 75.4.2, and 75.13.2 and Derived Categories of Schemes, Lemma 36.3.5 and 36.10.2. Moreover, pushforward and derived pullback agree with the corresponding operations on derived categories of modules, see Derived Categories of Spaces, Remark 75.6.3 and Derived Categories of Schemes, Lemmas 36.3.7 and 36.3.8. This reduces us to the statement formulated in the next paragraph. (To be sure these references show the object $M'$ lies $D_\mathit{QCoh}(\mathcal{O}_{Y'})$ as this is a triangulated subcategory of $D(\mathcal{O}_{Y'})$.)

Given a diagram of rings as above and a triple $(M, K', \alpha )$ where $M \in D(A)$, $K' \in D(B')$ are pseudo-coherent and $\alpha : M \otimes _ A^\mathbf {L} B \to K' \otimes _{B'}^\mathbf {L} B$ is an isomorphism suppose we have distinguished triangle

in $D(A')$. Goal: show that the induced maps $M' \otimes _{A'}^\mathbf {L} A \to M$ and $M' \otimes _{A'}^\mathbf {L} B' \to K'$ are isomorphisms. To do this, choose a bounded above complex $E^\bullet$ of finite free $A$-modules representing $M$. Since $(B', I)$ is a henselian pair (More on Algebra, Lemma 15.11.2) with $B = B'/I$ we may apply More on Algebra, Lemma 15.75.9 to see that there exists a bounded above complex $P^\bullet$ of free $B'$-modules such that $\alpha$ is represented by an isomorphism $E^\bullet \otimes _ A B \cong P^\bullet \otimes _{B'} B$. Then we can consider the short exact sequence

of complexes of $B'$-modules. More on Algebra, Lemma 15.6.9 implies $L^\bullet$ is a bounded above complex of finite projective $A'$-modules (in fact it is rather easy to show directly that $L^ n$ is finite free in our case) and that we have $L^\bullet \otimes _{A'} A = E^\bullet$ and $L^\bullet \otimes _{A'} B' = P^\bullet$. The short exact sequence gives a distinguished triangle

in $D(A')$ (Derived Categories, Section 13.12) which is isomorphic to the given distinguished triangle by general properties of triangulated categories (Derived Categories, Section 13.4). In other words, $L^\bullet$ represents $M'$ compatibly with the given maps. Thus the maps $M' \otimes _{A'}^\mathbf {L} A \to M$ and $M' \otimes _{A'}^\mathbf {L} B' \to K'$ are isomorphisms because we just saw that the corresponding thing is true for $L^\bullet$. $\square$