Remark

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

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

(1) Theorem 2 allows us, for every submodule F' of F, to identify Hom(E, F') with a sub- $\mathbf{Z}$ -module of Hom(E, F). When this identification is made, then, for every family $(M_\lambda)$ of submodules of F

\operatorname{Hom}(E, \bigcap_\lambda M_\lambda) = \bigcap_\lambda \operatorname{Hom}(E, M_\lambda)

for if $u \in \operatorname{Hom}(E, F)$ belongs to each of the $\operatorname{Hom}(E, M_\lambda)$ , then, for all $x \in E$ , $u(x) \in M_\lambda$ for all $\lambda$ and hence $u$ maps E into $\bigcap_\lambda M_\lambda$ .

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

Cited by

Nothing.

Cites

Theorem 2