Lemma 89.10.8. Consider a commutative diagram in a predeformation category $\mathcal{F}$

$\vcenter { \xymatrix{ y \ar[r] \ar[d] & x_2 \ar[d]^{a_2} \\ x_1 \ar[r]^{a_1} & x } } \quad \text{lying over} \vcenter { \xymatrix{ A_1 \times _ A A_2 \ar[r] \ar[d] & A_2 \ar[d]^{f_2} \\ A_1 \ar[r]^{f_1} & A } }$

in $\mathcal{C}_\Lambda$ where $f_2 : A_2 \to A$ is a small extension. Assume there is a map $h : A_1 \to A_2$ such that $f_2 = f_1 \circ h$. Let $I = \mathop{\mathrm{Ker}}(f_2)$. Consider the ring map

$g : A_1 \times _ A A_2 \longrightarrow k[I] = k \oplus I, \quad (u, v) \longmapsto \overline{u} \oplus (v - h(u))$

Choose a pushforward $y \to g_*y$. Assume $\mathcal{F}$ satisfies (S2). If there exists a morphism $x_1 \to g_*y$, then there exists a morphism $b: x_1 \to x_2$ such that $a_1 = a_2 \circ b$.

Proof. Note that $\text{id}_{A_1} \times g : A_1 \times _ A A_2 \to A_1 \times _ k k[I]$ is an isomorphism and that $k[I] \cong k[\epsilon ]$. Hence we have a diagram

$\vcenter { \xymatrix{ y \ar[r] \ar[d] & g_*y \ar[d] \\ x_1 \ar[r] & x_0 } } \quad \text{lying over}\quad \vcenter { \xymatrix{ A_1 \times _ k k[\epsilon ] \ar[r] \ar[d] & k[\epsilon ] \ar[d] \\ A_1 \ar[r] & k. } }$

where $x_0$ is an object of $\mathcal{F}$ lying over $k$ (every object of $\mathcal{F}$ has a unique morphism to $x_0$, see discussion following Definition 89.6.2). If we have a morphism $x_1 \to g_*y$ then Lemma 89.10.7 provides us with a section $s : x_1 \to y$ of the map $y \to x_1$. Composing this with the map $y \to x_2$ we obtain $b : x_1 \to x_2$ which has the property that $a_1 = a_2 \circ b$ because the diagram of the lemma commutes and because $s$ is a section. $\square$

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