Lemma 100.26.1. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Let $x \in |\mathcal{X}|$. Consider commutative diagrams

\[ \vcenter { \xymatrix{ U \ar[d]_ a \ar[r]_ h & V \ar[d]^ b \\ \mathcal{X} \ar[r]^ f & \mathcal{Y} } } \quad \text{with points} \vcenter { \xymatrix{ u \in |U| \ar[d] \\ x \in |\mathcal{X}| } } \]

where $U$ and $V$ are algebraic spaces, $b$ is flat, and $(a, h) : U \to \mathcal{X} \times _\mathcal {Y} V$ is flat. The following are equivalent

$h$ is flat at $u$ for one diagram as above,

$h$ is flat at $u$ for every diagram as above.

**Proof.**
Suppose we are given a second diagram $U', V', u', a', b', h'$ as in the lemma. Then we can consider

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

By Properties of Stacks, Lemma 99.4.3 there is a point $u'' \in |U \times _\mathcal {X} U'|$ mapping to $u$ and $u'$. If $h$ is flat at $u$, then the base change $U \times _ V (V \times _\mathcal {Y} V') \to V \times _\mathcal {Y} V'$ is flat at any point over $u$, see Morphisms of Spaces, Lemma 66.31.3. On the other hand, the morphism

\[ U \times _\mathcal {X} U' \to U \times _\mathcal {X} (\mathcal{X} \times _\mathcal {Y} V') = U \times _\mathcal {Y} V' = U \times _ V (V \times _\mathcal {Y} V') \]

is flat as a base change of $(a', h')$, see Lemma 100.25.3. Composing and using Morphisms of Spaces, Lemma 66.31.4 we conclude that $U \times _\mathcal {X} U' \to V \times _\mathcal {Y} V'$ is flat at $u''$. Then we can use composition by the flat map $V \times _\mathcal {Y} V' \to V'$ to conclude that $U \times _\mathcal {X} U' \to V'$ is flat at $u''$. Finally, since $U \times _\mathcal {X} U' \to U'$ is flat at $u''$ and $u''$ maps to $u'$ we conclude that $U' \to V'$ is flat at $u'$ by Morphisms of Spaces, Lemma 66.31.5.
$\square$

## Comments (0)