Abstract
It is well known that for each \(n\ge 0\), there is a continuous map \( f :S^{n}\rightarrow \partial \bigtriangleup ^{n+1}\) with the disjoint support property. Since \(S^{n}\) and \(\partial \bigtriangleup ^{n+1}\) are homeomorphic, it is natural to ask whether or not there is a homeomorphism \( h :S^{n}\rightarrow \partial \bigtriangleup ^{n+1}\) with the disjoint support property. In this paper, we prove that there is no such homeomorphism. Further, we also prove the following. Let \(K_{n}\) be any triangulation of \(S^{2}\) having n faces, \(n\ge 4\). We prove that for triangulation \(K_{6}\) of \(S^{2}\) having 6 faces, there exists a continuous map \( f :S^{2}\rightarrow \vert K_{6}\vert \) having the disjoint support property and there is no homeomorphism from \(S^{2}\) to \(\vert K_{6}\vert \) having the disjoint support property. For each \(n\ge 8\), we prove that \(S^{2}\) has at least two non-isomorphic triangulations \(L_{n}\) and \(K_{n}\) where the first one admits a continuous map \( f :S^{2}\rightarrow \vert L_{n}\vert \) with the disjoint support property and no homeomorphism with the disjoint support property. For \(K_{n}\), we prove that there is a homeomorphism \( h :S^{2}\rightarrow \vert K_{n}\vert \) with the disjoint support property.
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The second author is an honorary scientist of the National Academy of Sciences, India.
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Choudhury, S.B., Deo, S. Continuous maps with the disjoint support property. Period Math Hung 83, 20–31 (2021). https://doi.org/10.1007/s10998-020-00360-z
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DOI: https://doi.org/10.1007/s10998-020-00360-z