Abstract

We call a value y=f(x)y=f(x) of a map f:X→Yf:X\to Y dimensionally regular if dim⁡X≤dim⁡(Y×f−1(y))\dim X\le \dim(Y\times f^{-1}(y)). It was shown in \cite{first-exotic} that if a map f:X→Yf:X\to Y between compact metric spaces does not have dimensionally regular values, then XX is a Boltyanskii compactum, i.e. a compactum satisfying the equality dim⁡(X×X)=2dim⁡X−1\dim(X\times X)=2\dim X-1. In this paper we prove that every Boltyanskii compactum XX of dimension dim⁡X≥6\dim X \geq 6 admits a map f:X→Yf:X\to Y without dimensionally regular values. Also we exhibit a 4-dimensional Boltyanskii compactum for which every map has a dimensionally regular value.

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