The Stacks project

Proof. Observe that the corresponding map

is a bijection, see Pushouts of Spaces, Lemma 81.6.1. Thus using the commutative diagram

we see that we may assume that $Y'$ is a scheme over $B'$. By Remark 99.16.7 we may replace $B$ by $Y'$ and $X$ by $X \times _ B Y'$. Thus we may assume $B = Y'$.

Assume $B = Y'$. We first prove fully faithfulness of our functor. To do this, let $\xi _1, \xi _2$ be two objects of $\mathcal{C}\! \mathit{omplexes}_{X/B}$ over $Y'$. Then we have to show that

is bijective. However, we already know that $\mathit{Isom}(\xi _1, \xi _2)$ is an algebraic space over $B = Y'$. Thus this bijectivity follows from Artin's Axioms, Lemma 98.4.1 (or the aforementioned Pushouts of Spaces, Lemma 81.6.1).

Essential surjectivity. Let $(E_ Y, E_{Z'}, \alpha )$ be a triple, where $E_ Y \in D(\mathcal{O}_ Y)$ and $E_{Z'} \in D(\mathcal{O}_{X_{Z'}})$ are objects such that $(Y, Y \to B, E_ Y)$ is an object of $\mathcal{C}\! \mathit{omplexes}_{X/B}$ over $Y$, such that $(Z', Z' \to B, E_{Z'})$ is an object of $\mathcal{C}\! \mathit{omplexes}_{X/B}$ over $Z'$, and $\alpha : L(X_ Z \to X_ Y)^*E_ Y \to L(X_ Z \to X_{Z'})^*E_{Z'}$ is an isomorphism in $D(\mathcal{O}_{Z'})$. That is to say

is an object of the target of the arrow of our lemma. Observe that the diagram

is a pushout with $X_ Z \to X_ Y$ affine and $X_ Z \to X_{Z'}$ a thickening (see Pushouts of Spaces, Lemma 81.6.7). Hence by Pushouts of Spaces, Lemma 81.8.1 we find an object $E_{Y'} \in D(\mathcal{O}_{X_{Y'}})$ together with isomorphisms $L(X_ Y \to X_{Y'})^*E_{Y'} \to E_ Y$ and $L(X_{Z'} \to X_{Y'})^*E_{Y'} \to E_ Z$ compatible with $\alpha$. Clearly, if we show that $E_{Y'}$ is $Y'$-perfect, then we are done, because property (2) of Lemma 99.16.2 is a property on points (and $Y$ and $Y'$ have the same points). This follows from More on Morphisms of Spaces, Lemma 76.54.4. $\square$