There is a very natural property of categories fibred in groupoids over $\mathcal{C}_\Lambda $ which is easy to check in practice and which implies Schlessinger's properties (S1) and (S2) we have introduced earlier.
Definition 90.16.1. Let $\mathcal{F}$ be a category cofibered in groupoids over $\mathcal C_\Lambda $. We say that $\mathcal{F}$ satisfies condition (RS) if for every diagram in $\mathcal{F}$
\[ \vcenter { \xymatrix{ & x_2 \ar[d] \\ x_1 \ar[r] & x } } \quad \text{lying over}\quad \vcenter { \xymatrix{ & A_2 \ar[d] \\ A_1 \ar[r] & A } } \]
in $\mathcal{C}_\Lambda $ with $A_2 \to A$ surjective, there exists a fiber product $x_1 \times _ x x_2$ in $\mathcal{F}$ such that the diagram
\[ \vcenter { \xymatrix{ x_1 \times _ x x_2 \ar[r] \ar[d] & x_2 \ar[d] \\ x_1 \ar[r] & x } } \quad \text{lies over}\quad \vcenter { \xymatrix{ A_1 \times _ A A_2 \ar[r] \ar[d] & A_2 \ar[d] \\ A_1 \ar[r] & A. } } \]
Lemma 90.16.2. Let $\mathcal{F}$ be a category cofibered in groupoids over $\mathcal{C}_\Lambda $ satisfying (RS). Given a commutative diagram in $\mathcal{F}$
\[ \vcenter { \xymatrix{ y \ar[r] \ar[d] & x_2 \ar[d] \\ x_1 \ar[r] & x } } \quad \text{lying over}\quad \vcenter { \xymatrix{ A_1 \times _ A A_2 \ar[r] \ar[d] & A_2 \ar[d] \\ A_1 \ar[r] & A. } } \]
with $A_2 \to A$ surjective, then it is a fiber square.
Proof.
Since $\mathcal{F}$ satisfies (RS), there exists a fiber product diagram
\[ \vcenter { \xymatrix{ x_1 \times _ x x_2 \ar[r] \ar[d] & x_2 \ar[d] \\ x_1 \ar[r] & x } } \quad \text{lying over}\quad \vcenter { \xymatrix{ A_1 \times _ A A_2 \ar[r] \ar[d] & A_2 \ar[d] \\ A_1 \ar[r] & A. } } \]
The induced map $y \to x_1 \times _ x x_2$ lies over $\text{id} : A_1 \times _ A A_1 \to A_1 \times _ A A_1$, hence it is an isomorphism.
$\square$
Lemma 90.16.3. Let $\mathcal{F}$ be a category cofibered in groupoids over $\mathcal C_\Lambda $. Then $\mathcal{F}$ satisfies (RS) if the condition in Definition 90.16.1 is assumed to hold only when $A_2 \to A$ is a small extension.
Proof.
Apply Lemma 90.3.3. The proof is similar to that of Lemma 90.8.2.
$\square$
Lemma 90.16.4. Let $\mathcal{F}$ be a category cofibered in groupoids over $\mathcal{C}_\Lambda $. The following are equivalent
$\mathcal{F}$ satisfies (RS),
the functor $\mathcal{F}(A_1 \times _ A A_2) \to \mathcal{F}(A_1) \times _{\mathcal{F}(A)} \mathcal{F}(A_2)$ see (90.10.1.1) is an equivalence of categories whenever $A_2 \to A$ is surjective, and
same as in (2) whenever $A_2 \to A$ is a small extension.
Proof.
Assume (1). By Lemma 90.16.2 we see that every object of $\mathcal{F}(A_1 \times _ A A_2)$ is of the form $x_1 \times _ x x_2$. Moreover
\[ \mathop{\mathrm{Mor}}\nolimits _{A_1 \times _ A A_2}(x_1 \times _ x x_2, y_1 \times _ y y_2) = \mathop{\mathrm{Mor}}\nolimits _{A_1}(x_1, y_1) \times _{\mathop{\mathrm{Mor}}\nolimits _ A(x, y)} \mathop{\mathrm{Mor}}\nolimits _{A_2}(x_2, y_2). \]
Hence we see that $\mathcal{F}(A_1 \times _ A A_2)$ is a $2$-fibre product of $\mathcal{F}(A_1)$ with $\mathcal{F}(A_2)$ over $\mathcal{F}(A)$ by Categories, Remark 4.31.5. In other words, we see that (2) holds.
The implication (2) $\Rightarrow $ (3) is immediate.
Assume (3). Let $q_1 : A_1 \to A$ and $q_2 : A_2 \to A$ be given with $q_2$ a small extension. We will use the description of the $2$-fibre product $\mathcal{F}(A_1) \times _{\mathcal{F}(A)} \mathcal{F}(A_2)$ from Categories, Remark 4.31.5. Hence let $y \in \mathcal{F}(A_1 \times _ A A_2)$ correspond to $(x_1, x_2, x, a_1 : x_1 \to x, a_2 : x_2 \to x)$. Let $z$ be an object of $\mathcal{F}$ lying over $C$. Then
\begin{align*} \mathop{\mathrm{Mor}}\nolimits _\mathcal {F}(z, y) & = \{ (f, \alpha ) \mid f : C \to A_1 \times _ A A_2, \alpha : f_*z \to y\} \\ & = \{ (f_1, f_2, \alpha _1, \alpha _2) \mid f_ i : C \to A_ i, \ \alpha _ i : f_{i, *}z \to x_ i, \\ & \quad \quad \quad \quad q_1 \circ f_1 = q_2 \circ f_2, \ q_{1, *} \alpha _1 = q_{2, *}\alpha _2\} \\ & = \mathop{\mathrm{Mor}}\nolimits _\mathcal {F}(z, x_1) \times _{\mathop{\mathrm{Mor}}\nolimits _\mathcal {F}(z, x)} \mathop{\mathrm{Mor}}\nolimits _\mathcal {F}(z, x_2) \end{align*}
whence $y$ is a fibre product of $x_1$ and $x_2$ over $x$. Thus we see that $\mathcal{F}$ satisfies (RS) in case $A_2 \to A$ is a small extension. Hence (RS) holds by Lemma 90.16.3.
$\square$
Lemma 90.16.6. Let $\mathcal{F}$ be a category cofibered in groupoids over $\mathcal{C}_\Lambda $. The condition (RS) for $\mathcal{F}$ implies both (S1) and (S2) for $\mathcal{F}$.
Proof.
Using the reformulation of Lemma 90.16.4 and the explanation of (S1) following Definition 90.10.1 it is immediate that (RS) implies (S1). This proves the first part of (S2). The second part of (S2) follows because Lemma 90.16.2 tells us that $y = x_1 \times _{d, x_0, e} x_2 = y'$ if $y, y'$ are as in the second part of the definition of (S2) in Definition 90.10.1. (In fact the morphism $y \to y'$ is compatible with both $a, a'$ and $c, c'$!)
$\square$
The following lemma is the analogue of Lemma 90.10.5. Recall that if $\mathcal{F}$ is a category cofibred in groupoids over $\mathcal{C}_\Lambda $ and $x$ is an object of $\mathcal{F}$ lying over $A$, then we denote $\text{Aut}_ A(x) = \mathop{\mathrm{Mor}}\nolimits _ A(x, x) = \mathop{\mathrm{Mor}}\nolimits _{\mathcal{F}(A)}(x, x)$. If $x' \to x$ is a morphism of $\mathcal{F}$ lying over $A' \to A$ then there is a well defined map of groups $\text{Aut}_{A'}(x') \to \text{Aut}_ A(x)$.
Lemma 90.16.7. Let $\mathcal{F}$ be a category cofibered in groupoids over $\mathcal{C}_\Lambda $ satisfying (RS). The following conditions are equivalent:
$\overline{\mathcal{F}}$ satisfies (RS).
Let $f_1: A_1 \to A$ and $f_2: A_2 \to A$ be ring maps in $\mathcal{C}_\Lambda $ with $f_2$ surjective. The induced map of sets of isomorphism classes
\[ \overline{\mathcal{F}(A_1) \times _{\mathcal{F}(A)} \mathcal{F}(A_2)} \to \overline{\mathcal{F}}(A_1) \times _{\overline{\mathcal{F}}(A)} \overline{\mathcal{F}}(A_2) \]
is injective.
For every morphism $x' \to x$ in $\mathcal{F}$ lying over a surjective ring map $A' \to A$, the map $\text{Aut}_{A'}(x') \to \text{Aut}_ A(x)$ is surjective.
For every morphism $x' \to x$ in $\mathcal{F}$ lying over a small extension $A' \to A$, the map $\text{Aut}_{A'}(x') \to \text{Aut}_ A(x)$ is surjective.
Proof.
We prove that (1) is equivalent to (2) and (2) is equivalent to (3). The equivalence of (3) and (4) follows from Lemma 90.3.3.
Let $f_1: A_1 \to A$ and $f_2: A_2 \to A$ be ring maps in $\mathcal{C}_\Lambda $ with $f_2$ surjective. By Remark 90.16.5 we see $\overline{\mathcal{F}}$ satisfies (RS) if and only if the map
\[ \overline{\mathcal{F}}(A_1 \times _ A A_2) \to \overline{\mathcal F}(A_1) \times _{\overline{\mathcal{F}}(A)} \overline{\mathcal{F}}(A_2) \]
is bijective for any such $f_1, f_2$. This map is at least surjective since that is the condition of (S1) and $\overline{\mathcal{F}}$ satisfies (S1) by Lemmas 90.16.6 and 90.10.5. Moreover, this map factors as
\[ \overline{\mathcal{F}}(A_1 \times _ A A_2) \longrightarrow \overline{\mathcal{F}(A_1) \times _{\mathcal{F}(A)} \mathcal{F}(A_2)} \longrightarrow \overline{\mathcal{F}}(A_1) \times _{\overline{\mathcal{F}}(A)} \overline{\mathcal{F}}(A_2), \]
where the first map is a bijection since
\[ \mathcal{F}(A_1 \times _ A A_2) \longrightarrow \mathcal{F}(A_1) \times _{\mathcal{F}(A)} \mathcal{F}(A_2) \]
is an equivalence by (RS) for $\mathcal{F}$. Hence (1) is equivalent to (2).
Assume (2) holds. Let $x' \to x$ be a morphism in $\mathcal{F}$ lying over a surjective ring map $f: A' \to A$. Let $a \in \text{Aut}_ A(x)$. The objects
\[ (x', x', a : x \to x), \ (x', x', \text{id} : x \to x) \]
of $\mathcal{F}(A') \times _{\mathcal{F}(A)} \mathcal{F}(A')$ have the same image in $\overline{\mathcal{F}}(A') \times _{\overline{\mathcal{F}}(A)} \overline{\mathcal{F}}(A')$. By (2) there exists maps $b_1, b_2 : x' \to x'$ such that
\[ \xymatrix{ x \ar[r]_ a \ar[d]_{f_*b_1} & x \ar[d]^{f_*b_2} \\ x \ar[r]^{\text{id}} & x } \]
commutes. Hence $b_2^{-1} \circ b_1 \in \text{Aut}_{A'}(x')$ has image $a \in \text{Aut}_ A(x)$. Hence (3) holds.
Assume (3) holds. Suppose
\[ (x_1, x_2, a : (f_1)_*x_1 \to (f_2)_*x_2), \ (x'_1, x'_2, a' : (f_1)_*x'_1 \to (f_2)_*x'_2) \]
are objects of $\mathcal{F}(A_1) \times _{\mathcal{F}(A)} \mathcal{F}(A_2)$ with the same image in $\overline{\mathcal{F}}(A_1) \times _{\overline{\mathcal{F}}(A)} \overline{\mathcal{F}}(A_2)$. Then there are morphisms $b_1: x_1 \to x'_1$ in $\mathcal{F}(A_1)$ and $b_2: x_2 \to x'_2$ in $\mathcal F(A_2)$. By (3) we can modify $b_2$ by an automorphism of $x_2$ over $A_2$ so that the diagram
\[ \xymatrix{ (f_1)_*x_1 \ar[r]_ a \ar[d]_{(f_1)_*b_1} & (f_2)_*x_2 \ar[d]^{(f_2)_*b_2} \\ (f_1)_*x'_1 \ar[r]^{a'} & (f_2)_*x'_2. } \]
commutes. This proves $(x_1, x_2, a) \cong (x'_1, x'_2, a')$ in $\overline{\mathcal{F}(A_1) \times _{\mathcal{F}(A)} \mathcal{F}(A_2)}$. Hence (2) holds.
$\square$
Finally we define the notion of a deformation category.
Definition 90.16.8. A deformation category is a predeformation category $\mathcal{F}$ satisfying (RS). A morphism of deformation categories is a morphism of categories over $\mathcal{C}_\Lambda $.
Example 90.16.10. A prorepresentable functor $F$ is a deformation functor. Namely, suppose $R \in \mathop{\mathrm{Ob}}\nolimits (\widehat{\mathcal{C}}_\Lambda )$ and $F(A) = \mathop{\mathrm{Mor}}\nolimits _{\widehat{\mathcal{C}}_\Lambda }(R, A)$. There is a unique morphism $R \to k$, so $F(k)$ is a one element set. Since
\[ \mathop{\mathrm{Hom}}\nolimits _\Lambda (R, A_1 \times _ A A_2) = \mathop{\mathrm{Hom}}\nolimits _\Lambda (R, A_1) \times _{\mathop{\mathrm{Hom}}\nolimits _\Lambda (R, A)} \mathop{\mathrm{Hom}}\nolimits _\Lambda (R, A_2) \]
the same is true for maps in $\widehat{\mathcal{C}}_\Lambda $ and we see that $F$ has (RS).
The following is one of our typical remarks on passing from a category cofibered in groupoids to the predeformation category at a point over $k$: it says that this process preserves (RS).
Lemma 90.16.11. Let $\mathcal{F}$ be a category cofibered in groupoids over $\mathcal{C}_\Lambda $. Let $x_0 \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{F}(k))$. Let $\mathcal{F}_{x_0}$ be the category cofibred in groupoids over $\mathcal{C}_\Lambda $ constructed in Remark 90.6.4. If $\mathcal{F}$ satisfies (RS), then so does $\mathcal{F}_{x_0}$. In particular, $\mathcal{F}_{x_0}$ is a deformation category.
Proof.
Any diagram as in Definition 90.16.1 in $\mathcal{F}_{x_0}$ gives rise to a diagram in $\mathcal{F}$ and the output of (RS) for this diagram in $\mathcal{F}$ can be viewed as an output for $\mathcal{F}_{x_0}$ as well.
$\square$
The following lemma is the analogue of the fact that $2$-fibre products of algebraic stacks are algebraic stacks.
Lemma 90.16.12. Let
\[ \xymatrix{ \mathcal{H} \times _\mathcal {F} \mathcal{G} \ar[r] \ar[d] & \mathcal{G} \ar[d]^ g \\ \mathcal{H} \ar[r]^ f & \mathcal{F} } \]
be $2$-fibre product of categories cofibered in groupoids over $\mathcal{C}_\Lambda $. If $\mathcal{F}, \mathcal{G}, \mathcal{H}$ all satisfy (RS), then $\mathcal{H} \times _\mathcal {F} \mathcal{G}$ satisfies (RS).
Proof.
If $A$ is an object of $\mathcal{C}_\Lambda $, then an object of the fiber category of $\mathcal{H} \times _\mathcal {F} \mathcal{G}$ over $A$ is a triple $(u, v, a)$ where $u \in \mathcal{H}(A)$, $v \in \mathcal{G}(A)$, and $a : f(u) \to g(v)$ is a morphism in $\mathcal{F}(A)$. Consider a diagram in $\mathcal{H} \times _\mathcal {F} \mathcal{G}$
\[ \vcenter { \xymatrix{ & (u_2, v_2, a_2) \ar[d] \\ (u_1, v_1, a_1) \ar[r] & (u, v, a) } } \quad \text{lying over}\quad \vcenter { \xymatrix{ & A_2 \ar[d] \\ A_1 \ar[r] & A } } \]
in $\mathcal{C}_\Lambda $ with $A_2 \to A$ surjective. Since $\mathcal{H}$ and $\mathcal{G}$ satisfy (RS), there are fiber products $u_1 \times _ u u_2$ and $v_1 \times _ v v_2$ lying over $A_1 \times _ A A_2$. Since $\mathcal{F}$ satisfies (RS), Lemma 90.16.2 shows
\[ \vcenter { \xymatrix{ f(u_1 \times _ u u_2) \ar[r] \ar[d] & f(u_2) \ar[d] \\ f(u_1) \ar[r] & f(u) } } \quad \text{and}\quad \vcenter { \xymatrix{ g(v_1 \times _ v v_2) \ar[r] \ar[d] & g(v_2) \ar[d] \\ g(v_1) \ar[r] & g(v) } } \]
are both fiber squares in $\mathcal{F}$. Thus we can view $a_1 \times _ a a_2$ as a morphism from $f(u_1 \times _ u u_2)$ to $g(v_1 \times _ v v_2)$ over $A_1 \times _ A A_2$. It follows that
\[ \xymatrix{ (u_1 \times _ u u_2, v_1 \times _ v v_2, a_{1} \times _ a a_2) \ar[d] \ar[r] & (u_2, v_2, a_2) \ar[d] \\ (u_1, v_1, a_1) \ar[r] & (u, v, a) } \]
is a fiber square in $\mathcal{H} \times _\mathcal {F} \mathcal{G}$ as desired.
$\square$
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