Lemma 105.8.6. Let $T \to T'$ be a first order thickening of affine schemes. Let $\mathcal{X}'$ be an algebraic stack over $T'$ whose structure morphism $\mathcal{X}' \to T'$ is smooth. Let $x : T \to \mathcal{X}'$ be a morphism over $T'$. Then there exists a morphsm $x' : T' \to \mathcal{X}'$ over $T'$ with $x'|_ T = x$.

Proof. We may apply the result of Lemma 105.7.4. Thus it suffices to construct a smooth surjective morphism $W' \to T'$ with $W'$ affine such that $x|_{T \times _{W'} T'}$ lifts to $W'$. (We urge the reader to find their own proof of this fact using the analogous result for algebraic spaces already established.) We choose a scheme $U'$ and a surjective smooth morphism $U' \to \mathcal{X}'$. Observe that $U' \to T'$ is smooth and that the projection $T \times _{\mathcal{X}'} U' \to T$ is surjective smooth. Choose an affine scheme $W$ and an étale morphism $W \to T \times _{\mathcal{X}'} U'$ such that $W \to T$ is surjective. Then $W \to T$ is a smooth morphism of affine schemes. After replacing $W$ by a disjoint union of principal affine opens, we may assume there exists a smooth morphism of affines $W' \to T'$ such that $W = T \times _{T'} W'$, see Algebra, Lemma 10.137.20. By More on Morphisms of Spaces, Lemma 75.19.6 we can find a morphism $W' \to U'$ over $T'$ lifting the given morphism $W \to U'$. This finishes the proof. $\square$

Comment #5424 by Will Chen on

In the proof: "esthablished"

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