Lemma 34.8.8. Let $T$ be a scheme.
If $T' \to T$ is an isomorphism then $\{ T' \to T\} $ is a ph covering of $T$.
If $\{ T_ i \to T\} _{i\in I}$ is a ph covering and for each $i$ we have a ph covering $\{ T_{ij} \to T_ i\} _{j\in J_ i}$, then $\{ T_{ij} \to T\} _{i \in I, j\in J_ i}$ is a ph covering.
If $\{ T_ i \to T\} _{i\in I}$ is a ph covering and $T' \to T$ is a morphism of schemes then $\{ T' \times _ T T_ i \to T'\} _{i\in I}$ is a ph covering.
Proof. Assertion (1) is clear.
Proof of (3). The base change $T_ i \times _ T T' \to T'$ is locally of finite type by Morphisms, Lemma 29.15.4. hence we only need to check the condition on affine opens. Let $U' \subset T'$ be an affine open subscheme. Since $U'$ is quasi-compact we can find a finite affine open covering $U' = U'_1 \cup \ldots \cup U'$ such that $U'_ j \to T$ maps into an affine open $U_ j \subset T$. Choose a standard ph covering $\{ U_{jl} \to U_ j\} _{l = 1, \ldots , n_ j}$ refining $\{ T_ i \times _ T U_ j \to U_ j\} $. By Lemma 34.8.2 the base change $\{ U_{jl} \times _{U_ j} U'_ j \to U'_ j\} $ is a standard ph covering. Note that $\{ U'_ j \to U'\} $ is a standard ph covering as well. By Lemma 34.8.3 the family $\{ U_{jl} \times _{U_ j} U'_ j \to U'\} $ can be refined by a standard ph covering. Since $\{ U_{jl} \times _{U_ j} U'_ j \to U'\} $ refines $\{ T_ i \times _ T U' \to U'\} $ we conclude.
Proof of (2). Composition preserves being locally of finite type, see Morphisms, Lemma 29.15.3. Hence we only need to check the condition on affine opens. Let $U \subset T$ be affine open. First we pick a standard ph covering $\{ U_ k \to U\} _{k = 1, \ldots , m}$ refining $\{ T_ i \times _ T U \to U\} $. Say the refinement is given by morphisms $U_ k \to T_{i_ k}$ over $T$. Then
is a ph covering by part (3). As $U_ k$ is affine, we can find a standard ph covering $\{ U_{ka} \to U_ k\} _{a = 1, \ldots , b_ k}$ refining this family. Then we apply Lemma 34.8.3 to see that $\{ U_{ka} \to U\} $ can be refined by a standard ph covering. Since $\{ U_{ka} \to U\} $ refines $\{ T_{ij} \times _ T U \to U\} $ this finishes the proof. $\square$
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