The Stacks project

Lemma 88.10.5. Let $\mathcal{F}$ be a category cofibered in groupoids over $\mathcal{C}_\Lambda $.

  1. If $\mathcal{F}$ satisfies (S1), then so does $\overline{\mathcal{F}}$.

  2. If $\mathcal{F}$ satisfies (S2), then so does $\overline{\mathcal{F}}$ provided at least one of the following conditions is satisfied

    1. $\mathcal{F}$ is a predeformation category,

    2. the category $\mathcal{F}(k)$ is a set or a setoid, or

    3. for any morphism $x_\epsilon \to x_0$ of $\mathcal{F}$ lying over $k[\epsilon ] \to k$ the pushforward map $\text{Aut}_{k[\epsilon ]}(x_\epsilon ) \to \text{Aut}_ k(x_0)$ is surjective.

Proof. Assume $\mathcal{F}$ has (S1). Suppose we have ring maps $f_ i : A_ i \to A$ in $\mathcal{C}_\Lambda $ with $f_2$ surjective. Let $x_ i \in \mathcal{F}(A_ i)$ such that the pushforwards $f_{1, *}(x_1)$ and $f_{2, *}(x_2)$ are isomorphic. Then we can denote $x$ an object of $\mathcal{F}$ over $A$ isomorphic to both of these and we obtain a diagram as in (S1). Hence we find an object $y$ of $\mathcal{F}$ over $A_1 \times _ A A_2$ whose pushforward to $A_1$, resp. $A_2$ is isomorphic to $x_1$, resp. $x_2$. In this way we see that (S1) holds for $\overline{\mathcal{F}}$.

Assume $\mathcal{F}$ has (S2). The first part of (S2) for $\overline{\mathcal{F}}$ follows as in the argument above. The second part of (S2) for $\overline{\mathcal{F}}$ signifies that the map

\[ \overline{\mathcal{F}}(A \times _ k k[\epsilon ]) \to \overline{\mathcal{F}}(A) \times _{\overline{\mathcal{F}}(k)} \overline{\mathcal{F}}(k[\epsilon ]) \]

is injective for any ring $A$ in $\mathcal{C}_\Lambda $. Suppose that $y, y' \in \mathcal{F}(A \times _ k k[\epsilon ])$. Using the axioms of cofibred categories we can choose commutative diagrams

\[ \vcenter { \xymatrix{ y \ar[r]_ c \ar[d]_ a & x_\epsilon \ar[d]^ e \\ x \ar[r]^ d & x_0 } } \quad \text{and}\quad \vcenter { \xymatrix{ y' \ar[r]_{c'} \ar[d]_{a'} & x'_\epsilon \ar[d]^{e'} \\ x' \ar[r]^{d'} & x'_0 } } \quad \text{lying over}\quad \vcenter { \xymatrix{ A \times _ k k[\epsilon ] \ar[d] \ar[r] & k[\epsilon ] \ar[d] \\ A \ar[r] & k } } \]

Assume that there exist isomorphisms $\alpha : x \to x'$ in $\mathcal{F}(A)$ and $\beta : x_\epsilon \to x'_\epsilon $ in $\mathcal{F}(k[\epsilon ])$. This also means there exists an isomorphism $\gamma : x_0 \to x'_0$ compatible with $\alpha $. To prove (S2) for $\overline{\mathcal{F}}$ we have to show that there exists an isomorphism $y \to y'$ in $\mathcal{F}(A \times _ k k[\epsilon ])$. By (S2) for $\mathcal{F}$ such a morphism will exist if we can choose the isomorphisms $\alpha $ and $\beta $ and $\gamma $ such that

\[ \xymatrix{ x \ar[d]^\alpha \ar[r] & x_0 \ar[d]^\gamma & x_\epsilon \ar[d]^\beta \ar[l]^ e \\ x' \ar[r] & x'_0 & x'_\epsilon \ar[l]_{e'} } \]

is commutative (because then we can replace $x$ by $x'$ and $x_\epsilon $ by $x'_\epsilon $ in the previous displayed diagram). The left hand square commutes by our choice of $\gamma $. We can factor $e' \circ \beta $ as $\gamma ' \circ e$ for some second map $\gamma ' : x_0 \to x'_0$. Now the question is whether we can arrange it so that $\gamma = \gamma '$? This is clear if $\mathcal{F}(k)$ is a set, or a setoid. Moreover, if $\text{Aut}_{k[\epsilon ]}(x_\epsilon ) \to \text{Aut}_ k(x_0)$ is surjective, then we can adjust the choice of $\beta $ by precomposing with an automorphism of $x_\epsilon $ whose image is $\gamma ^{-1} \circ \gamma '$ to make things work. $\square$


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 06I0. Beware of the difference between the letter 'O' and the digit '0'.