Lemma 48.8.2. Suppose we have a diagram (48.4.0.1) where f and g are tor independent. Let K \in D_\mathit{QCoh}(\mathcal{O}_ Y). The diagram
\xymatrix{ L(g')^*(Lf^*K \otimes ^\mathbf {L}_{\mathcal{O}_ X} a(\mathcal{O}_ Y)) \ar[r] \ar[d] & L(g')^*a(K) \ar[d] \\ L(f')^*Lg^*K \otimes _{\mathcal{O}_{X'}}^\mathbf {L} a'(\mathcal{O}_{Y'}) \ar[r] & a'(Lg^*K) }
commutes where the horizontal arrows are the maps (48.8.0.1) for K and Lg^*K and the vertical maps are constructed using Cohomology, Remark 20.28.3 and (48.5.0.1).
Proof.
In this proof we will write f_* for Rf_* and f^* for Lf^*, etc, and we will write \otimes for \otimes ^\mathbf {L}_{\mathcal{O}_ X}, etc. Let us write (48.8.0.1) as the composition
\begin{align*} f^*K \otimes a(\mathcal{O}_ Y) & \to a(f_*(f^*K \otimes a(\mathcal{O}_ Y))) \\ & \leftarrow a(K \otimes f_*a(\mathcal{O}_ K)) \\ & \to a(K \otimes \mathcal{O}_ Y) \\ & \to a(K) \end{align*}
Here the first arrow is the unit \eta _ f, the second arrow is a applied to Cohomology, Equation (20.54.2.1) which is an isomorphism by Derived Categories of Schemes, Lemma 36.22.1, the third arrow is a applied to \text{id}_ K \otimes \text{Tr}_ f, and the fourth arrow is a applied to the isomorphism K \otimes \mathcal{O}_ Y = K. The proof of the lemma consists in showing that each of these maps gives rise to a commutative square as in the statement of the lemma. For \eta _ f and \text{Tr}_ f this is Lemmas 48.7.2 and 48.7.1. For the arrow using Cohomology, Equation (20.54.2.1) this is Cohomology, Remark 20.54.5. For the multiplication map it is clear. This finishes the proof.
\square
Comments (0)