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arXiv 2503.03439math.CTmath.CO
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Lawvere 的第四个开放问题:对称单纯集拓扑斯中的层级

Lawvere's fourth open problem: Levels in the topos of symmetric simplicial sets

Ryuya Hora, Yuhi Kamio, Yuki Maehara

AI总结:

本文解决了 Lawvere 提出的拓扑斯理论第四个开放问题,计算了对称单纯集拓扑斯中的 Aufhebung 关系,证明其层级关系为 2l-1,与单纯集情形一致。

摘要:

在单纯集的拓扑斯中,针对给定的自然数 $n$,提出以下问题是合理的:使得 $n$-骨架性蕴含 $m$-余骨架性的最小值 $m$ 是多少?这是 Lawvere 意义上的 Aufhebung 关系的一个实例,Lawvere 为任意 Grothendieck 拓扑斯 $\mathcal{E}$(而非 $\mathbf{sSet}$)引入了这一概念,并以层级/本质子拓扑斯替代维度。我们计算了对称单纯集拓扑斯的这一 Aufhebung 关系。特别地,我们证明对于标记为 $l\geq 3$ 的层级,该关系由 $2l-1$ 给出,这与先前已知的单纯集情形一致。这一结果为 Lawvere 于 2009 年提出的拓扑斯理论中七个开放问题中的第四个提供了解决方案。

英文摘要:

In the topos of simplicial sets, it makes sense to ask the following question about a given natural number $n$: what is the minimum value $m$ such that $n$-skeletality implies $m$-coskeletality? This is an instance of the Aufhebung relation in the sense of Lawvere, who introduced this notion for an arbitrary Grothendieck topos $\mathcal{E}$ in place of $\mathbf{sSet}$, and levels/essential subtopoi in place of dimensions. We compute this Aufhebung relation for the topos of symmetric simplicial sets. In particular, we show that it is given by $2l-1$ for the level labelled by $l\geq 3$, which coincides with the previously known case of simplicial sets. This result provides a solution to the fourth of the seven open problems in topos theory posed by Lawvere in 2009.

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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