Deprecated: $wgMWOAuthSharedUserIDs=false is deprecated, set $wgMWOAuthSharedUserIDs=true, $wgMWOAuthSharedUserSource='local' instead [Called from MediaWiki\HookContainer\HookContainer::run in /var/www/html/w/includes/HookContainer/HookContainer.php at line 135] in /var/www/html/w/includes/Debug/MWDebug.php on line 372
On piecewise confluent mappings - MaRDI portal

On piecewise confluent mappings (Q2714186)

From MaRDI portal





scientific article; zbMATH DE number 1603991
Language Label Description Also known as
English
On piecewise confluent mappings
scientific article; zbMATH DE number 1603991

    Statements

    0 references
    0 references
    12 June 2001
    0 references
    confluent mapping
    0 references
    end continuum
    0 references
    On piecewise confluent mappings (English)
    0 references
    The authors consider the following problem of W. T. Ingram, which is problem 35 from the Houston Problem Book, and some variations of the problem [\textit{H. Cook, W. T. Ingram} and \textit{A. Lelek}, Lect. Notes Pure Appl. Math. 170, 365-398 (1995; Zbl 0828.54001)]. ``Suppose \(f\) is a continuous mapping of a continuum \(X\) onto a continuum \(Y\), \(Y=H\cup K\) is a decomposition of \(Y\) into subcontinua \(H\) and \(K\), \(f\mid f^{-1}(H)\) and \(f\mid f^{-1}(K)\) are confluent, and \(H\cap K\) is a continuum which does not cut \(Y\) and is an end continuum of both \(H\) and \(K\). Is \(f\) confluent?''NEWLINENEWLINENEWLINEThe authors show if one uses one definition of end continuum the conditions in the problem cannot be satisfied. However, if one uses the following definition of end continuum, then the answer to the problem is ``yes''. A subcontinuum \(P\) of a continuum \(X\) is said to be an end continuum in \(X\) provided that for every two subcontinua \(K\) and \(L\) of \(X\) the condition \(P\subset K\cap L\) implies that either \(K\subset L\) or \(L\subset K\).
    0 references

    Identifiers

    0 references
    0 references
    0 references
    0 references