Corollary 1

Tag 05R6, permanent. Cite it as [Bourbaki, Tag 05R6].

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

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

\begin{array}{ccc} \operatorname{Hom}(\lim E_\alpha, F) & \xrightarrow{d_F} & \lim \operatorname{Hom}(E_\alpha, F) \\ \downarrow \operatorname{Hom}(1_E, v) & & \downarrow \lim \overline{v}_\alpha \\ \operatorname{Hom}(\lim E_\alpha, F') & \xrightarrow{d_{F'}} & \lim \operatorname{Hom}(E_\alpha, F') \end{array}

is commutative.

For all $u \in \operatorname{Hom}(\lim E_\alpha, F)$ , $d_{F'}(v \circ u) = (v \circ u \circ f_\alpha)$ by definition and the commutativity of diagram (6) then 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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