Remark

Tag 0DU3, permanent. Cite it as [Bourbaki, Tag 0DU3].

If $f, g \in \mathrm{Hom}_A(M, N)$ , one does not have $I(f + g) = I(f) + I(g)$ . However, these two morphisms are homotopic by X, p. 49, prop. 3 bis.

If M is a right A-module, one puts $I(M) = I(M^\circ)^\circ$ ; this is called the canonical injective resolution of M, and one has

I(M^\circ) = I(M)^\circ .

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

Cited by

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Cites

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