Lemma 101.16.1. Let $\mathcal{P}$ be a property of morphisms of algebraic spaces which is smooth local on the source-and-target. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Consider commutative diagrams

$\xymatrix{ U \ar[d]_ a \ar[r]_ h & V \ar[d]^ b \\ \mathcal{X} \ar[r]^ f & \mathcal{Y} }$

where $U$ and $V$ are algebraic spaces and the vertical arrows are smooth. The following are equivalent

1. for any diagram as above such that in addition $U \to \mathcal{X} \times _\mathcal {Y} V$ is smooth the morphism $h$ has property $\mathcal{P}$, and

2. for some diagram as above with $a : U \to \mathcal{X}$ surjective the morphism $h$ has property $\mathcal{P}$.

If $\mathcal{X}$ and $\mathcal{Y}$ are representable by algebraic spaces, then this is also equivalent to $f$ (as a morphism of algebraic spaces) having property $\mathcal{P}$. If $\mathcal{P}$ is also preserved under any base change, and fppf local on the base, then for morphisms $f$ which are representable by algebraic spaces this is also equivalent to $f$ having property $\mathcal{P}$ in the sense of Properties of Stacks, Section 100.3.

Proof. Let us prove the implication (1) $\Rightarrow$ (2). Pick an algebraic space $V$ and a surjective and smooth morphism $V \to \mathcal{Y}$. Pick an algebraic space $U$ and a surjective and smooth morphism $U \to \mathcal{X} \times _\mathcal {Y} V$. Note that $U \to \mathcal{X}$ is surjective and smooth as well, as a composition of the base change $\mathcal{X} \times _\mathcal {Y} V \to \mathcal{X}$ and the chosen map $U \to \mathcal{X} \times _\mathcal {Y} V$. Hence we obtain a diagram as in (1). Thus if (1) holds, then $h : U \to V$ has property $\mathcal{P}$, which means that (2) holds as $U \to \mathcal{X}$ is surjective.

Conversely, assume (2) holds and let $U, V, a, b, h$ be as in (2). Next, let $U', V', a', b', h'$ be any diagram as in (1). Picture

$\xymatrix{ U \ar[d] \ar[r]_ h & V \ar[d] \\ \mathcal{X} \ar[r]^ f & \mathcal{Y} } \quad \quad \xymatrix{ U' \ar[d] \ar[r]_{h'} & V' \ar[d] \\ \mathcal{X} \ar[r]^ f & \mathcal{Y} }$

To show that (2) implies (1) we have to prove that $h'$ has $\mathcal{P}$. To do this consider the commutative diagram

$\xymatrix{ U \ar[dd]^ h & U \times _\mathcal {X} U' \ar[d] \ar[l] \ar@/^6ex/[dd]^{(h, h')} \ar[r] & U' \ar[dd]^{h'} \\ & U \times _\mathcal {Y} V' \ar[lu] \ar[d] & \\ V & V \times _\mathcal {Y} V' \ar[l] \ar[r] & V' }$

of algebraic spaces. Note that the horizontal arrows are smooth as base changes of the smooth morphisms $V \to \mathcal{Y}$, $V' \to \mathcal{Y}$, $U \to \mathcal{X}$, and $U' \to \mathcal{X}$. Note that

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

is cartesian, hence the left vertical arrow is smooth as $U', V', a', b', h'$ is as in (1). Since $\mathcal{P}$ is smooth local on the target by Descent on Spaces, Lemma 74.20.2 part (2) we see that the base change $U \times _\mathcal {Y} V' \to V \times _\mathcal {Y} V'$ has $\mathcal{P}$. Since $\mathcal{P}$ is smooth local on the source by Descent on Spaces, Lemma 74.20.2 part (1) we can precompose by the smooth morphism $U \times _\mathcal {X} U' \to U \times _\mathcal {Y} V'$ and conclude $(h, h')$ has $\mathcal{P}$. Since $V \times _\mathcal {Y} V' \to V'$ is smooth we conclude $U \times _\mathcal {X} U' \to V'$ has $\mathcal{P}$ by Descent on Spaces, Lemma 74.20.2 part (3). Finally, since $U \times _ X U' \to U'$ is surjective and smooth and $\mathcal{P}$ is smooth local on the source (same lemma) we conclude that $h'$ has $\mathcal{P}$. This finishes the proof of the equivalence of (1) and (2).

If $\mathcal{X}$ and $\mathcal{Y}$ are representable, then Descent on Spaces, Lemma 74.20.3 applies which shows that (1) and (2) are equivalent to $f$ having $\mathcal{P}$.

Finally, suppose $f$ is representable, and $U, V, a, b, h$ are as in part (2) of the lemma, and that $\mathcal{P}$ is preserved under arbitrary base change. We have to show that for any scheme $Z$ and morphism $Z \to \mathcal{X}$ the base change $Z \times _\mathcal {Y} \mathcal{X} \to Z$ has property $\mathcal{P}$. Consider the diagram

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

Note that the top horizontal arrow is a base change of $h$ and hence has property $\mathcal{P}$. The left vertical arrow is smooth and surjective and the right vertical arrow is smooth. Thus Descent on Spaces, Lemma 74.20.3 kicks in and shows that $Z \times _\mathcal {Y} \mathcal{X} \to Z$ has property $\mathcal{P}$. $\square$

Comment #7206 by Anonymous on

A pedantic comment: in the lemma statement, should "Consider commutative diagrams..." be "Consider $2$-commutative diagrams..."? Similarly at Section 0CIE. Another option might be to add to the abuse of notation list in Section 04XE.

For example, in Section 04XE, one could add the following line to "Here are our conventions regarding morphisms of algebraic stacks:"

(3) We may abuse notation and say that a diagram of algebraic stacks commutes if the diagram is $2$-commutative in the $2$-category of algebraic stacks.

Comment #7463 by Anonymous on

Typo in my previous comment, should have been Section 04XA instead of 04XE.

Comment #7614 by on

Very good! OK, I have decided to go with your second option and added this as possible abuse of language here. But I would be interested in somebody carefully going through all the possible cases and add "2-" everywhere and then removing this additional abuse. What do people think?

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