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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 mapping cylinder ： ウィキペディア英語版 mapping cylinder In mathematics, specifically algebraic topology, the mapping cylinder of a function $f$ between topological spaces $X$ and $Y$ is the quotient :$M\_f\; =\; ((()\backslash times\; X)\; \backslash amalg\; Y)\backslash ,/\backslash ,\backslash sim$ where the union is disjoint, and ∼ is the equivalence relation generated by :$(0,x)\backslash sim\; f(x)\backslash quad\backslash textx\backslash in\; X.$ That is, the mapping cylinder $M\_f$ is obtained by gluing one end of $X\backslash times()$ to $Y$ via the map $f$. Notice that the "top" of the cylinder $\backslash \backslash times\; X$ is homeomorphic to $X$, while the "bottom" is the space $f(X)\backslash subset\; Y$. See 〔Algebraic Topology by Allen Hatcher. Page 2〕 for more details. ==Basic properties== The bottom ''Y'' is a deformation retract of $M\_f$. The projection $M\_f\; \backslash to\; Y$ splits (via $Y\; \backslash ni\; y\; \backslash mapsto\; y\; \backslash in\; Y\; \backslash subset\; M\_f$), and a deformation retraction $R$ is given by: :$R:\; M\_f\; \backslash times\; I\; \backslash rightarrow\; M\_f$ :$((),s)\; \backslash mapsto\; (t,x)$ (where points in $Y$ stay fixed, which is well-defined, because $()=(0,x)$ for all $s$). The map $f:X\; \backslash to\; Y$ is a homotopy equivalence if and only if the "top" $\backslash \backslash times\; X$ is a strong deformation retract of $M\_f$. A proof can be found in.〔Algebraic Topology by Allen Hatcher. Corollary 0.16〕 An explicit formula for the strong deformation retraction is produced in.〔A Short Note on Mapping Cylinders by A. Aguado〕 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「mapping cylinder」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.