Lemma 89.20.3. Let $f : \mathcal{F} \to \mathcal{G}$ be a map of deformation categories. Let $x_0 \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{F}(k))$ with image $y_0 \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{G}(k))$. If

1. the map $T\mathcal{F} \to T\mathcal{G}$ is surjective, and

2. for every small extension $A' \to A$ in $\mathcal{C}_\Lambda$ and $x \in \mathcal{F}(A)$ with image $y \in \mathcal{G}(A)$ if there is a lift of $y$ to $A'$, then there is a lift of $x$ to $A'$,

then $\mathcal{F} \to \mathcal{G}$ is smooth (and vice versa).

Proof. Let $A' \to A$ be a small extension. Let $x \in \mathcal{F}(A)$. Let $y' \to f(x)$ be a morphism in $\mathcal{G}$ over $A' \to A$. Consider the functor $\text{Lift}(A', x) \to \text{Lift}(A', f(x))$ induced by $f$. We have to show that there exists an object $x' \to x$ of $\text{Lift}(A', x)$ mapping to $y' \to f(x)$, see Lemma 89.8.2. By condition (2) we know that $\text{Lift}(A', x)$ is not the empty category. By condition (2) and Lemma 89.17.5 we conlude that the map on isomorphism classes is surjective as desired. $\square$

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