hcju/stacksqa huggingface dataset:
We will use Lemma \ref{lemma-descending-properties-morphisms} to prove this. Assumptions (1) and (2) of that lemma follow from Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-finite-presentation-local}. Let $Z' \to Z$ be a surjective flat morphism of affine schemes over $S$. Let $f : X \to Z$ be a morphism of algebraic spaces, and assume that the base change $f' : Z' \times_Z X \to Z'$ is locally of finite presentation. We have to show that $f$ is locally of finite presentation. Let $U$ be a scheme and let $U \to X$ be surjective and \'etale. By Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-finite-presentation-local} again, it is enough to show that $U \to Z$ is locally of finite presentation. Since $f'$ is locally of finite presentation, and since $Z' \times_Z U$ is a scheme \'etale over $Z' \times_Z X$ we conclude (by the same lemma again) that $Z' \times_Z U \to Z'$ is locally of finite presentation. As $\{Z' \to Z\}$ is an fpqc covering we conclude that $U \to Z$ is locally of finite presentation by Descent, Lemma \ref{descent-lemma-descending-property-locally-finite-presentation} as desired.
We will use Lemma \ref{lemma-descending-properties-morphisms} to prove this. Assumptions (1) and (2) of that lemma follow from Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-flat-local}. Let $Z' \to Z$ be a surjective flat morphism of affine schemes over $S$. Let $f : X \to Z$ be a morphism of algebraic spaces, and assume that the base change $f' : Z' \times_Z X \to Z'$ is flat. We have to show that $f$ is flat. Let $U$ be a scheme and let $U \to X$ be surjective and \'etale. By Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-flat-local} again, it is enough to show that $U \to Z$ is flat. Since $f'$ is flat, and since $Z' \times_Z U$ is a scheme \'etale over $Z' \times_Z X$ we conclude (by the same lemma again) that $Z' \times_Z U \to Z'$ is flat. As $\{Z' \to Z\}$ is an fpqc covering we conclude that $U \to Z$ is flat by Descent, Lemma \ref{descent-lemma-descending-property-flat} as desired.
We will use Lemma \ref{lemma-descending-properties-morphisms} to prove this. Assumptions (1) and (2) of that lemma follow from Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-closed-immersion-local}. Consider a cartesian diagram $$ \xymatrix{ X' \ar[r] \ar[d] & X \ar[d] \\ Z' \ar[r] & Z } $$ of algebraic spaces over $S$ where $Z' \to Z$ is a surjective flat morphism of affine schemes, and $X' \to Z'$ is an open immersion. We have to show that $X \to Z$ is an open immersion. Note that $|X'| \subset |Z'|$ corresponds to an open subscheme $U' \subset Z'$ (isomorphic to $X'$) with the property that $\text{pr}_0^{-1}(U') = \text{pr}_1^{-1}(U')$ as open subschemes of $Z' \times_Z Z'$. Hence there exists an open subscheme $U \subset Z$ such that $X' = (Z' \to Z)^{-1}(U)$, see Descent, Lemma \ref{descent-lemma-open-fpqc-covering}. By Properties of Spaces, Proposition \ref{spaces-properties-proposition-sheaf-fpqc} we see that $X$ satisfies the sheaf condition for the fpqc topology. Now we have the fpqc covering $\mathcal{U} = \{U' \to U\}$ and the element $U' \to X' \to X \in \check{H}^0(\mathcal{U}, X)$. By the sheaf condition we obtain a morphism $U \to X$ such that $$ \xymatrix{ U' \ar[r] \ar[d]^{\cong} \ar@/_3ex/[dd] & U \ar[d] \ar@/^3ex/[dd] \\ X' \ar[r] \ar[d] & X \ar[d] \\ Z' \ar[r] & Z } $$ is commutative. On the other hand, we know that for any scheme $T$ over $S$ and $T$-valued point $T \to X$ the composition $T \to X \to Z$ is a morphism such that $Z' \times_Z T \to Z'$ factors through $U'$. Clearly this means that $T \to Z$ factors through $U$. In other words the map of sheaves $U \to X$ is bijective and we win.
We will use Lemma \ref{lemma-descending-properties-morphisms} to prove this. Assumptions (1) and (2) of that lemma follow from Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-affine-local}. Let $Z' \to Z$ be a surjective flat morphism of affine schemes over $S$. Let $f : X \to Z$ be a morphism of algebraic spaces, and assume that the base change $f' : Z' \times_Z X \to Z'$ is affine. Let $X'$ be a scheme representing $Z' \times_Z X$. We obtain a canonical isomorphism $$ \varphi : X' \times_Z Z' \longrightarrow Z' \times_Z X' $$ since both schemes represent the algebraic space $Z' \times_Z Z' \times_Z X$. This is a descent datum for $X'/Z'/Z$, see Descent, Definition \ref{descent-definition-descent-datum} (verification omitted, compare with Descent, Lemma \ref{descent-lemma-descent-data-sheaves}). Since $X' \to Z'$ is affine this descent datum is effective, see Descent, Lemma \ref{descent-lemma-affine}. Thus there exists a scheme $Y \to Z$ over $Z$ and an isomorphism $\psi : Z' \times_Z Y \to X'$ compatible with descent data. Of course $Y \to Z$ is affine (by construction or by Descent, Lemma \ref{descent-lemma-descending-property-affine}). Note that $\mathcal{Y} = \{Z' \times_Z Y \to Y\}$ is a fpqc covering, and interpreting $\psi$ as an element of $X(Z' \times_Z Y)$ we see that $\psi \in \check{H}^0(\mathcal{Y}, X)$. By the sheaf condition for $X$ with respect to this covering (see Properties of Spaces, Proposition \ref{spaces-properties-proposition-sheaf-fpqc}) we obtain a morphism $Y \to X$. By construction the base change of this to $Z'$ is an isomorphism, hence an isomorphism by Lemma \ref{lemma-descending-property-isomorphism}. This proves that $X$ is representable by an affine scheme and we win.
We will use Lemma \ref{lemma-descending-properties-morphisms} to prove this. Assumptions (1) and (2) of that lemma follow from Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-closed-immersion-local}. Consider a cartesian diagram $$ \xymatrix{ X' \ar[r] \ar[d] & X \ar[d] \\ Z' \ar[r] & Z } $$ of algebraic spaces over $S$ where $Z' \to Z$ is a surjective flat morphism of affine schemes, and $X' \to Z'$ is a closed immersion. We have to show that $X \to Z$ is a closed immersion. The morphism $X' \to Z'$ is affine. Hence by Lemma \ref{lemma-descending-property-affine} we see that $X$ is a scheme and $X \to Z$ is affine. It follows from Descent, Lemma \ref{descent-lemma-descending-property-closed-immersion} that $X \to Z$ is a closed immersion as desired.
A base change of a separated morphism is separated, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-separated}. Hence the direct implication in Definition \ref{definition-property-morphisms-local}. \medskip\noindent Let $\{Y_i \to Y\}_{i \in I}$ be an fpqc covering of algebraic spaces over $S$. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume each base change $X_i := Y_i \times_Y X \to Y_i$ is separated. This means that each of the morphisms $$ \Delta_i : X_i \longrightarrow X_i \times_{Y_i} X_i = Y_i \times_Y (X \times_Y X) $$ is a closed immersion. The base change of a fpqc covering is an fpqc covering, see Topologies on Spaces, Lemma \ref{spaces-topologies-lemma-fpqc} hence $\{Y_i \times_Y (X \times_Y X) \to X \times_Y X\}$ is an fpqc covering of algebraic spaces. Moreover, each $\Delta_i$ is the base change of the morphism $\Delta : X \to X \times_Y X$. Hence it follows from Lemma \ref{lemma-descending-property-closed-immersion} that $\Delta$ is a closed immersion, i.e., $f$ is separated.
We will use Lemma \ref{lemma-descending-properties-morphisms} to prove this. Assumptions (1) and (2) of that lemma follow from Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-quasi-affine-local}. Let $Z' \to Z$ be a surjective flat morphism of affine schemes over $S$. Let $f : X \to Z$ be a morphism of algebraic spaces, and assume that the base change $f' : Z' \times_Z X \to Z'$ is quasi-affine. Let $X'$ be a scheme representing $Z' \times_Z X$. We obtain a canonical isomorphism $$ \varphi : X' \times_Z Z' \longrightarrow Z' \times_Z X' $$ since both schemes represent the algebraic space $Z' \times_Z Z' \times_Z X$. This is a descent datum for $X'/Z'/Z$, see Descent, Definition \ref{descent-definition-descent-datum} (verification omitted, compare with Descent, Lemma \ref{descent-lemma-descent-data-sheaves}). Since $X' \to Z'$ is quasi-affine this descent datum is effective, see Descent, Lemma \ref{descent-lemma-quasi-affine}. Thus there exists a scheme $Y \to Z$ over $Z$ and an isomorphism $\psi : Z' \times_Z Y \to X'$ compatible with descent data. Of course $Y \to Z$ is quasi-affine (by construction or by Descent, Lemma \ref{descent-lemma-descending-property-quasi-affine}). Note that $\mathcal{Y} = \{Z' \times_Z Y \to Y\}$ is a fpqc covering, and interpreting $\psi$ as an element of $X(Z' \times_Z Y)$ we see that $\psi \in \check{H}^0(\mathcal{Y}, X)$. By the sheaf condition for $X$ (see Properties of Spaces, Proposition \ref{spaces-properties-proposition-sheaf-fpqc}) we obtain a morphism $Y \to X$. By construction the base change of this to $Z'$ is an isomorphism, hence an isomorphism by Lemma \ref{lemma-descending-property-isomorphism}. This proves that $X$ is representable by a quasi-affine scheme and we win.
We will use Lemma \ref{lemma-descending-properties-morphisms} to prove this. Assumptions (1) and (2) of that lemma follow from Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-closed-immersion-local} and \ref{spaces-morphisms-lemma-quasi-compact-local}. Consider a cartesian diagram $$ \xymatrix{ X' \ar[r] \ar[d] & X \ar[d] \\ Z' \ar[r] & Z } $$ of algebraic spaces over $S$ where $Z' \to Z$ is a surjective flat morphism of affine schemes, and $X' \to Z'$ is a quasi-compact immersion. We have to show that $X \to Z$ is a closed immersion. The morphism $X' \to Z'$ is quasi-affine. Hence by Lemma \ref{lemma-descending-property-quasi-affine} we see that $X$ is a scheme and $X \to Z$ is quasi-affine. It follows from Descent, Lemma \ref{descent-lemma-descending-property-quasi-compact-immersion} that $X \to Z$ is a quasi-compact immersion as desired.
We have already seen that ``quasi-compact'' is fpqc local on the base, see Lemma \ref{lemma-descending-property-quasi-compact}. Hence it is enough to prove the lemma for ``locally quasi-finite''. We will use Lemma \ref{lemma-descending-properties-morphisms} to prove this. Assumptions (1) and (2) of that lemma follow from Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-quasi-finite-local}. Let $Z' \to Z$ be a surjective flat morphism of affine schemes over $S$. Let $f : X \to Z$ be a morphism of algebraic spaces, and assume that the base change $f' : Z' \times_Z X \to Z'$ is locally quasi-finite. We have to show that $f$ is locally quasi-finite. Let $U$ be a scheme and let $U \to X$ be surjective and \'etale. By Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-quasi-finite-local} again, it is enough to show that $U \to Z$ is locally quasi-finite. Since $f'$ is locally quasi-finite, and since $Z' \times_Z U$ is a scheme \'etale over $Z' \times_Z X$ we conclude (by the same lemma again) that $Z' \times_Z U \to Z'$ is locally quasi-finite. As $\{Z' \to Z\}$ is an fpqc covering we conclude that $U \to Z$ is locally quasi-finite by Descent, Lemma \ref{descent-lemma-descending-property-quasi-finite} as desired.
We will use Lemma \ref{lemma-descending-properties-morphisms} to prove this. Assumptions (1) and (2) of that lemma follow from Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-syntomic-local}. Let $Z' \to Z$ be a surjective flat morphism of affine schemes over $S$. Let $f : X \to Z$ be a morphism of algebraic spaces, and assume that the base change $f' : Z' \times_Z X \to Z'$ is syntomic. We have to show that $f$ is syntomic. Let $U$ be a scheme and let $U \to X$ be surjective and \'etale. By Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-syntomic-local} again, it is enough to show that $U \to Z$ is syntomic. Since $f'$ is syntomic, and since $Z' \times_Z U$ is a scheme \'etale over $Z' \times_Z X$ we conclude (by the same lemma again) that $Z' \times_Z U \to Z'$ is syntomic. As $\{Z' \to Z\}$ is an fpqc covering we conclude that $U \to Z$ is syntomic by Descent, Lemma \ref{descent-lemma-descending-property-syntomic} as desired.
We will use Lemma \ref{lemma-descending-properties-morphisms} to prove this. Assumptions (1) and (2) of that lemma follow from Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-smooth-local}. Let $Z' \to Z$ be a surjective flat morphism of affine schemes over $S$. Let $f : X \to Z$ be a morphism of algebraic spaces, and assume that the base change $f' : Z' \times_Z X \to Z'$ is smooth. We have to show that $f$ is smooth. Let $U$ be a scheme and let $U \to X$ be surjective and \'etale. By Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-smooth-local} again, it is enough to show that $U \to Z$ is smooth. Since $f'$ is smooth, and since $Z' \times_Z U$ is a scheme \'etale over $Z' \times_Z X$ we conclude (by the same lemma again) that $Z' \times_Z U \to Z'$ is smooth. As $\{Z' \to Z\}$ is an fpqc covering we conclude that $U \to Z$ is smooth by Descent, Lemma \ref{descent-lemma-descending-property-smooth} as desired.
We will use Lemma \ref{lemma-descending-properties-morphisms} to prove this. Assumptions (1) and (2) of that lemma follow from Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-unramified-local}. Let $Z' \to Z$ be a surjective flat morphism of affine schemes over $S$. Let $f : X \to Z$ be a morphism of algebraic spaces, and assume that the base change $f' : Z' \times_Z X \to Z'$ is unramified. We have to show that $f$ is unramified. Let $U$ be a scheme and let $U \to X$ be surjective and \'etale. By Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-unramified-local} again, it is enough to show that $U \to Z$ is unramified. Since $f'$ is unramified, and since $Z' \times_Z U$ is a scheme \'etale over $Z' \times_Z X$ we conclude (by the same lemma again) that $Z' \times_Z U \to Z'$ is unramified. As $\{Z' \to Z\}$ is an fpqc covering we conclude that $U \to Z$ is unramified by Descent, Lemma \ref{descent-lemma-descending-property-unramified} as desired.
We will use Lemma \ref{lemma-descending-properties-morphisms} to prove this. Assumptions (1) and (2) of that lemma follow from Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-etale-local}. Let $Z' \to Z$ be a surjective flat morphism of affine schemes over $S$. Let $f : X \to Z$ be a morphism of algebraic spaces, and assume that the base change $f' : Z' \times_Z X \to Z'$ is \'etale. We have to show that $f$ is \'etale. Let $U$ be a scheme and let $U \to X$ be surjective and \'etale. By Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-etale-local} again, it is enough to show that $U \to Z$ is \'etale. Since $f'$ is \'etale, and since $Z' \times_Z U$ is a scheme \'etale over $Z' \times_Z X$ we conclude (by the same lemma again) that $Z' \times_Z U \to Z'$ is \'etale. As $\{Z' \to Z\}$ is an fpqc covering we conclude that $U \to Z$ is \'etale by Descent, Lemma \ref{descent-lemma-descending-property-etale} as desired.
Let $f : X \to Y$ be a morphism of algebraic spaces. Let $\{Y_i \to Y\}$ be an fpqc covering, and assume each of the base changes $f_i : X_i \to Y_i$ of $f$ is a monomorphism. We have to show that $f$ is a monomorphism. \medskip\noindent First proof. Note that $f$ is a monomorphism if and only if $\Delta : X \to X \times_Y X$ is an isomorphism. By applying this to $f_i$ we see that each of the morphisms $$ \Delta_i : X_i \longrightarrow X_i \times_{Y_i} X_i = Y_i \times_Y (X \times_Y X) $$ is an isomorphism. The base change of an fpqc covering is an fpqc covering, see Topologies on Spaces, Lemma \ref{spaces-topologies-lemma-fpqc} hence $\{Y_i \times_Y (X \times_Y X) \to X \times_Y X\}$ is an fpqc covering of algebraic spaces. Moreover, each $\Delta_i$ is the base change of the morphism $\Delta : X \to X \times_Y X$. Hence it follows from Lemma \ref{lemma-descending-property-isomorphism} that $\Delta$ is an isomorphism, i.e., $f$ is a monomorphism. \medskip\noindent Second proof. Let $V$ be a scheme, and let $V \to Y$ be a surjective \'etale morphism. If we can show that $V \times_Y X \to V$ is a monomorphism, then it follows that $X \to Y$ is a monomorphism. Namely, given any cartesian diagram of sheaves $$ \vcenter{ \xymatrix{ \mathcal{F} \ar[r]_a \ar[d]_b & \mathcal{G} \ar[d]^c \\ \mathcal{H} \ar[r]^d & \mathcal{I} } } \quad \quad \mathcal{F} = \mathcal{H} \times_\mathcal{I} \mathcal{G} $$ if $c$ is a surjection of sheaves, and $a$ is injective, then also $d$ is injective. This reduces the problem to the case where $Y$ is a scheme. Moreover, in this case we may assume that the algebraic spaces $Y_i$ are schemes also, since we can always refine the covering to place ourselves in this situation, see Topologies on Spaces, Lemma \ref{spaces-topologies-lemma-refine-fpqc-schemes}. \medskip\noindent Assume $\{Y_i \to Y\}$ is an fpqc covering of schemes. Let $a, b : T \to X$ be two morphisms such that $f \circ a = f \circ b$. We have to show that $a = b$. Since $f_i$ is a monomorphism we see that $a_i = b_i$, where $a_i, b_i : Y_i \times_Y T \to X_i$ are the base changes. In particular the compositions $Y_i \times_Y T \to T \to X$ are equal. Since $\{Y_i \times_Y T \to T\}$ is an fpqc covering we deduce that $a = b$ from Properties of Spaces, Proposition \ref{spaces-properties-proposition-sheaf-fpqc}.
Let $f : X \to Y$ be a morphism of algebraic spaces. Let $\{Y_i \to Y\}_{i \in I}$ be an fppf covering of $Y$. Let $f_i : X_i \to Y_i$ be the base change of $f$. \medskip\noindent If $f$ is an immersion, then each $f_i$ is an immersion by Spaces, Lemma \ref{spaces-lemma-base-change-immersions}. This proves the direct implication in Definition \ref{definition-property-morphisms-local}. \medskip\noindent Conversely, assume each $f_i$ is an immersion. By Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-immersions-monomorphisms} this implies each $f_i$ is separated. By Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-immersion-quasi-finite} this implies each $f_i$ is locally quasi-finite. Hence we see that $f$ is locally quasi-finite and separated, by applying Lemmas \ref{lemma-descending-property-separated} and \ref{lemma-descending-property-quasi-finite}. By Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-locally-quasi-finite-separated-representable} this implies that $f$ is representable! \medskip\noindent By Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-closed-immersion-local} it suffices to show that for every scheme $Z$ and morphism $Z \to Y$ the base change $Z \times_Y X \to Z$ is an immersion. By Topologies on Spaces, Lemma \ref{spaces-topologies-lemma-refine-fppf-schemes} we can find an fppf covering $\{Z_i \to Z\}$ by schemes which refines the pullback of the covering $\{Y_i \to Y\}$ to $Z$. Hence we see that $Z \times_Y X \to Z$ (which is a morphism of schemes according to the result of the preceding paragraph) becomes an immersion after pulling back to the members of an fppf (by schemes) of $Z$. Hence $Z \times_Y X \to Z$ is an immersion by the result for schemes, see Descent, Lemma \ref{descent-lemma-descending-fppf-property-immersion}.
A base change of a locally separated morphism is locally separated, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-base-change-separated}. Hence the direct implication in Definition \ref{definition-property-morphisms-local}. \medskip\noindent Let $\{Y_i \to Y\}_{i \in I}$ be an fppf covering of algebraic spaces over $S$. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume each base change $X_i := Y_i \times_Y X \to Y_i$ is locally separated. This means that each of the morphisms $$ \Delta_i : X_i \longrightarrow X_i \times_{Y_i} X_i = Y_i \times_Y (X \times_Y X) $$ is an immersion. The base change of a fppf covering is an fppf covering, see Topologies on Spaces, Lemma \ref{spaces-topologies-lemma-fppf} hence $\{Y_i \times_Y (X \times_Y X) \to X \times_Y X\}$ is an fppf covering of algebraic spaces. Moreover, each $\Delta_i$ is the base change of the morphism $\Delta : X \to X \times_Y X$. Hence it follows from Lemma \ref{lemma-descending-fppf-property-immersion} that $\Delta$ is a immersion, i.e., $f$ is locally separated.
We write everything out completely. \medskip\noindent Proof of (1). Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $\{X_i \to X\}_{i \in I}$ be a smooth covering of $X$. If each composition $h_i : X_i \to Y$ has $\mathcal{P}$, then for each $|x| \in X$ we can find an $i \in I$ and a point $x_i \in |X_i|$ mapping to $x$. Then $(X_i, x_i) \to (X, x)$ is a smooth morphism of pairs, and $\text{id}_Y : Y \to Y$ is a smooth morphism, and $h_i$ is as in part (3) of Definition \ref{definition-local-source-target}. Thus we see that $f$ has $\mathcal{P}$. Conversely, if $f$ has $\mathcal{P}$ then each $X_i \to Y$ has $\mathcal{P}$ by Definition \ref{definition-local-source-target} part (1). \medskip\noindent Proof of (2). Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $\{Y_i \to Y\}_{i \in I}$ be a smooth covering of $Y$. Write $X_i = Y_i \times_Y X$ and $h_i : X_i \to Y_i$ for the base change of $f$.  If each  $h_i : X_i \to Y_i$ has $\mathcal{P}$, then for each $x \in |X|$ we pick an $i \in I$ and a point $x_i \in |X_i|$ mapping to $x$. Then $(X_i, x_i) \to (X, x)$ is a smooth morphism of pairs, $Y_i \to Y$ is smooth, and $h_i$ is as in part (3) of Definition \ref{definition-local-source-target}. Thus we see that $f$ has $\mathcal{P}$. Conversely, if $f$ has $\mathcal{P}$, then each $X_i \to Y_i$ has $\mathcal{P}$ by Definition \ref{definition-local-source-target} part (2). \medskip\noindent Proof of (3). Assume $f : X \to Y$ has $\mathcal{P}$ and $g : Y \to Z$ is smooth. For every $x \in |X|$ we can think of $(X, x) \to (X, x)$ as a smooth morphism of pairs, $Y \to Z$ is a smooth morphism, and $h = f$ is as in part (3) of Definition \ref{definition-local-source-target}. Thus we see that $g \circ f$ has $\mathcal{P}$.
The equivalence of (a) and (b) is part of Definition \ref{definition-local-source-target}. The equivalence of (a) and (e) is Lemma \ref{lemma-local-source-target-implies} part (2). The equivalence of (a) and (f) is Lemma \ref{lemma-local-source-target-implies} part (1). As (a) is now equivalent to (e) and (f) it follows that (a) equivalent to (g). \medskip\noindent It is clear that (c) implies (b). If (b) holds, then for any $x \in |X|$ we can choose a smooth morphism of pairs $a_x : (U_x, u_x) \to (X, x)$, a smooth morphism $b_x : V_x \to Y$, and a morphism $h_x : U_x \to V_x$ such that $f \circ a_x = b_x \circ h_x$ and $h_x$ has $\mathcal{P}$. Then $h = \coprod h_x : \coprod U_x \to \coprod V_x$ with $a = \coprod a_x$ and $b = \coprod b_x$ is a diagram as in (c). (Note that $h$ has property $\mathcal{P}$ as $\{V_x \to \coprod V_x\}$ is a smooth covering and $\mathcal{P}$ is smooth local on the target.) Thus (b) is equivalent to (c). \medskip\noindent Now we know that (a), (b), (c), (e), (f), and (g) are equivalent. Suppose (a) holds. Let $U, V, a, b, h$ be as in (d). Then $X \times_Y V \to V$ has $\mathcal{P}$ as $\mathcal{P}$ is stable under smooth base change, whence $U \to V$ has $\mathcal{P}$ as $\mathcal{P}$ is stable under precomposing with smooth morphisms. Conversely, if (d) holds, then setting $U = X$ and $V = Y$ we see that $f$ has $\mathcal{P}$.