Proposition 11

— If $M' \xrightarrow{u} M \xrightarrow{v} M''$ is an exact sequence of A-modules of finite length, then the sequence $D(M'') \xrightarrow{D(v)} D(M) \xrightarrow{D(u)} D(M')$ is exact [^1].

Let us show first of all that given an exact sequence

(1)

the corresponding sequence

is exact; indeed we know that the sequence

is exact (II, p. 227, Th. 1); on the other hand, it follows from (1) that

(II, p. 212, Prop. 16); by Prop. 10 we thus have

in other words $\operatorname{long}(D(M')) = \operatorname{long}(D(M)/D(M''))$. Since $D(M)/D(M'')$ is naturally identified with a submodule of $D(M')$, it must be equal to $D(M')$, which proves our assertion.

This implies immediately that if $u : M' \to M$ is injective, then $D(u) : D(M) \to D(M')$ is surjective; the conclusion then follows from II, p. 199, remark 4.

For every A-module M, let $\mathfrak{S}(M)$ denote the set of submodules of M. For every submodule N of M (resp. every submodule N' of D(M)), let $N^0$ (resp. ${N'}^0$) denote the submodule of $D(M)$ (resp. M) consisting of those $x' \in D(M)$ (resp. $x \in M$) such that $(y, x') = 0$ for all $y \in N$ (resp. $(x, y') = 0$ for all $y' \in N'$).