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翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク Tensor-hom adjunction ： ウィキペディア英語版 Tensor-hom adjunction In mathematics, the tensor-hom adjunction is that the tensor product and Hom functors $-\; \backslash otimes\; X$ and $\backslash operatorname(X,-)$ form an adjoint pair: :$\backslash operatorname(Y\; \backslash otimes\; X,\; Z)\; \backslash cong\; \backslash operatorname(Y,\backslash operatorname(X,Z)).$ This is made more precise below. The order "tensor-hom adjunction" is because tensor is the left adjoint, while hom is the right adjoint. ==General Statement== Say ''R'' and ''S'' are (possibly noncommutative) rings, and consider the right module categories (an analogous statement holds for left modules): :$\backslash mathcal\; =\; \backslash mathrm\_R\; \backslash quad\; \backslash text\; \backslash quad\; \backslash mathcal\; =\; \backslash mathrm\_S.$ Fix an (''R'',''S'') bimodule ''X'' and define functors ''F'': ''C'' → ''D'' and ''G'': ''D'' → ''C'' as follows: :$F(Y)\; =\; Y\; \backslash otimes\_R\; X\; \backslash quad\; \backslash text\; Y\; \backslash in\; \backslash mathcal$ :$G(Z)\; =\; \backslash operatorname\_S\; (X,\; Z)\; \backslash quad\; \backslash text\; Z\; \backslash in\; \backslash mathcal$ Then ''F'' is left adjoint to ''G''. This means there is a natural isomorphism :$\backslash operatorname\_S\; (Y\; \backslash otimes\_R\; X,\; Z)\; \backslash cong\; \backslash operatorname\_R\; (Y\; ,\; \backslash operatorname\_S\; (X,\; Z)).$ This is actually an isomorphism of abelian groups. More precisely, if ''Y'' is an (''A'', ''R'') bimodule and ''Z'' is a (''B'', ''S'') bimodule, then this is an isomorphism of (''B'', ''A'') bimodules. This is one of the motivating examples of the structure in a closed bicategory.〔 〕 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「Tensor-hom adjunction」の詳細全文を読む スポンサード リンク 翻訳と辞書 : 翻訳のためのインターネットリソース Copyright(C) kotoba.ne.jp 1997-2016. All Rights Reserved.