# The Stacks Project

## Tag: 036M

This tag has label descent-section-fppf-local-source, it is called Properties of morphisms local in the fppf topology on the source in the Stacks project and it points to

The corresponding content:

### 31.24. Properties of morphisms local in the fppf topology on the source

Here are some properties of morphisms that are fppf local on the source.

Lemma 31.24.1. The property $\mathcal{P}(f)=$''$f$ is locally of finite presentation'' is fppf local on the source.

Proof. Being locally of finite presentation is Zariski local on the source and the target, see Morphisms, Lemma 25.22.2. It is a property which is preserved under composition, see Morphisms, Lemma 25.22.3. This proves (1), (2) and (3) of Lemma 31.22.3. The final condition (4) is Lemma 31.10.1. Hence we win. $\square$

Lemma 31.24.2. The property $\mathcal{P}(f)=$''$f$ is locally of finite type'' is fppf local on the source.

Proof. Being locally of finite type is Zariski local on the source and the target, see Morphisms, Lemma 25.16.2. It is a property which is preserved under composition, see Morphisms, Lemma 25.16.3, and a flat morphism locally of finite presentation is locally of finite type, see Morphisms, Lemma 25.22.8. This proves (1), (2) and (3) of Lemma 31.22.3. The final condition (4) is Lemma 31.10.2. Hence we win. $\square$

Lemma 31.24.3. The property $\mathcal{P}(f)=$''$f$ is open'' is fppf local on the source.

Proof. Being an open morphism is clearly Zariski local on the source and the target. It is a property which is preserved under composition, see Morphisms, Lemma 25.24.3, and a flat morphism of finite presentation is open, see Morphisms, Lemma 25.26.9 This proves (1), (2) and (3) of Lemma 31.22.3. The final condition (4) follows from Morphisms, Lemma 25.26.10. Hence we win. $\square$

Lemma 31.24.4. The property $\mathcal{P}(f)=$''$f$ is universally open'' is fppf local on the source.

Proof. Let $f : X \to Y$ be a morphism of schemes. Let $\{X_i \to X\}_{i \in I}$ be an fppf covering. Denote $f_i : X_i \to X$ the compositions. We have to show that $f$ is universally open if and only if each $f_i$ is universally open. If $f$ is universally open, then also each $f_i$ is universally open since the maps $X_i \to X$ are universally open and compositions of universally open morphisms are universally open (Morphisms, Lemmas 25.26.9 and 25.24.3). Conversely, assume each $f_i$ is universally open. Let $Y' \to Y$ be a morphism of schemes. Denote $X' = Y' \times_Y X$ and $X'_i = Y' \times_Y X_i$. Note that $\{X_i' \to X'\}_{i \in I}$ is an fppf covering also. The morphisms $f'_i : X_i' \to Y'$ are open by assumption. Hence by the Lemma 31.24.3 above we conclude that $f' : X' \to Y'$ is open as desired. $\square$

\section{Properties of morphisms local in the fppf topology on the source}
\label{section-fppf-local-source}

\noindent
Here are some properties of morphisms that are fppf local on the source.

\begin{lemma}
\label{lemma-locally-finite-presentation-fppf-local-source}
The property $\mathcal{P}(f)=$$f$ is locally of finite presentation''
is fppf local on the source.
\end{lemma}

\begin{proof}
Being locally of finite presentation is Zariski local on the source
and the target, see Morphisms,
Lemma \ref{morphisms-lemma-locally-finite-presentation-characterize}.
It is a property which is preserved under composition, see
Morphisms, Lemma \ref{morphisms-lemma-composition-finite-presentation}.
This proves
(1), (2) and (3) of Lemma \ref{lemma-properties-morphisms-local-source}.
The final condition (4) is
Lemma \ref{lemma-flat-finitely-presented-permanence-algebra}. Hence we win.
\end{proof}

\begin{lemma}
\label{lemma-locally-finite-type-fppf-local-source}
The property $\mathcal{P}(f)=$$f$ is locally of finite type''
is fppf local on the source.
\end{lemma}

\begin{proof}
Being locally of finite type is Zariski local on the source
and the target, see Morphisms,
Lemma \ref{morphisms-lemma-locally-finite-type-characterize}.
It is a property which is preserved under composition, see
Morphisms, Lemma \ref{morphisms-lemma-composition-finite-type}, and
a flat morphism locally of finite presentation is locally of finite type, see
Morphisms, Lemma \ref{morphisms-lemma-finite-presentation-finite-type}.
This proves
(1), (2) and (3) of Lemma \ref{lemma-properties-morphisms-local-source}.
The final condition (4) is
Lemma \ref{lemma-finite-type-local-source-fppf-algebra}. Hence we win.
\end{proof}

\begin{lemma}
\label{lemma-open-fppf-local-source}
The property $\mathcal{P}(f)=$$f$ is open''
is fppf local on the source.
\end{lemma}

\begin{proof}
Being an open morphism is clearly Zariski local on the source and the target.
It is a property which is preserved under composition, see
Morphisms, Lemma \ref{morphisms-lemma-composition-open}, and
a flat morphism of finite presentation is open, see
Morphisms, Lemma \ref{morphisms-lemma-fppf-open}
This proves
(1), (2) and (3) of Lemma \ref{lemma-properties-morphisms-local-source}.
The final condition (4) follows from
Morphisms, Lemma \ref{morphisms-lemma-fpqc-quotient-topology}.
Hence we win.
\end{proof}

\begin{lemma}
\label{lemma-universally-open-fppf-local-source}
The property $\mathcal{P}(f)=$$f$ is universally open''
is fppf local on the source.
\end{lemma}

\begin{proof}
Let $f : X \to Y$ be a morphism of schemes.
Let $\{X_i \to X\}_{i \in I}$ be an fppf covering.
Denote $f_i : X_i \to X$ the compositions.
We have to show that $f$ is universally open if and only if
each $f_i$ is universally open. If $f$ is universally open,
then also each $f_i$ is universally open since the maps
$X_i \to X$ are universally open and compositions
of universally open morphisms are universally open
(Morphisms, Lemmas \ref{morphisms-lemma-fppf-open}
and \ref{morphisms-lemma-composition-open}).
Conversely, assume each $f_i$ is universally open.
Let $Y' \to Y$ be a morphism of schemes.
Denote $X' = Y' \times_Y X$ and $X'_i = Y' \times_Y X_i$.
Note that $\{X_i' \to X'\}_{i \in I}$ is an fppf covering also.
The morphisms $f'_i : X_i' \to Y'$ are open by assumption.
Hence by the Lemma \ref{lemma-open-fppf-local-source}
above we conclude that $f' : X' \to Y'$ is open as desired.
\end{proof}


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