Lemma 114.15.1. Let $S$ be a scheme. Let $X$ be a reasonable algebraic space over $S$. Then $|X|$ is Kolmogorov (see Topology, Definition 5.8.6).

## 114.15 Very reasonable algebraic spaces

Material that is somewhat obsolete.

**Proof.**
Follows from the definitions and Decent Spaces, Lemma 67.12.3.
$\square$

In the rest of this section we make some remarks about very reasonable algebraic spaces. If there exists a scheme $U$ and a surjective, étale, quasi-compact morphism $U \to X$, then $X$ is very reasonable, see Decent Spaces, Lemma 67.4.7.

Lemma 114.15.2. A scheme is very reasonable.

**Proof.**
This is true because the identity map is a quasi-compact, surjective étale morphism.
$\square$

Lemma 114.15.3. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. If there exists a Zariski open covering $X = \bigcup X_ i$ such that each $X_ i$ is very reasonable, then $X$ is very reasonable.

**Proof.**
This is case $(\epsilon )$ of Decent Spaces, Lemma 67.5.2.
$\square$

Lemma 114.15.4. An algebraic space which is Zariski locally quasi-separated is very reasonable. In particular any quasi-separated algebraic space is very reasonable.

**Proof.**
This is one of the implications of Decent Spaces, Lemma 67.5.1.
$\square$

Lemma 114.15.5. Let $S$ be a scheme. Let $X$, $Y$ be algebraic spaces over $S$. Let $Y \to X$ be a representable morphism. If $X$ is very reasonable, so is $Y$.

**Proof.**
This is case $(\epsilon )$ of Decent Spaces, Lemma 67.5.3.
$\square$

Remark 114.15.6. Very reasonable algebraic spaces form a strictly larger collection than Zariski locally quasi-separated algebraic spaces. Consider an algebraic space of the form $X = [U/G]$ (see Spaces, Definition 64.14.4) where $G$ is a finite group acting without fixed points on a non-quasi-separated scheme $U$. Namely, in this case $U \times _ X U = U \times G$ and clearly both projections to $U$ are quasi-compact, hence $X$ is very reasonable. On the other hand, the diagonal $U \times _ X U \to U \times U$ is not quasi-compact, hence this algebraic space is not quasi-separated. Now, take $U$ the infinite affine space over a field $k$ of characteristic $\not= 2$ with zero doubled, see Schemes, Example 26.21.4. Let $0_1, 0_2$ be the two zeros of $U$. Let $G = \{ +1, -1\} $, and let $-1$ act by $-1$ on all coordinates, and by switching $0_1$ and $0_2$. Then $[U/G]$ is very reasonable but not Zariski locally quasi-separated (details omitted).

Warning: The following lemma should be used with caution, as the schemes $U_ i$ in it are not necessarily separated or even quasi-separated.

Lemma 114.15.7. Let $S$ be a scheme. Let $X$ be a very reasonable algebraic space over $S$. There exists a set of schemes $U_ i$ and morphisms $U_ i \to X$ such that

each $U_ i$ is a quasi-compact scheme,

each $U_ i \to X$ is étale,

both projections $U_ i \times _ X U_ i \to U_ i$ are quasi-compact, and

the morphism $\coprod U_ i \to X$ is surjective (and étale).

**Proof.**
Decent Spaces, Definition 67.6.1 says that there exist $U_ i \to X$ such that (2), (3) and (4) hold. Fix $i$, and set $R_ i = U_ i \times _ X U_ i$, and denote $s, t : R_ i \to U_ i$ the projections. For any affine open $W \subset U_ i$ the open $W' = t(s^{-1}(W)) \subset U_ i$ is a quasi-compact $R_ i$-invariant open (see Groupoids, Lemma 39.19.2). Hence $W'$ is a quasi-compact scheme, $W' \to X$ is étale, and $W' \times _ X W' = s^{-1}(W') = t^{-1}(W')$ so both projections $W' \times _ X W' \to W'$ are quasi-compact. This means the family of $W' \to X$, where $W \subset U_ i$ runs through the members of affine open coverings of the $U_ i$ gives what we want.
$\square$

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