Lemma 86.19.4. 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 representable by algebraic spaces,

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, $U \to X$ is surjective étale, and $U \to V$ is representable by algebraic spaces,

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 the vertical arrows are representable by algebraic spaces, the morphism $U \to V$ is representable by algebraic spaces,

4. there exists a covering $\{ Y_ j \to Y\}$ as in Definition 86.11.1 and for each $j$ a covering $\{ X_{ji} \to Y_ j \times _ Y X\}$ as in Definition 86.11.1 such that $X_{ji} \to Y_ j$ is representable by algebraic spaces for each $j$ and $i$,

5. there exist a covering $\{ X_ i \to X\}$ as in Definition 86.11.1 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, such that $X_ i \to Y_ i$ is representable by algebraic spaces, and

Proof. It is clear that (1) implies (2) because we can take $U = X$ and $V = Y$. Conversely, (2) implies (1) by Bootstrap, Lemma 79.11.4 applied to $U \to X \to Y$.

Assume (1) is true and consider a diagram as in (3). Then $U \to Y$ is representable by algebraic spaces (as the composition $U \to X \to Y$, see Bootstrap, Lemma 79.3.8) and factors through $V$. Thus $U \to V$ is representable by algebraic spaces by Lemma 86.19.3.

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 86.15.3. 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 86.11.1 and for each $j$ any covering $\{ X_{ji} \to Y_ j \times _ Y X\}$ as in Definition 86.11.1. Then $X_{ij} \to Y_ j$ is representable by algebraic spaces by (3) and we see that (4) is true. This concludes the proof. $\square$

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