The Stacks project

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. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0DYP. Beware of the difference between the letter 'O' and the digit '0'.