Bourbaki

Remark

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

Algebra, chapter IX, § 1. Formes sesquilinéaires

If the left adjoint u∗u^* of u:E→E′u : E \to E' is everywhere defined, and if sΦ′s_{\Phi'} and dΦ′d_{\Phi'} are injective, formula (30) shows that uu is the right adjoint of u∗u^*.

It follows from (30) that, if u1u_1 and u2u_2 are two homomorphisms from EE into E′E' admitting everywhere defined adjoints, and if cc is an element of the center of AA, one has

{(u1+u2)∗=u1∗+u2∗;1∗=1;(cu1)∗=c.u1∗ when Φ and Φ′ are bilinear ;(cu1)∗=cj′.u1∗ when Φ and Φ′ are sesquilinear. \begin{cases} (u_1 + u_2)^* = u_1^* + u_2^* ; 1^* = 1 ; \\ (cu_1)^* = c . u_1^* \text{ when } \Phi \text{ and } \Phi' \text{ are bilinear ;} \\ (cu_1)^* = c^{j'} . u_1^* \text{ when } \Phi \text{ and } \Phi' \text{ are sesquilinear.} \end{cases}

Moreover, if E′′E'' is a third AA-left module, F′′F'' a third AA-right module (resp. left module), Φ′′\Phi'' a bilinear form (resp. sesquilinear for JJ) on E′′×F′′E'' \times F'', and if u′u' is a homomorphism from E′E' into E′′E'' admitting an everywhere defined (left) adjoint, one has

(u′∘u)∗=u∗∘u′∗.(u' \circ u)^* = u^* \circ {u'}^*.

In particular, if uu is an isomorphism of EE onto E′E', and if the adjoints u∗u^* and (u−1)∗(u^{-1})^* are everywhere defined, u∗u^* is an isomorphism of F′F' onto FF, and one has (u∗)−1=(u−1)∗(u^*)^{-1} = (u^{-1})^*. Analogous properties for right adjoints.

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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Translated with machine help, not yet reviewed by a person.