Lemma 66.18.10. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$ which is locally of finite type. Let $y \in |Y|$. Let $k$ be a field and let $\mathop{\mathrm{Spec}}(k) \to Y$ be in the equivalence class defining $y$. Set $X_ k = \mathop{\mathrm{Spec}}(k) \times _ Y X$ and let $F = f^{-1}(\{ y\} )$ with the induced topology from $|X|$. Consider the following conditions

$F$ is finite,

$F$ is a discrete topological space,

$\dim (F) = 0$,

$|X_ k|$ is a finite set,

$|X_ k|$ is a discrete space,

$\dim (|X_ k|) = 0$,

$\dim (X_ k) = 0$,

$f$ is quasi-finite at all points of $|X|$ lying over $y$.

Then we have

\[ \xymatrix{ (0ACL) & (0ACP) \ar@{=>}[l] \ar@{=>}[r]_{f\text{ decent}} & (0ACQ) \ar@{<=>}[r] & (0ACR) \ar@{<=>}[r] & (0ACS) \ar@{<=>}[r] & (0ACT) } \]

If $Y$ is decent, then conditions (2) and (3) are equivalent to each other and to conditions (5), (6), (7), and (8). If $Y$ and $X$ are decent, then (1) implies all the other conditions.

**Proof.**
By Lemma 66.18.8 conditions (5), (6), and (7) are equivalent to each other and to the condition that $X_ k \to \mathop{\mathrm{Spec}}(k)$ is locally quasi-finite. Thus by Morphisms of Spaces, Lemma 65.27.2 they are also equivalent to (8). If $f$ is decent, then $X_ k$ is a decent algebraic space and Lemma 66.18.8 shows that (4) implies (5).

The map $|X_ k| \to F$ is surjective by Properties of Spaces, Lemma 64.4.3 and we see (4) $\Rightarrow $ (1).

If $Y$ is decent, then we can pick a quasi-compact monomorphism $\mathop{\mathrm{Spec}}(k') \to Y$ in the equivalence class of $y$. In this case Lemma 66.18.6 tells us that $|X_{k'}| \to F$ is a homeomorphism. Combined with the arguments given above this implies the remaining statements of the lemma; details omitted.
$\square$

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