Lemma 37.57.6. Let $f : X \to S$ be a morphism of schemes which is locally of finite type. Let $E$ be an object of $D(\mathcal{O}_ X)$. Fix $m \in \mathbf{Z}$. The following are equivalent

1. $E$ is $m$-pseudo-coherent relative to $S$,

2. for every affine opens $U \subset X$ and $V \subset S$ with $f(U) \subset V$ the equivalent conditions of Lemma 37.57.1 are satisfied for the pair $(U \to V, E|_ U)$.

Moreover, if this is true, then for every open subschemes $U \subset X$ and $V \subset S$ with $f(U) \subset V$ the restriction $E|_ U$ is $m$-pseudo-coherent relative to $V$.

Proof. The final statement is clear from the equivalence of (1) and (2). It is also clear that (2) implies (1). Assume (1) holds. Let $S = \bigcup V_ i$ and $f^{-1}(V_ i) = \bigcup U_{ij}$ be affine open coverings as in Definition 37.57.2. Let $U \subset X$ and $V \subset S$ be as in (2). By Lemma 37.57.5 it suffices to find a standard open covering $U = \bigcup U_ k$ of $U$ such that the equivalent conditions of Lemma 37.57.1 are satisfied for the pairs $(U_ k \to V, E|_{U_ k})$. In other words, for every $u \in U$ it suffices to find a standard affine open $u \in U' \subset U$ such that the equivalent conditions of Lemma 37.57.1 are satisfied for the pair $(U' \to V, E|_{U'})$. Pick $i$ such that $f(u) \in V_ i$ and then pick $j$ such that $u \in U_{ij}$. By Schemes, Lemma 26.11.5 we can find $v \in V' \subset V \cap V_ i$ which is standard affine open in $V'$ and $V_ i$. Then $f^{-1}V' \cap U$, resp. $f^{-1}V' \cap U_{ij}$ are standard affine opens of $U$, resp. $U_{ij}$. Applying the lemma again we can find $u \in U' \subset f^{-1}V' \cap U \cap U_{ij}$ which is standard affine open in both $f^{-1}V' \cap U$ and $f^{-1}V' \cap U_{ij}$. Thus $U'$ is also a standard affine open of $U$ and $U_{ij}$. By Lemma 37.57.4 the assumption that the equivalent conditions of Lemma 37.57.1 are satisfied for the pair $(U_{ij} \to V_ i, E|_{U_{ij}})$ implies that the equivalent conditions of Lemma 37.57.1 are satisfied for the pair $(U' \to V, E|_{U'})$. $\square$

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