We still have to develop the general machinery needed to say what it means for a morphism of algebraic stacks to have a given property at a point. For the moment the following lemma is sufficient.
Lemma 101.26.1. 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 100.4.3 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 67.31.3. 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 101.25.3. Composing and using Morphisms of Spaces, Lemma 67.31.4 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 67.31.5.
\square
Comments (0)