# Endliche Mengen/Monotone Abbildungen/Standard/1/Aufgabe/Lösung

Zur Navigation springen Zur Suche springen

a)

 ${\displaystyle {}j}$ ${\displaystyle {}0}$ ${\displaystyle {}1}$ ${\displaystyle {}2}$ ${\displaystyle {}3}$ ${\displaystyle {}4}$ ${\displaystyle {}D_{3}(j)}$ ${\displaystyle {}0}$ ${\displaystyle {}1}$ ${\displaystyle {}2}$ ${\displaystyle {}4}$ ${\displaystyle {}5}$

b)

 ${\displaystyle {}j}$ ${\displaystyle {}0}$ ${\displaystyle {}1}$ ${\displaystyle {}2}$ ${\displaystyle {}3}$ ${\displaystyle {}4}$ ${\displaystyle {}5}$ ${\displaystyle {}6}$ ${\displaystyle {}S_{3}(j)}$ ${\displaystyle {}0}$ ${\displaystyle {}1}$ ${\displaystyle {}2}$ ${\displaystyle {}3}$ ${\displaystyle {}3}$ ${\displaystyle {}4}$ ${\displaystyle {}5}$

c) Wir behaupten

${\displaystyle {}\varphi =D_{3}\circ D_{1}\circ S_{3}\circ S_{1}\,.}$

Die Komposition hat für die Elemente ${\displaystyle {}0,1,2,3,4,5}$ jeweils den folgenden Effekt:

${\displaystyle 0\mapsto 0\mapsto 0\mapsto 0\mapsto 0,}$
${\displaystyle 1\mapsto 1\mapsto 1\mapsto 2\mapsto 2,}$
${\displaystyle 2\mapsto 1\mapsto 1\mapsto 2\mapsto 2,}$
${\displaystyle 3\mapsto 2\mapsto 2\mapsto 3\mapsto 4,}$
${\displaystyle 4\mapsto 3\mapsto 3\mapsto 4\mapsto 5,}$
${\displaystyle 5\mapsto 4\mapsto 3\mapsto 4\mapsto 5.}$
Das Gesamtergebnis stimmt also mit ${\displaystyle {}\varphi }$ überein.
Zur gelösten Aufgabe