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 (20.50.2.1) 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 (20.50.2.1) this is Cohomology, Remark 20.50.5. For the multiplication map it is clear. This finishes the proof.
$\square$

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