Lemma 89.10.7. Let $p: \mathcal{F} \to \mathcal{C}_\Lambda$ be a category cofibered in groupoids. Consider a diagram of $\mathcal{F}$

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

in $\mathcal{C}_\Lambda$. Assume $\mathcal{F}$ satisfies (S2). Then there exists a morphism $s : x \to y$ with $a \circ s = \text{id}_ x$ if and only if there exists a morphism $s_\epsilon : x \to x_\epsilon$ with $e \circ s_\epsilon = d$.

Proof. The “only if” direction is clear. Conversely, assume there exists a morphism $s_\epsilon : x \to x_\epsilon$ with $e \circ s_\epsilon = d$. Note that $p(s_\epsilon ) : A \to k[\epsilon ]$ is a ring map compatible with the map $A \to k$. Hence we obtain

$\sigma = (\text{id}_ A, p(s_\epsilon )) : A \to A \times _ k k[\epsilon ].$

Choose a pushforward $x \to \sigma _*x$. By construction we can factor $s_\epsilon$ as $x \to \sigma _*x \to x_\epsilon$. Moreover, as $\sigma$ is a section of $A \times _ k k[\epsilon ] \to A$, we get a morphism $\sigma _*x \to x$ such that $x \to \sigma _*x \to x$ is $\text{id}_ x$. Because $e \circ s_\epsilon = d$ we find that the diagram

$\xymatrix{ \sigma _*x \ar[r] \ar[d] & x_\epsilon \ar[d]_ e \\ x \ar[r]^ d & x_0 }$

is commutative. Hence by (S2) we obtain a morphism $\sigma _*x \to y$ such that $\sigma _*x \to y \to x$ is the given map $\sigma _*x \to x$. The solution to the problem is now to take $a : x \to y$ equal to the composition $x \to \sigma _*x \to y$. $\square$

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