Example

(3) Applying Proposition 16 to the case $n = 1$, $M_1 = M$, $P = A$, then $H = M^*$, the dual of M, and it is seen that for a K-derivation $(d_0, d, d')$ of $(A, M, M)$ is derived a K-derivation $(d_0, d^*, d^*)$ of $(A, M^*, M^*)$ such that

(38) $d_0\langle m, m^*\rangle = \langle dm, m^*\rangle + \langle m, d^*m^*\rangle$

for $m \in M$ and $m^* \in M^*$. The K-linear mapping of $M \oplus M^*$ into itself which is equal to $d$ on $M$ and to $d^*$ on $M^*$ then satisfies condition (29) and there is therefore a $K$-derivation $D$ of the $A$-algebra $T(M \oplus M^*)$, which reduces to $d_0$ on $A$, to $d$ on $M$ and to $d^*$ on $M^*$. The restriction $d_J^I$ of $D$ to the sub-$A$-module $T_J^I(M)$ of $T(M \oplus M^*)$ ($§ 5$, no. 6) is a $K$-endomorphism of $T_J^I(M)$ such that $(d_0, d_J^I, d_J^I)$ is a $K$-derivation of $(A, T_J^I(M), T_J^I(M))$. Moreover, for $i \in I, j \in J$, if we write $I' = I - \{i\}, J' = J - \{j\}$, it is immediately verified that for the contraction $c_j^i$ ($§ 5$, no. 6)

(4) Let $M_i$ ($1 \leq i \leq 3$) be three $A$-modules and, for each $i$, suppose that $(d_0, d_i, d_i)$ is a derivation of $(A, M_i, M_i)$; applying Proposition 16 again for $n = 1$, a derivation $(d_0, d_{ij}, d_{ij})$ of $(A, H_{ij}, H_{ij})$, where $H_{ij} = \operatorname{Hom}_A(M_i, M_j)$, is derived for each ordered pair $(i, j)$. With this notation, for $u \in \operatorname{Hom}_A(M_1, M_2)$ and $v \in \operatorname{Hom}_A(M_2, M_3)$,

as is immediately verified from the definitions.