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arXiv 1507.07752math.AT
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关于Ganea映射的截面类别

About sectional category of the Ganea maps

Jean-Paul Doeraene

AI总结:

本文研究了Ganea映射的截面类别及其相对类别,通过引入相对类别阶数的概念,推导了截面类别与相对类别之间的不等式关系,并给出了严格性的例子。

摘要:

我们首先计算了James截面类别(secat)of Ganea映射g_k相对于任何映射f的截面类别:我们证明secat(g_k)是secat(f)/(k+1)的整数部分。接着我们计算g_k的相对类别(relcat)。为此,我们引入了映射的相对类别阶数(relcat_k)并证明relcat(g_k)是relcat_k(f)/(k+1)的整数部分。然后我们建立了任何阶数的secat和relcat之间的不等式:我们证明secat(f) <= relcat_k(f) <= secat(f) + k + 1和relcat_k(f) <= relcat_(k+1)(f) <= relcat_k(f) + 1。我们给出了显示这些不等式可能严格成立的例子。

英文摘要:

We first compute the James' sectional category (secat) of the Ganea map g_k of any map f in terms of the sectional category of f: We show that secat(g_k) is the integer part of secat(f)/(k+1). Next we compute the relative category (relcat) of g_k. In order to do this, we introduce the relative category of order k (relcat_k) of a map and show that relcat(g_k) is the integer part of relcat_k(f)/(k+1). Then we establish some inequalities linking secat and relcat of any order: We show that secat(f) <= relcat_k(f) <= secat(f) + k + 1 and relcat_k(f) <= relcat_(k+1)(f) <= relcat_k(f) + 1. We give examples that show that these inequalities may be strict.

机构、总结、中文摘要均由AI分析而来,可能存在错误,请以论文正文为准。

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