MyWikiBiz, Author Your Legacy — Saturday September 06, 2025
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, 13:42, 29 July 2008
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| \end{matrix}</math> | | \end{matrix}</math> |
| | | |
− | We may now use the features in <math>\operatorname{d}\mathcal{X} = \{ \operatorname{d}x_i \} = \{ \operatorname{d}x_1, \ldots, \operatorname{d}x_n \}</math> to classify the paths of <math>(\mathbb{B} \to X)</math> by way of the pairs in <math>X^2.\!</math> If ''X'' <math>\cong</math> '''B'''<sup>''n''</sup> then a path in ''X'' has the form ''q'' : ('''B''' → '''B'''<sup>''n''</sup>) <math>\cong</math> '''B'''<sup>''n''</sup> × '''B'''<sup>''n''</sup> <math>\cong</math> '''B'''<sup>2''n''</sup> <math>\cong</math> ('''B'''<sup>2</sup>)<sup>''n''</sup>. Intuitively, we want to map this ('''B'''<sup>2</sup>)<sup>''n''</sup> onto ''D''<sup>''n''</sup> by mapping each component '''B'''<sup>2</sup> onto a copy of '''D'''. But in our current situation "'''D'''" is just a name we give, or an accidental quality we attribute, to coefficient values in '''B''' when they are attached to features in d<font face="lucida calligraphy">X</font>. | + | We may now use the features in <math>\operatorname{d}\mathcal{X} = \{ \operatorname{d}x_i \} = \{ \operatorname{d}x_1, \ldots, \operatorname{d}x_n \}</math> to classify the paths of <math>(\mathbb{B} \to X)</math> by way of the pairs in <math>X^2.\!</math> If <math>X \cong \mathbb{B}^n,</math> then a path <math>q\!</math> in <math>X\!</math> has the following form: |
| + | |
| + | : <math>\begin{matrix} |
| + | q : (\mathbb{B} \to \mathbb{B}^n) & \cong & \mathbb{B}^n \times \mathbb{B}^n & \cong & \mathbb{B}^{2n} & \cong & (\mathbb{B}^2)^n. |
| + | \end{matrix}</math> |
| + | |
| + | Intuitively, we want to map this ('''B'''<sup>2</sup>)<sup>''n''</sup> onto ''D''<sup>''n''</sup> by mapping each component '''B'''<sup>2</sup> onto a copy of '''D'''. But in our current situation "'''D'''" is just a name we give, or an accidental quality we attribute, to coefficient values in '''B''' when they are attached to features in d<font face="lucida calligraphy">X</font>. |
| | | |
| Therefore, define d''x''<sub>''i''</sub> : ''X''<sup>2</sup> → '''B''' such that: | | Therefore, define d''x''<sub>''i''</sub> : ''X''<sup>2</sup> → '''B''' such that: |