Inequivalent Cantor sets in \(R^{3}\) whose complements have the same fundamental group (Q2839374)

In the paper under review, the authors respond affirmatively to a question appearing in Open Problems in Topology II, namely, can two different (rigid) Cantor sets have complements with the same fundamental group? In a previous paper, \textit{D. J. Garity, D. Repovš} and \textit{M. Željko} [Proc. Am. Math. Soc. 134, No. 8, 2447--2456 (2006; Zbl 1165.54309)] give a positive answer to the above question by showing that there exist uncountably many inequivalent rigid wild Cantor sets in \(\mathbb{R}^3\) with simply connected complement.NEWLINENEWLINEThe main theorem of this paper is the following: Let \(C\) be a Cantor set in \(\mathbb{R}^3\). Suppose there is some integer \(N\geq 3\) such that there are only finitely many points in \(C\) of local genus \(N\). Then there are uncountable many inequivalent Cantor sets \(C_\alpha\) in \(\mathbb{R}^3\) with complement having the same fundamental group as the complement of \(C\). As an application the following result is obtained: Let \(M\) be an open 3-manifold with end point (Freudenthal) compactification \(M^*\) such that \(M^* \cong S^3\) and such that \(M^*- M\) is a Cantor set \(C\). If the Cantor set \(C\) has an associated integer \(N\geq 3\) such that only finitely many points of \(C\) have local genus \(N\), then there are uncountable many open 3-manifolds not homeomorphic to \(M\) that have the same fundamental group as \(M\). The main results are proved using analysis of local genus of points of Cantor sets, a construction for producing rigid Cantor sets with simply connected complement, and manifold decomposition theory. The paper finishes with 8 open questions emerging from the techniques used there.