Corollary

Tag 05R0, permanent. Cite it as [Bourbaki, Tag 05R0].

Algebra, chapter II, § 6. Inverse and direct limits of modules

For every $A$ -module homomorphism $v: F \to F'$ , the*

\bar{v}_\alpha = \operatorname{Hom}(v, 1_{E_\alpha}): \operatorname{Hom}(F', E_\alpha) \to \operatorname{Hom}(F, E_\alpha)

form an inverse system of $\mathbf{Z}$ -linear mappings and the diagram

(3)

\begin{array}{ccc} \operatorname{Hom}(F', \lim_{\leftarrow} E_\alpha) & \xrightarrow{l_{F'}} & \lim_{\leftarrow} \operatorname{Hom}(F', E_\alpha) \\ \downarrow \operatorname{Hom}(v, 1_E) & & \downarrow \lim_{\leftarrow} \bar{v}_\alpha \\ \operatorname{Hom}(F, \lim_{\leftarrow} E_\alpha) & \xrightarrow{l_F} & \lim_{\leftarrow} \operatorname{Hom}(F, E_\alpha) \end{array}

is commutative.

For all $u \in \operatorname{Hom}(F', \lim_{\leftarrow} E_\alpha)$ , $l_F(u \circ v) = (f_\alpha \circ u \circ v)$ by definition and the commutativity of diagram (3) follows immediately from the definitions.

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

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Cites

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