This section is the analogue of Morphisms, Section 29.28. The formulations in this section are a bit awkward since we do not have local rings of algebraic spaces at points.
Lemma 67.34.1. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $x \in |X|$. Assume $f$ is locally of finite type. Then we have
where the notation is as in Definition 67.33.1.
Proof. This follows immediately from Morphisms, Lemma 29.28.1 applied to $h : U \to V$ and $u \in U$ as in Lemma 67.22.5. $\square$
Lemma 67.34.2. Let $S$ be a scheme. Let $f : X \to Y$ and $g : Y \to Z$ be morphisms of algebraic spaces over $S$. Let $x \in |X|$ and set $y = f(x)$. Assume $f$ and $g$ locally of finite type. Then
equality holds in (1) if for some morphism $\mathop{\mathrm{Spec}}(k) \to Z$ from the spectrum of a field in the class of $g(f(x)) = g(y)$ the morphism $X_ k \to Y_ k$ is flat at $x$, for example if $f$ is flat at $x$,
Proof. Choose a diagram
with $U, V, W$ schemes and vertical arrows étale and surjective. (See Spaces, Lemma 65.11.6.) Choose $u \in U$ mapping to $x$. Set $v, w$ equal to the images of $u$ in $V, W$. Apply Morphisms, Lemma 29.28.2 to the top row and the points $u, v, w$. Details omitted. $\square$
Lemma 67.34.3. Let $S$ be a scheme. Let
be a fibre product diagram of algebraic spaces over $S$. Let $x' \in |X'|$. Set $x = g'(x')$. Assume $f$ locally of finite type. Then
we have
and the common value is $\geq 0$,
given $x$ and $y' \in |Y'|$ mapping to the same $y \in |Y|$ there exists a choice of $x'$ such that the integer in (2) is $0$.
Proof. Choose a surjective étale morphism $V \to Y$ with $V$ a scheme. Choose a surjective étale morphism $U \to V \times _ Y X$ with $U$ a scheme. Choose a surjective étale morphism $V' \to V \times _ Y Y'$ with $V'$ a scheme. Set $U' = V' \times _ V U$. Then the induced morphism $U' \to X'$ is also surjective and étale (argument omitted). Choose $u' \in U'$ mapping to $x'$. At this point parts (1) and (2) follow by applying Morphisms, Lemma 29.28.3 to the diagram of schemes involving $U', U, V', V$ and the point $u'$. To prove (3) first choose $v \in V$ mapping to $y$. Then using Properties of Spaces, Lemma 66.4.3 we can choose $v' \in V'$ mapping to $y'$ and $v$ and $u \in U$ mapping to $x$ and $v$. Finally, according to Morphisms, Lemma 29.28.3 we can choose $u' \in U'$ mapping to $v'$ and $u$ such that the integer is zero. Then taking $x' \in |X'|$ the image of $u'$ works. $\square$
Lemma 67.34.4. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Let $n \geq 0$. Assume $f$ is locally of finite type. The set
is open in $|X|$.
Proof. Choose a diagram
where $U$ and $V$ are schemes and the vertical arrows are surjective and étale, see Spaces, Lemma 65.11.6. By Morphisms, Lemma 29.28.4 the set $U_ n$ of points where $h$ has relative dimension $\leq n$ is open in $U$. By our definition of relative dimension for morphisms of algebraic spaces at points we see that $U_ n = a^{-1}(W_ n)$. The lemma follows by definition of the topology on $|X|$. $\square$
Lemma 67.34.5. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$ Let $n \geq 0$. Assume $f$ is locally of finite presentation. The open
of Lemma 67.34.4 is retrocompact in $|X|$. (See Topology, Definition 5.12.1.)
Proof. Choose a diagram
where $U$ and $V$ are schemes and the vertical arrows are surjective and étale, see Spaces, Lemma 65.11.6. In the proof of Lemma 67.34.4 we have seen that $a^{-1}(W_ n) = U_ n$ is the corresponding set for the morphism $h$. By Morphisms, Lemma 29.28.6 we see that $U_ n$ is retrocompact in $U$. The lemma follows by definition of the topology on $|X|$, compare with Properties of Spaces, Lemma 66.5.5 and its proof. $\square$
Lemma 67.34.6. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume $f$ is locally of finite type. Then $f$ is locally quasi-finite if and only if $f$ has relative dimension $0$ at each $x \in |X|$.
Proof. Choose a diagram
where $U$ and $V$ are schemes and the vertical arrows are surjective and étale, see Spaces, Lemma 65.11.6. The definitions imply that $h$ is locally quasi-finite if and only if $f$ is locally quasi-finite, and that $f$ has relative dimension $0$ at all $x \in |X|$ if and only if $h$ has relative dimension $0$ at all $u \in U$. Hence the result follows from the result for $h$ which is Morphisms, Lemma 29.29.5. $\square$
Lemma 67.34.7. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. Assume $f$ is locally of finite type. Then there exists a canonical open subspace $X' \subset X$ such that $f|_{X'} : X' \to Y$ is locally quasi-finite, and such that the relative dimension of $f$ at any $x \in |X|$, $x \not\in |X'|$ is $\geq 1$. Formation of $X'$ commutes with arbitrary base change.
Proof. Combine Lemmas 67.34.4, 67.34.6, and 67.34.3. $\square$
Lemma 67.34.8. Let $S$ be a scheme. Consider a cartesian diagram
where $X \to Y$ is a morphism of algebraic spaces over $S$ which is locally of finite type and where $k$ is a field over $S$. Let $z \in |F|$ be such that $\dim _ z(F) = 0$. Then, after replacing $X$ by an open subspace containing $p(z)$, the morphism
is locally quasi-finite.
Proof. Let $X' \subset X$ be the open subspace over which $f$ is locally quasi-finite found in Lemma 67.34.7. Since the formation of $X'$ commutes with arbitrary base change we see that $z \in X' \times _ Y \mathop{\mathrm{Spec}}(k)$. Hence the lemma is clear. $\square$
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