The Stacks project

Proof. Write $Y = \mathop{\mathrm{Spec}}(A)$ and $Y' = \mathop{\mathrm{Spec}}(A')$. As a base change of an affine morphism, the morphism $g'$ is affine. Let $M$ be a perfect generator for $D_\mathit{QCoh}(\mathcal{O}_ X)$, see Derived Categories of Schemes, Theorem 36.15.3. Then $L(g')^*M$ is a generator for $D_\mathit{QCoh}(\mathcal{O}_{X'})$, see Derived Categories of Schemes, Remark 36.16.4. Hence it suffices to show that (48.5.0.1) induces an isomorphism

of global hom complexes, see Cohomology, Section 20.44, as this will imply the cone of $L(g')^*a(K) \to a'(Lg^*K)$ is zero. The structure of the proof is as follows: we will first show that these Hom complexes are isomorphic and in the last part of the proof we will show that the isomorphism is induced by (48.6.2.1).

The left hand side. Because $M$ is perfect, the canonical map

is an isomorphism by Derived Categories of Schemes, Lemma 36.22.6. We can combine this with the isomorphism $R\mathop{\mathrm{Hom}}\nolimits _ Y(Rf_*M, K) = R\mathop{\mathrm{Hom}}\nolimits _ X(M, a(K))$ of Lemma 48.3.10 to get that the left hand side equals $R\mathop{\mathrm{Hom}}\nolimits _ Y(Rf_*M, K) \otimes ^\mathbf {L}_ A A'$.

The right hand side. Here we first use the isomorphism

of Lemma 48.3.10. Then we use the base change map $Lg^*Rf_*M \to Rf'_*L(g')^*M$ is an isomorphism by Derived Categories of Schemes, Lemma 36.22.5. Hence we may rewrite this as $R\mathop{\mathrm{Hom}}\nolimits _{Y'}(Lg^*Rf_*M, Lg^*K)$. Since $Y$, $Y'$ are affine and $K$, $Rf_*M$ are in $D_\mathit{QCoh}(\mathcal{O}_ Y)$ (Derived Categories of Schemes, Lemma 36.4.1) we have a canonical map

in $D(A')$. This is the arrow More on Algebra, Equation (15.99.1.1) where we have used Derived Categories of Schemes, Lemmas 36.3.5 and 36.10.8 to translate back and forth into algebra.

1. If $f$ is flat and of finite presentation, the complex $Rf_*M$ is perfect on $Y$ by Derived Categories of Schemes, Lemma 36.30.4 and $\beta$ is an isomorphism by More on Algebra, Lemma 15.99.2 part (1).

2. If $f$ is perfect and $Y$ Noetherian, the complex $Rf_*M$ is perfect on $Y$ by More on Morphisms, Lemma 37.61.13 and $\beta$ is an isomorphism as before.

3. If $g$ has finite tor dimension and $Y$ is Noetherian, the complex $Rf_*M$ is pseudo-coherent on $Y$ (Derived Categories of Schemes, Lemmas 36.11.3 and 36.10.3) and $\beta$ is an isomorphism by More on Algebra, Lemma 15.99.2 part (4).

We conclude that we obtain the same answer as in the previous paragraph.

In the rest of the proof we show that the identifications of the left and right hand side of (48.6.2.1) given in the second and third paragraph are in fact given by (48.6.2.1). To make our formulas manageable we will use $(-, -)_ X = R\mathop{\mathrm{Hom}}\nolimits _ X(-, -)$, use $- \otimes A'$ in stead of $- \otimes _ A^\mathbf {L} A'$, and we will abbreviate $g^* = Lg^*$ and $f_* = Rf_*$. Consider the following commutative diagram

The arrows labeled $\alpha$ are the maps from Derived Categories of Schemes, Lemma 36.22.6 for the diagram with corners $X', X, Y', Y$. The upper part of the diagram is commutative as the horizontal arrows are functorial in the entries. The middle vertical arrows come from the invertible transformation $g'_* \circ a' \to a \circ g_*$ of Lemma 48.4.1 and therefore the middle square is commutative. Going down the left hand side is (48.6.2.1). The upper horizontal arrows provide the identifications used in the second paragraph of the proof. The lower horizontal arrows including $\beta$ provide the identifications used in the third paragraph of the proof. Given $E \in D(A)$, $E' \in D(A')$, and $c : E \to E'$ in $D(A)$ we will denote $\mu _ c : E \otimes A' \to E'$ the map induced by $c$ and the adjointness of restriction and base change; if $c$ is clear we write $\mu = \mu _ c$, i.e., we drop $c$ from the notation. The map $\mu$ in the diagram is of this form with $c$ given by the identification $(M, g'_*a(g^*K))_ X = ((g')^*M, a'(g^*K))_{X'}$ ; the triangle involving $\mu$ is commutative by Derived Categories of Schemes, Remark 36.22.7.

Observe that

is commutative by the very definition of the transformation $g'_* \circ a' \to a \circ g_*$. Letting $\mu '$ be as above corresponding to the identification $(f_*M, g_*g^*K)_ X = (g^*f_*M, g^*K)_{Y'}$, then the hexagon commutes as well. Thus it suffices to show that $\beta$ is equal to the composition of $(f_*M, K)_ Y \otimes A' \to (f_*M, g_*g^*K)_ X \otimes A'$ and $\mu '$. To do this, it suffices to prove the two induced maps $(f_*M, K)_ Y \to (g^*f_*M, g^*K)_{Y'}$ are the same. In other words, it suffices to show the diagram

commutes for all $E, K \in D(A)$. Since this is how $\beta$ is constructed in More on Algebra, Section 15.99 the proof is complete. $\square$