hcju/stacksqa huggingface dataset:
The problem is local on $S$ and $Y$. Hence we may assume that $S$ and $Y$ are affine. Since $f$ is flat and locally of finite presentation, we see that $f$ is open (Morphisms, Lemma \ref{morphisms-lemma-fppf-open}). Hence, since $Y$ is quasi-compact, there exist finitely many affine opens $X_i \subset X$ such that $Y = \bigcup f(X_i)$. Clearly we may replace $X$ by $\coprod X_i$, and hence we may assume $X$ is affine as well. In this case the lemma is equivalent to Lemma \ref{lemma-flat-finitely-presented-permanence-algebra} (resp. Lemma \ref{lemma-finite-type-local-source-fppf-algebra}) above.
Assume (1) and that $p$ is smooth. By Lemma \ref{lemma-flat-finitely-presented-permanence} we see that $q$ is locally of finite presentation. By Morphisms, Lemma \ref{morphisms-lemma-flat-permanence} we see that $q$ is flat. Hence now it suffices to show that the fibres of $q$ are smooth, see Morphisms, Lemma \ref{morphisms-lemma-smooth-flat-smooth-fibres}. Apply Varieties, Lemma \ref{varieties-lemma-flat-under-smooth} to the flat surjective morphisms $X_s \to Y_s$ for $s \in S$ to conclude. We omit the proof of the \'etale case.
By Lemma \ref{lemma-flat-finitely-presented-permanence} we see that $q$ is of finite presentation. By Morphisms, Lemma \ref{morphisms-lemma-flat-permanence} we see that $q$ is flat. By Morphisms, Lemma \ref{morphisms-lemma-syntomic-locally-standard-syntomic} it now suffices to show that the local rings of the fibres of $Y \to S$ and the fibres of $X \to Y$ are local complete intersection rings. To do this we may take the fibre of $X \to Y \to S$ at a point $s \in S$, i.e., we may assume $S$ is the spectrum of a field. Pick a point $x \in X$ with image $y \in Y$ and consider the ring map $$ \mathcal{O}_{Y, y} \longrightarrow \mathcal{O}_{X, x} $$ This is a flat local homomorphism of local Noetherian rings. The local ring $\mathcal{O}_{X, x}$ is a complete intersection. Thus may use Avramov's result, see Divided Power Algebra, Lemma \ref{dpa-lemma-avramov}, to conclude that both $\mathcal{O}_{Y, y}$ and $\mathcal{O}_{X, x}/\mathfrak m_y\mathcal{O}_{X, x}$ are complete intersection rings.
Let $\Spec(C) \subset Z$ be an affine open and let $\Spec(B) \subset Y$ be an affine open which maps into $\Spec(C)$. The assumption on $X \to Y$ implies we can find a standard affine fppf covering $\{\Spec(B_j) \to \Spec(B)\}$ and lifts $x_j : \Spec(B_j) \to X$. Since $\Spec(B_j)$ is quasi-compact we can find finitely many affine opens $\Spec(A_i) \subset X$ lying over $\Spec(B)$ such that the image of each $x_j$ is contained in the union $\bigcup \Spec(A_i)$. Hence after replacing each $\Spec(B_j)$ by a standard affine Zariski coverings of itself we may assume we have a standard affine fppf covering $\{\Spec(B_i) \to \Spec(B)\}$ such that each $\Spec(B_i) \to Y$ factors through an affine open $\Spec(A_i) \subset X$ lying over $\Spec(B)$. In other words, we have ring maps $C \to B \to A_i \to B_i$ for each $i$. Note that we can also consider $$ C \to B \to A = \prod A_i \to B' = \prod B_i $$ and that the ring map $B \to \prod B_i$ is faithfully flat and of finite presentation. \medskip\noindent The case $P = flat$. In this case we know that $C \to A$ is flat and we have to prove that $C \to B$ is flat. Suppose that $N \to N' \to N''$ is an exact sequence of $C$-modules. We want to show that $N \otimes_C B \to N' \otimes_C B \to N'' \otimes_C B$ is exact. Let $H$ be its cohomology and let $H'$ be the cohomology of $N \otimes_C B' \to N' \otimes_C B' \to N'' \otimes_C B'$. As $B \to B'$ is flat we know that $H' = H \otimes_B B'$. On the other hand $N \otimes_C A \to N' \otimes_C A \to N'' \otimes_C A$ is exact hence has zero cohomology. Hence the map $H \to H'$ is zero (as it factors through the zero module). Thus $H' = 0$. As $B \to B'$ is faithfully flat we conclude that $H = 0$ as desired. \medskip\noindent The case $P = locally\ finite\ type$. In this case we know that $C \to A$ is of finite type and we have to prove that $C \to B$ is of finite type. Because $B \to B'$ is of finite presentation (hence of finite type) we see that $A \to B'$ is of finite type, see Algebra, Lemma \ref{algebra-lemma-compose-finite-type}. Therefore $C \to B'$ is of finite type and we conclude by Lemma \ref{lemma-finite-type-local-source-fppf-algebra}. \medskip\noindent The case $P = locally\ finite\ presentation$. In this case we know that $C \to A$ is of finite presentation and we have to prove that $C \to B$ is of finite presentation. Because $B \to B'$ is of finite presentation and $B \to A$ of finite type we see that $A \to B'$ is of finite presentation, see Algebra, Lemma \ref{algebra-lemma-compose-finite-type}. Therefore $C \to B'$ is of finite presentation and we conclude by Lemma \ref{lemma-flat-finitely-presented-permanence-algebra}.
This follows almost immediately from the definition of a $\tau$-covering, see Topologies, Definition \ref{topologies-definition-fpqc-covering} \ref{topologies-definition-fppf-covering} \ref{topologies-definition-etale-covering} \ref{topologies-definition-smooth-covering}, or \ref{topologies-definition-syntomic-covering} and Topologies, Lemma \ref{topologies-lemma-fpqc-affine}, \ref{topologies-lemma-fppf-affine}, \ref{topologies-lemma-etale-affine}, \ref{topologies-lemma-smooth-affine}, or \ref{topologies-lemma-syntomic-affine}. Details omitted.
We will use Lemma \ref{lemma-descending-properties}. First we note that ``being locally Noetherian'' is local in the Zariski topology. This is clear from the definition, see Properties, Definition \ref{properties-definition-noetherian}. Next, we show that if $S' \to S$ is a flat, finitely presented morphism of affines and $S$ is locally Noetherian, then $S'$ is locally Noetherian. This is Morphisms, Lemma \ref{morphisms-lemma-finite-type-noetherian}. Finally, we have to show that if $S' \to S$ is a surjective flat, finitely presented morphism of affines and $S'$ is locally Noetherian, then $S$ is locally Noetherian. This follows from Algebra, Lemma \ref{algebra-lemma-descent-Noetherian}. Thus (1), (2) and (3) of Lemma \ref{lemma-descending-properties} hold and we win.
We will use Lemma \ref{lemma-descending-properties}. First we note that ``being Jacobson'' is local in the Zariski topology. This is Properties, Lemma \ref{properties-lemma-locally-jacobson}. Next, we show that if $S' \to S$ is a flat, finitely presented morphism of affines and $S$ is Jacobson, then $S'$ is Jacobson. This is Morphisms, Lemma \ref{morphisms-lemma-Jacobson-universally-Jacobson}. Finally, we have to show that if $f : S' \to S$ is a surjective flat, finitely presented morphism of affines and $S'$ is Jacobson, then $S$ is Jacobson. Say $S = \Spec(A)$ and $S' = \Spec(B)$ and $S' \to S$ given by $A \to B$. Then $A \to B$ is finitely presented and faithfully flat. Moreover, the ring $B$ is Jacobson, see Properties, Lemma \ref{properties-lemma-locally-jacobson}. \medskip\noindent By Algebra, Lemma \ref{algebra-lemma-fppf-fpqf} there exists a diagram $$ \xymatrix{ B \ar[rr] & & B' \\ & A \ar[ru] \ar[lu] & } $$ with $A \to B'$ finitely presented, faithfully flat and quasi-finite. In particular, $B \to B'$ is finite type, and we see from Algebra, Proposition \ref{algebra-proposition-Jacobson-permanence} that $B'$ is Jacobson. Hence we may assume that $A \to B$ is quasi-finite as well as faithfully flat and of finite presentation. \medskip\noindent Assume $A$ is not Jacobson to get a contradiction. According to Algebra, Lemma \ref{algebra-lemma-characterize-jacobson} there exists a nonmaximal prime $\mathfrak p \subset A$ and an element $f \in A$, $f \not \in \mathfrak p$ such that $V(\mathfrak p) \cap D(f) = \{\mathfrak p\}$. \medskip\noindent This leads to a contradiction as follows. First let $\mathfrak p \subset \mathfrak m$ be a maximal ideal of $A$. Pick a prime $\mathfrak m' \subset B$ lying over $\mathfrak m$ (exists because $A \to B$ is faithfully flat, see Algebra, Lemma \ref{algebra-lemma-ff-rings}). As $A \to B$ is flat, by going down see Algebra, Lemma \ref{algebra-lemma-flat-going-down}, we can find a prime $\mathfrak q \subset \mathfrak m'$ lying over $\mathfrak p$. In particular we see that $\mathfrak q$ is not maximal. Hence according to Algebra, Lemma \ref{algebra-lemma-characterize-jacobson} again the set $V(\mathfrak q) \cap D(f)$ is infinite (here we finally use that $B$ is Jacobson). All points of $V(\mathfrak q) \cap D(f)$ map to $V(\mathfrak p) \cap D(f) = \{\mathfrak p\}$. Hence the fibre over $\mathfrak p$ is infinite. This contradicts the fact that $A \to B$ is quasi-finite (see Algebra, Lemma \ref{algebra-lemma-quasi-finite} or more explicitly Morphisms, Lemma \ref{morphisms-lemma-quasi-finite}). Thus the lemma is proved.
We will use Lemma \ref{lemma-descending-properties}. First we note that $\mathcal{P}$ is local in the Zariski topology. Next, we show that if $T \to S$ is a flat, finitely presented morphism of affines and $S$ has a finite number of irreducible components, then so does $T$. Namely, since $T \to S$ is flat, the generic points of $T$ map to the generic points of $S$, see Morphisms, Lemma \ref{morphisms-lemma-generalizations-lift-flat}. Hence it suffices to show that for $s \in S$ the fibre $T_s$ has a finite number of generic points. Note that $T_s$ is an affine scheme of finite type over $\kappa(s)$, see Morphisms, Lemma \ref{morphisms-lemma-base-change-finite-type}. Hence $T_s$ is Noetherian and has a finite number of irreducible components (Morphisms, Lemma \ref{morphisms-lemma-finite-type-noetherian} and Properties, Lemma \ref{properties-lemma-Noetherian-irreducible-components}). Finally, we have to show that if $T \to S$ is a surjective flat, finitely presented morphism of affines and $T$ has a finite number of irreducible components, then so does $S$. In this case the arguments above show that every generic point of $S$ is the image of a generic point of $T$ and the result is clear. Thus (1), (2) and (3) of Lemma \ref{lemma-descending-properties} hold and we win.
We will check (1), (2) and (3) of Lemma \ref{lemma-descending-properties}. As a syntomic morphism is flat of finite presentation (Morphisms, Lemmas \ref{morphisms-lemma-syntomic-flat} and \ref{morphisms-lemma-syntomic-locally-finite-presentation}) we have already checked this for ``being locally Noetherian'' in the proof of Lemma \ref{lemma-Noetherian-local-fppf}. We will use this without further mention in the proof. First we note that $\mathcal{P}$ is local in the Zariski topology. This is clear from the definition, see Cohomology of Schemes, Definition \ref{coherent-definition-depth}. Next, we show that if $S' \to S$ is a syntomic morphism of affines and $S$ has $\mathcal{P}$, then $S'$ has $\mathcal{P}$. This is Algebra, Lemma \ref{algebra-lemma-Sk-goes-up} (use Morphisms, Lemma \ref{morphisms-lemma-syntomic-characterize} and Algebra, Definition \ref{algebra-definition-lci} and Lemma \ref{algebra-lemma-lci-CM}). Finally, we show that if $S' \to S$ is a surjective syntomic morphism of affines and $S'$ has $\mathcal{P}$, then $S$ has $\mathcal{P}$. This is Algebra, Lemma \ref{algebra-lemma-descent-Sk}. Thus (1), (2) and (3) of Lemma \ref{lemma-descending-properties} hold and we win.
We will use Lemma \ref{lemma-descending-properties}. First we note that ``being reduced'' is local in the Zariski topology. This is clear from the definition, see Schemes, Definition \ref{schemes-definition-reduced}. Next, we show that if $S' \to S$ is a smooth morphism of affines and $S$ is reduced, then $S'$ is reduced. This is Algebra, Lemma \ref{algebra-lemma-reduced-goes-up}. Finally, we show that if $S' \to S$ is a surjective smooth morphism of affines and $S'$ is reduced, then $S$ is reduced. This is Algebra, Lemma \ref{algebra-lemma-descent-reduced}. Thus (1), (2) and (3) of Lemma \ref{lemma-descending-properties} hold and we win.
We will use Lemma \ref{lemma-descending-properties}. First we show ``being normal'' is local in the Zariski topology. This is clear from the definition, see Properties, Definition \ref{properties-definition-normal}. Next, we show that if $S' \to S$ is a smooth morphism of affines and $S$ is normal, then $S'$ is normal. This is Algebra, Lemma \ref{algebra-lemma-normal-goes-up}. Finally, we show that if $S' \to S$ is a surjective smooth morphism of affines and $S'$ is normal, then $S$ is normal. This is Algebra, Lemma \ref{algebra-lemma-descent-normal}. Thus (1), (2) and (3) of Lemma \ref{lemma-descending-properties} hold and we win.
We will check (1), (2) and (3) of Lemma \ref{lemma-descending-properties}. As a smooth morphism is flat of finite presentation (Morphisms, Lemmas \ref{morphisms-lemma-smooth-flat} and \ref{morphisms-lemma-smooth-locally-finite-presentation}) we have already checked this for ``being locally Noetherian'' in the proof of Lemma \ref{lemma-Noetherian-local-fppf}. We will use this without further mention in the proof. First we note that $\mathcal{P}$ is local in the Zariski topology. This is clear from the definition, see Properties, Definition \ref{properties-definition-Rk}. Next, we show that if $S' \to S$ is a smooth morphism of affines and $S$ has $\mathcal{P}$, then $S'$ has $\mathcal{P}$. This is Algebra, Lemmas \ref{algebra-lemma-Rk-goes-up} (use Morphisms, Lemma \ref{morphisms-lemma-smooth-characterize}, Algebra, Lemmas \ref{algebra-lemma-base-change-smooth} and \ref{algebra-lemma-characterize-smooth-over-field}). Finally, we show that if $S' \to S$ is a surjective smooth morphism of affines and $S'$ has $\mathcal{P}$, then $S$ has $\mathcal{P}$. This is Algebra, Lemma \ref{algebra-lemma-descent-Rk}. Thus (1), (2) and (3) of Lemma \ref{lemma-descending-properties} hold and we win.
We will check (1), (2) and (3) of Lemma \ref{lemma-descending-properties}. First we note that being Nagata is local in the Zariski topology. This is Properties, Lemma \ref{properties-lemma-locally-nagata}. Next, we show that if $S' \to S$ is a smooth morphism of affines and $S$ is Nagata, then $S'$ is Nagata. This is Morphisms, Lemma \ref{morphisms-lemma-finite-type-nagata}. Finally, we show that if $S' \to S$ is a surjective smooth morphism of affines and $S'$ is Nagata, then $S$ is Nagata. This is Algebra, Lemma \ref{algebra-lemma-descent-nagata}. Thus (1), (2) and (3) of Lemma \ref{lemma-descending-properties} hold and we win.
This lemma reduces to the following algebra statement: If $A \to B$ is a faithfully flat, finitely presented ring homomorphism with $B$ Noetherian and regular, then $A$ is Noetherian and regular. We see that $A$ is Noetherian by Algebra, Lemma \ref{algebra-lemma-descent-Noetherian} and regular by Algebra, Lemma \ref{algebra-lemma-flat-under-regular}.
In the statement $\dim_u(U)$ is the dimension of $U$ at $u$ as defined in Topology, Definition \ref{topology-definition-Krull} as the minimum of the Krull dimensions of open neighbourhoods of $u$ in $U$. Similarly for $\dim_v(V)$. \medskip\noindent Let us show that $\dim_v(V) \geq \dim_u(U)$. Let $V'$ be an open neighbourhood of $v$ in $V$. Then there exists an open neighbourhood $U'$ of $u$ in $U$ contained in $f^{-1}(V')$ such that $\dim_u(U) = \dim(U')$. Suppose that $Z_0 \subset Z_1 \subset \ldots \subset Z_n$ is a chain of irreducible closed subschemes of $U'$. If $\xi_i \in Z_i$ is the generic point then we have specializations $\xi_n \leadsto \xi_{n - 1} \leadsto \ldots \leadsto \xi_0$. This gives specializations $f(\xi_n) \leadsto f(\xi_{n - 1}) \leadsto \ldots \leadsto f(\xi_0)$ in $V'$. Note that $f(\xi_j) \not = f(\xi_i)$ if $i \not = j$ as the fibres of $f$ are discrete (see Morphisms, Lemma \ref{morphisms-lemma-etale-over-field}). Hence we see that $\dim(V') \geq n$. The inequality $\dim_v(V) \geq \dim_u(U)$ follows formally. \medskip\noindent Let us show that $\dim_u(U) \geq \dim_v(V)$. Let $U'$ be an open neighbourhood of $u$ in $U$. Note that $V' = f(U')$ is an open neighbourhood of $v$ by Morphisms, Lemma \ref{morphisms-lemma-fppf-open}. Hence $\dim(V') \geq \dim_v(V)$. Pick a chain $Z_0 \subset Z_1 \subset \ldots \subset Z_n$ of irreducible closed subschemes of $V'$. Let $\xi_i \in Z_i$ be the generic point, so we have specializations $\xi_n \leadsto \xi_{n - 1} \leadsto \ldots \leadsto \xi_0$. Since $\xi_0 \in f(U')$ we can find a point $\eta_0 \in U'$ with $f(\eta_0) = \xi_0$. Consider the map of local rings $$ \mathcal{O}_{V', \xi_0} \longrightarrow \mathcal{O}_{U', \eta_0} $$ which is a flat local ring map by Morphisms, Lemma \ref{morphisms-lemma-etale-flat}. Note that the points $\xi_i$ correspond to primes of the ring on the left by Schemes, Lemma \ref{schemes-lemma-specialize-points}. Hence by going down (see Algebra, Section \ref{algebra-section-going-up}) for the displayed ring map we can find a sequence of specializations $\eta_n \leadsto \eta_{n - 1} \leadsto \ldots \leadsto \eta_0$ in $U'$ mapping to the sequence $\xi_n \leadsto \xi_{n - 1} \leadsto \ldots \leadsto \xi_0$ under $f$. This implies that $\dim_u(U) \geq \dim_v(V)$.
The algebraic statement we are asked to prove is the following: If $A \to B$ is an \'etale ring map and $\mathfrak q$ is a prime of $B$ lying over $\mathfrak p \subset A$, then $\dim(A_{\mathfrak p}) = \dim(B_{\mathfrak q})$. This is More on Algebra, Lemma \ref{more-algebra-lemma-dimension-etale-extension}.
The algebraic statement we are asked to prove is the following: If $A \to B$ is an \'etale ring map and $\mathfrak q$ is a prime of $B$ lying over $\mathfrak p \subset A$, then $A_{\mathfrak p}$ is regular if and only if $B_{\mathfrak q}$ is regular. This is More on Algebra, Lemma \ref{more-algebra-lemma-regular-etale-extension}.
Consider the union $W$ of the images $g(Y') \subset Y$ of morphisms $g : Y' \to Y$ with the properties: \begin{enumerate} \item $g$ is flat and locally of finite presentation, syntomic, smooth, or \'etale, and \item the base change $Y' \times_{g, Y} X \to Y'$ has property $\mathcal{P}$. \end{enumerate} Since such a morphism $g$ is open (see Morphisms, Lemma \ref{morphisms-lemma-fppf-open}) we see that $W \subset Y$ is an open subset of $Y$. Since $\mathcal{P}$ is local in the $\tau$ topology the restriction $X_W \to W$ has property $\mathcal{P}$ because we are given a covering $\{Y' \to W\}$ of $W$ such that the pullbacks have $\mathcal{P}$. This proves the existence and proves that $W(f)$ has property (1). To see property (2) note that $W(f') \supset g^{-1}(W(f))$ because $\mathcal{P}$ is stable under base change by flat and locally of finite presentation, syntomic, smooth, or \'etale morphisms, see Lemma \ref{lemma-pullback-property-local-target}. On the other hand, if $Y'' \subset Y'$ is an open such that $X_{Y''} \to Y''$ has property $\mathcal{P}$, then $Y'' \to Y$ factors through $W$ by construction, i.e., $Y'' \subset g^{-1}(W(f))$. This proves (2). Assertion (3) follows from (2) because each morphism $Y_i \to Y$ is flat and locally of finite presentation, syntomic, smooth, or \'etale by our definition of a $\tau$-covering.
A base change of a quasi-compact morphism is quasi-compact, see Schemes, Lemma \ref{schemes-lemma-quasi-compact-preserved-base-change}. Being quasi-compact is Zariski local on the base, see Schemes, Lemma \ref{schemes-lemma-quasi-compact-affine}. Finally, let $S' \to S$ be a flat surjective morphism of affine schemes, and let $f : X \to S$ be a morphism. Assume that the base change $f' : X' \to S'$ is quasi-compact. Then $X'$ is quasi-compact, and $X' \to X$ is surjective. Hence $X$ is quasi-compact. This implies that $f$ is quasi-compact. Therefore Lemma \ref{lemma-descending-properties-morphisms} applies and we win.
Any base change of a quasi-separated morphism is quasi-separated, see Schemes, Lemma \ref{schemes-lemma-separated-permanence}. Being quasi-separated is Zariski local on the base (from the definition or by Schemes, Lemma \ref{schemes-lemma-characterize-quasi-separated}). Finally, let $S' \to S$ be a flat surjective morphism of affine schemes, and let $f : X \to S$ be a morphism. Assume that the base change $f' : X' \to S'$ is quasi-separated. This means that $\Delta' : X' \to X'\times_{S'} X'$ is quasi-compact. Note that $\Delta'$ is the base change of $\Delta : X \to X \times_S X$ via $S' \to S$. By Lemma \ref{lemma-descending-property-quasi-compact} this implies $\Delta$ is quasi-compact, and hence $f$ is quasi-separated. Therefore Lemma \ref{lemma-descending-properties-morphisms} applies and we win.
A base change of a universally closed morphism is universally closed by definition. Being universally closed is Zariski local on the base (from the definition or by Morphisms, Lemma \ref{morphisms-lemma-universally-closed-local-on-the-base}). Finally, let $S' \to S$ be a flat surjective morphism of affine schemes, and let $f : X \to S$ be a morphism. Assume that the base change $f' : X' \to S'$ is universally closed. Let $T \to S$ be any morphism. Consider the diagram $$ \xymatrix{ X' \ar[d] & S' \times_S T \times_S X \ar[d] \ar[r] \ar[l] & T \times_S X \ar[d] \\ S' & S' \times_S T \ar[r] \ar[l] & T } $$ in which both squares are cartesian. Thus the assumption implies that the middle vertical arrow is closed. The right horizontal arrows are flat, quasi-compact and surjective (as base changes of $S' \to S$). Hence a subset of $T$ is closed if and only if its inverse image in $S' \times_S T$ is closed, see Morphisms, Lemma \ref{morphisms-lemma-fpqc-quotient-topology}. An easy diagram chase shows that the right vertical arrow is closed too, and we conclude $X \to S$ is universally closed. Therefore Lemma \ref{lemma-descending-properties-morphisms} applies and we win.
A base change of a separated morphism is separated, see Schemes, Lemma \ref{schemes-lemma-separated-permanence}. Being separated is Zariski local on the base (from the definition or by Schemes, Lemma \ref{schemes-lemma-characterize-separated}). Finally, let $S' \to S$ be a flat surjective morphism of affine schemes, and let $f : X \to S$ be a morphism. Assume that the base change $f' : X' \to S'$ is separated. This means that $\Delta' : X' \to X'\times_{S'} X'$ is a closed immersion, hence universally closed. Note that $\Delta'$ is the base change of $\Delta : X \to X \times_S X$ via $S' \to S$. By Lemma \ref{lemma-descending-property-universally-closed} this implies $\Delta$ is universally closed. Since it is an immersion (Schemes, Lemma \ref{schemes-lemma-diagonal-immersion}) we conclude $\Delta$ is a closed immersion. Hence $f$ is separated. Therefore Lemma \ref{lemma-descending-properties-morphisms} applies and we win.
A base change of a universally injective morphism is universally injective (this is formal). Being universally injective is Zariski local on the base; this is clear from the definition. Finally, let $S' \to S$ be a flat surjective morphism of affine schemes, and let $f : X \to S$ be a morphism. Assume that the base change $f' : X' \to S'$ is universally injective. Let $K$ be a field, and let $a, b : \Spec(K) \to X$ be two morphisms such that $f \circ a = f \circ b$. As $S' \to S$ is surjective and by the discussion in Schemes, Section \ref{schemes-section-points} there exists a field extension $K \subset K'$ and a morphism $\Spec(K') \to S'$ such that the following solid diagram commutes $$ \xymatrix{ \Spec(K') \ar[rrd] \ar@{-->}[rd]_{a', b'} \ar[dd] \\  & X' \ar[r] \ar[d] & S' \ar[d] \\ \Spec(K) \ar[r]^{a, b} & X \ar[r] & S } $$ As the square is cartesian we get the two dotted arrows $a'$, $b'$ making the diagram commute. Since $X' \to S'$ is universally injective we get $a' = b'$, by Morphisms, Lemma \ref{morphisms-lemma-universally-injective}. Clearly this forces $a = b$ (by the discussion in Schemes, Section \ref{schemes-section-points}). Therefore Lemma \ref{lemma-descending-properties-morphisms} applies and we win. \medskip\noindent An alternative proof would be to use the characterization of a universally injective morphism as one whose diagonal is surjective, see Morphisms, Lemma \ref{morphisms-lemma-universally-injective}. The lemma then follows from the fact that the property of being surjective is fpqc local on the base, see Lemma \ref{lemma-descending-property-surjective}. (Hint: use that the base change of the diagonal is the diagonal of the base change.)