Lemma 90.10.4. Let $\mathcal{F}$ be a category cofibred in groupoids over $\mathcal{C}_\Lambda$. If $\mathcal{F}$ satisfies (S2), then the condition of (S2) also holds when $k[\epsilon ]$ is replaced by $k[V]$ for any finite dimensional $k$-vector space $V$.

Proof. In the case that $\mathcal{F}$ is cofibred in sets, i.e., corresponds to a functor $F : \mathcal{C}_\Lambda \to \textit{Sets}$ this follows from the description of (S2) for $F$ in Remark 90.10.3 and the fact that $k[V] \cong k[\epsilon ] \times _ k \ldots \times _ k k[\epsilon ]$ with $\dim _ k V$ factors. The case of functors is what we will use in the rest of this chapter.

We prove the general case by induction on $\dim (V)$. If $\dim (V) = 1$, then $k[V] \cong k[\epsilon ]$ and the result holds by assumption. If $\dim (V) > 1$ we write $V = V' \oplus k\epsilon$. Pick a diagram

$\vcenter { \xymatrix{ & x_ V \ar[d] \\ x \ar[r] & x_0 } } \quad \text{lying over}\quad \vcenter { \xymatrix{ & k[V] \ar[d] \\ A \ar[r] & k } }$

Choose a morphism $x_ V \to x_{V'}$ lying over $k[V] \to k[V']$ and a morphism $x_ V \to x_\epsilon$ lying over $k[V] \to k[\epsilon ]$. Note that the morphism $x_ V \to x_0$ factors as $x_ V \to x_{V'} \to x_0$ and as $x_ V \to x_\epsilon \to x_0$. By induction hypothesis we can find a diagram

$\vcenter { \xymatrix{ y' \ar[d] \ar[r] & x_{V'} \ar[d] \\ x \ar[r] & x_0 } } \quad \text{lying over}\quad \vcenter { \xymatrix{ A \times _ k k[V'] \ar[d] \ar[r] & k[V'] \ar[d] \\ A \ar[r] & k } }$

This gives us a commutative diagram

$\vcenter { \xymatrix{ & x_\epsilon \ar[d] \\ y' \ar[r] & x_0 } } \quad \text{lying over}\quad \vcenter { \xymatrix{ & k[\epsilon ] \ar[d] \\ A \times _ k k[V'] \ar[r] & k } }$

Hence by (S2) we get a commutative diagram

$\vcenter { \xymatrix{ y \ar[d] \ar[r] & x_\epsilon \ar[d] \\ y' \ar[r] & x_0 } } \quad \text{lying over}\quad \vcenter { \xymatrix{ (A \times _ k k[V']) \times _ k k[\epsilon ] \ar[d] \ar[r] & k[\epsilon ] \ar[d] \\ A \times _ k k[V'] \ar[r] & k } }$

Note that $(A \times _ k k[V']) \times _ k k[\epsilon ] = A \times _ k k[V' \oplus k\epsilon ] = A \times _ k k[V]$. We claim that $y$ fits into the correct commutative diagram. To see this we let $y \to y_ V$ be a morphism lying over $A \times _ k k[V] \to k[V]$. We can factor the morphisms $y \to y' \to x_{V'}$ and $y \to x_\epsilon$ through the morphism $y \to y_ V$ (by the axioms of categories cofibred in groupoids). Hence we see that both $y_ V$ and $x_ V$ fit into commutative diagrams

$\vcenter { \xymatrix{ y_ V \ar[r] \ar[d] & x_\epsilon \ar[d] \\ x_{V'} \ar[r] & x_0 } } \quad \text{and}\quad \vcenter { \xymatrix{ x_ V \ar[r] \ar[d] & x_\epsilon \ar[d] \\ x_{V'} \ar[r] & x_0 } }$

and hence by the second part of (S2) there exists an isomorphism $y_ V \to x_ V$ compatible with $y_ V \to x_{V'}$ and $x_ V \to x_{V'}$ and in particular compatible with the maps to $x_0$. The composition $y \to y_ V \to x_ V$ then fits into the required commutative diagram

$\vcenter { \xymatrix{ y \ar[r] \ar[d] & x_ V \ar[d] \\ x \ar[r] & x_0 } } \quad \text{lying over}\quad \vcenter { \xymatrix{ A \times _ k k[V] \ar[d] \ar[r] & k[V] \ar[d] \\ A \ar[r] & k } }$

In this way we see that the first part of $(S2)$ holds with $k[\epsilon ]$ replaced by $k[V]$.

To prove the second part suppose given two commutative diagrams

$\vcenter { \xymatrix{ y \ar[r] \ar[d] & x_ V \ar[d] \\ x \ar[r] & x_0 } } \quad \text{and}\quad \vcenter { \xymatrix{ y' \ar[r] \ar[d] & x_ V \ar[d] \\ x \ar[r] & x_0 } } \quad \text{lying over}\quad \vcenter { \xymatrix{ A \times _ k k[V] \ar[d] \ar[r] & k[V] \ar[d] \\ A \ar[r] & k } }$

We will use the morphisms $x_ V \to x_{V'} \to x_0$ and $x_ V \to x_\epsilon \to x_0$ introduced in the first paragraph of the proof. Choose morphisms $y \to y_{V'}$ and $y' \to y'_{V'}$ lying over $A \times _ k k[V] \to A \times _ k k[V']$. The axioms of a cofibred category imply we can find commutative diagrams

$\vcenter { \xymatrix{ y_{V'} \ar[r] \ar[d] & x_{V'} \ar[d] \\ x \ar[r] & x_0 } } \quad \text{and}\quad \vcenter { \xymatrix{ y'_{V'} \ar[r] \ar[d] & x_{V'} \ar[d] \\ x \ar[r] & x_0 } } \quad \text{lying over}\quad \vcenter { \xymatrix{ A \times _ k k[V'] \ar[d] \ar[r] & k[V'] \ar[d] \\ A \ar[r] & k } }$

By induction hypothesis we obtain an isomorphism $b : y_{V'} \to y'_{V'}$ compatible with the morphisms $y_{V'} \to x$ and $y'_{V'} \to x$, in particular compatible with the morphisms to $x_0$. Then we have commutative diagrams

$\vcenter { \xymatrix{ y \ar[r] \ar[d] & x_\epsilon \ar[d] \\ y'_{V'} \ar[r] & x_0 } } \quad \text{and}\quad \vcenter { \xymatrix{ y' \ar[r] \ar[d] & x_\epsilon \ar[d] \\ y'_{V'} \ar[r] & x_0 } } \quad \text{lying over}\quad \vcenter { \xymatrix{ A \times _ k k[\epsilon ] \ar[d] \ar[r] & k[\epsilon ] \ar[d] \\ A \ar[r] & k } }$

where the morphism $y \to y'_{V'}$ is the composition $y \to y_{V'} \xrightarrow {b} y'_{V'}$ and where the morphisms $y \to x_\epsilon$ and $y' \to x_\epsilon$ are the compositions of the maps $y \to x_ V$ and $y' \to x_ V$ with the morphism $x_ V \to x_\epsilon$. Then the second part of (S2) guarantees the existence of an isomorphism $y \to y'$ compatible with the maps to $y'_{V'}$, in particular compatible with the maps to $x$ (because $b$ was compatible with the maps to $x$). $\square$

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