The Stacks project

Lemma 105.8.7 (Infinitesimal lifting criterion). Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. The following are equivalent:

  1. The morphism $f$ is smooth.

  2. The morphism $f$ is locally of finite presentation and formally smooth.

Proof. Assume $f$ is smooth. Then $f$ is locally of finite presentation by Morphisms of Stacks, Lemma 100.33.5. Hence it suffices given a diagram (105.8.1.1) and a $\gamma : y \circ i \to f \circ x$ to find a dotted arrow (see Lemma 105.8.2). Forming fibre products we obtain

\[ \xymatrix{ T \ar[d] \ar[r] & T' \times _\mathcal {Y} \mathcal{X} \ar[d] \ar[r] & \mathcal{X} \ar[d] \\ T' \ar[r] & T' \ar[r] & \mathcal{Y} } \]

Thus we see it is sufficient to find a dotted arrow in the left square. Since $T' \times _\mathcal {Y} \mathcal{X} \to T'$ is smooth (Morphisms of Stacks, Lemma 100.33.3) existence of a dotted arrow in the left square is guaranteed by Lemma 105.8.6.

Conversely, suppose that $f$ is locally of finite presentation and formally smooth. Choose a scheme $U$ and a surjective smooth morphism $U \to \mathcal{X}$. Then $a : U \to \mathcal{X}$ and $b : U \to \mathcal{Y}$ are representable by algebraic spaces and locally of finite presentation (use Morphisms of Stacks, Lemma 100.27.2 and the fact seen above that a smooth morphism is locally of finite presentation). We will apply the general principle of Algebraic Stacks, Lemma 93.10.9 with as input the equivalence of More on Morphisms of Spaces, Lemma 75.19.6 and simultaneously use the translation of Criteria for Representability, Lemma 96.6.3. We first apply this to $a$ to see that $a$ is formally smooth on objects. Next, we use that $f$ is formally smooth on objects by assumption (see Lemma 105.8.2) and Criteria for Representability, Lemma 96.6.2 to see that $b = f \circ a$ is formally smooth on objects. Then we apply the principle once more to conclude that $b$ is smooth. This means that $f$ is smooth by the definition of smoothness for morphisms of algebraic stacks and the proof is complete. $\square$


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