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
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
This gives us a commutative diagram
Hence by (S2) we get a commutative diagram
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
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
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
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
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
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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