Example

Let $B$ be a $k$-algebra, $N$ a $B$-module. Endow the $k$-module $B \oplus N$ with the structure of a $k$-algebra defined by $(b, x)(b', x') = (bb', bx' + b'x)$ (cf. A, III, p. 127), so that $N$ is a square-zero ideal of $B \oplus N$. Let $\varphi : A \to B$ be a homomorphism of $k$-algebras. Then the liftings of $\varphi$ to $B \oplus N$ are the mappings $x \mapsto (\varphi(x), \delta(x))$, where $\delta$ runs through the set of $k$-derivations of $A$ into $N$ (cf. loc. cit., prop. 12).

Let $\Omega_k(A)$ be the module of $k$-differentials of the ring $A$, and let $d : A \longrightarrow \Omega_k(A)$ be the universal $k$-derivation (A, III, p. 133 and 134); let us recall (loc. cit.) that for every $A$-module $M$, the mapping $v \mapsto v \circ d$ is an $A$-linear isomorphism of $\mathrm{Hom}_A(\Omega_k(A), M)$ onto the $A$-module of $k$-derivations of $A$ into $M$.

Let J be an ideal of A. By A, III, p. 137, one has an exact sequence of $A/J$-linear mappings

where $\bar{d}$ is the homomorphism induced by passing to quotients from the restriction of $d$ to J.

Let $\rho : A \to A/J^2$ and $\pi : A/J^2 \to A/J$ denote the canonical surjections. Let $v : (A/J) \otimes_A \Omega_k(A) \longrightarrow J/J^2$ be a $k$-linear mapping; to it one associates a $k$-linear mapping $H_v : A \to A/J^2$ by putting $H_v(x) = \rho(x) - v(1 \otimes dx)$. If $v$ is a retraction of $\bar{d}$, $H_v$ vanishes on J, hence induces by passing to the quotient a $k$-linear mapping $h_v : A/J \to A/J^2$. On the other hand, given a $k$-linear mapping $h : A/J \to A/J^2$, let us denote by $\psi_h : A/J \oplus J/J^2 \longrightarrow A/J^2$ the mapping $(x, y) \mapsto h(x) + y$.