Lemma 1

Tag 05NX, permanent. Cite it as [Bourbaki, Tag 05NX].

Algebra, chapter II, § 2. Modules of linear mappings. Duality

Let M, N be two supplementary submodules in an A-module E and i : M → E, j : N → E the canonical injections. Then the diagram

\begin{array}{ccc} M \oplus N & \xrightarrow{c_M \oplus c_N} & M** \oplus N** \\ i + j \downarrow & & \uparrow t_{i_1} + t_{j_1} \\ E & \xrightarrow{c_E} & E** \end{array}

(28)

is commutative.

By definition, for $x \in M, y \in N, z^* \in E*$ ,

\begin{align*} \langle c_E(i(x) + j(y)), z^* \rangle &= \langle i(x) + j(y), z^* \rangle \\ &= \langle i(x), z^* \rangle + \langle j(y), z^* \rangle \\ &= \langle x, t_i(z^*) \rangle + \langle y, t_j(z^*) \rangle \\ &= \langle c_M(x), t_i(z^*) \rangle + \langle c_N(y), t_j(z^*) \rangle \\ &= \langle t_i(c_M(x)) + t_j(c_N(y)), z^* \rangle. \end{align*}

This being so, if E is a free module (resp. a free module with a finite basis), $c_E$ is injective (resp. bijective); on the other hand, it follows from no. 6, Proposition 10, that $t_i \oplus t_j$ is bijective; the commutativity of diagram (28) then implies that $c_M \oplus c_N$ is injective (resp. bijective) and so therefore are $c_M$ and $c_N$ (§ 1, no. 6, Corollary 1 to Proposition 7), whence the corollary, taking account of no. 2, Proposition 4.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Lemma 1

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