Lemma 87.24.6. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of formal algebraic spaces over $S$. The following are equivalent:

1. the morphism $f$ is locally of finite type,

2. there exists a commutative diagram

$\xymatrix{ U \ar[d] \ar[r] & V \ar[d] \\ X \ar[r] & Y }$

where $U$, $V$ are formal algebraic spaces, the vertical arrows are representable by algebraic spaces and étale, $U \to X$ is surjective, and $U \to V$ is locally of finite type,

3. for any commutative diagram

$\xymatrix{ U \ar[d] \ar[r] & V \ar[d] \\ X \ar[r] & Y }$

where $U$, $V$ are formal algebraic spaces and vertical arrows representable by algebraic spaces and étale, the morphism $U \to V$ is locally of finite type,

4. there exists a covering $\{ Y_ j \to Y\}$ as in Definition 87.11.1 and for each $j$ a covering $\{ X_{ji} \to Y_ j \times _ Y X\}$ as in Definition 87.11.1 such that $X_{ji} \to Y_ j$ is locally of finite type for each $j$ and $i$,

5. there exist a covering $\{ X_ i \to X\}$ as in Definition 87.11.1 and for each $i$ a factorization $X_ i \to Y_ i \to Y$ where $Y_ i$ is an affine formal algebraic space, $Y_ i \to Y$ is representable by algebraic spaces and étale, such that $X_ i \to Y_ i$ is locally of finite type, and

Proof. In each of the 5 cases the morphism $f : X \to Y$ is representable by algebraic spaces, see Lemma 87.19.4. We will use this below without further mention.

It is clear that (1) implies (2) because we can take $U = X$ and $V = Y$. Conversely, assume given a diagram as in (2). Let $T$ be a scheme and let $T \to Y$ be a morphism. Then we can consider

$\xymatrix{ U \times _ Y T \ar[d] \ar[r] & V \times _ Y T \ar[d] \\ X \times _ Y T \ar[r] & T }$

The vertical arrows are étale and the top horizontal arrow is locally of finite type as base changes of such morphisms. Hence by Morphisms of Spaces, Lemma 67.23.4 we conclude that $X \times _ Y T \to T$ is locally of finite type. In other words (1) holds.

Assume (1) is true and consider a diagram as in (3). Then $U \to Y$ is locally of finite type (as the composition $U \to X \to Y$, see Bootstrap, Lemma 80.4.3). Let $T$ be a scheme and let $T \to V$ be a morphism. Then the projection $T \times _ V U \to T$ factors as

$T \times _ V U = (T \times _ Y U) \times _{(V \times _ Y V)} V \to T \times _ Y U \to T$

The second arrow is locally of finite type (as a base change of the composition $U \to X \to Y$) and the first is the base change of the diagonal $V \to V \times _ Y V$ which is locally of finite type by Lemma 87.15.5.

It is clear that (3) implies (2). Thus now (1) – (3) are equivalent.

Observe that the condition in (4) makes sense as the fibre product $Y_ j \times _ Y X$ is a formal algebraic space by Lemma 87.15.3. It is clear that (4) implies (5).

Assume $X_ i \to Y_ i \to Y$ as in (5). Then we set $V = \coprod Y_ i$ and $U = \coprod X_ i$ to see that (5) implies (2).

Finally, assume (1) – (3) are true. Thus we can choose any covering $\{ Y_ j \to Y\}$ as in Definition 87.11.1 and for each $j$ any covering $\{ X_{ji} \to Y_ j \times _ Y X\}$ as in Definition 87.11.1. Then $X_{ij} \to Y_ j$ is locally of finite type by (3) and we see that (4) is true. This concludes the proof. $\square$

Comment #1964 by Brian Conrad on

The sentence preceding this in the main text should end with a colon (to convey that its justification is this result), and in the paragraph near the end of the proof beginning with "Assume (1) in true" the map $T \times_Y U \rightarrow U$ should have target $T$ (occurs twice).

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