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By Martin Goldstern

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Extra resources for Algebra, WS 2009

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Sind f, g beide Isomorphismen, so ist auch g ◦ f ein Isomorphismus. b) Ist f Isomorphismus von A nach A∗ , so ist f −1 Isomorphismus von A∗ nach A. ¨ Beweis. Ubung. Bilder und (vollst¨ andige) Urbilder von Unteralgebren bei Homomorphismen sind wieder ¨ Unteralgebren (Ubung). 6 Satz. Sei (H, ·) eine Halbgruppe, (H ∗ , ·) ein Gruppoid und f : H → H ∗ ein Homomorphismus. Dann ist die Unteralgebra (f (H), ·) von (H ∗ , ·) eine Halbgruppe. Beweis. Seien x, y, z ∈ f (H). Dann gibt es a, b, c ∈ H mit f (a) = x, f (b) = y und f (c) = z.

2 Beispiel. K = {1, 2}, A1 = (A1 , ·, e, −1 ), A2 = (A2 , +, 0, −) seien Gruppen. Dann wird in A1 × A2 = (A1 × A2 , ◦, (e, 0), ) folgendermaßen gerechnet: (a1 , a2 ) ◦ (b1 , b2 ) = (a1 b1 , a2 + b2 ), 41 (a1 , a2 ) = (a−1 1 , −a2 ). Es gilt: A1 ×A2 ist eine Gruppe. Assoziativgesetz: ((a1 , a2 )◦(b1 , b2 ))◦(c1 , c2 ) = (a1 b1 c1 , a2 + b2 + c2 ) = (a1 , a2 ) ◦ ((b1 , b2 ) ◦ (c1 , c2 )); (e, 0) ist neutrales Element: (e, 0) ◦ (a1 , a2 ) = (ea1 , 0 + a2 ) = (a1 , a2 ) = (a1 e, a2 + 0) = (a1 , a2 ) ◦ (e, 0); (a1 , a2 ) ist Inverses von (a1 , a2 ): (a1 , a2 ) ◦ −1 (a1 , a2 ) = (a1 , a2 ) ◦ (a−1 1 , −a2 ) = (a1 a1 , a2 + (−a2 )) = (e, 0), analog (a1 , a2 ) ◦ (a1 , a2 ) = (e, 0).

Weiters sei k : i∈I Ai → i∈I Ai /∼ die kanonische Abbildung, die jeder Familie (ai )i∈I ¨ ihre ∼-Aquivalenzklasse zuordnet. F¨ ur b1 = b2 gibt es ein i mit b1 , b2 ∈ Ai , daher stimmen ψ(b1 ) und ψ(b2 ) an der Stelle i (sowie auch an allen folgenden j ≥ i) nicht u ¨berein, somit ist ψ(b1 ) ∼ ψ(b2 ), also (k ◦ ψ)(b1 ) = (k ◦ ψ)(b2 ). Also ist k ◦ ψ injektiv. 13 −→ −→ Das Bild von lim Ai in i∈I Ai /∼ ist also eine Unteralgebra von i∈I Ai /∼, die zu lim Ai isomorph ist. i∈I i∈I Nach dem gerade bewiesenen Satz l¨ asst sich also ein direkter Limes als Kombination von Produkt, Quotient und Unteralgebra schreiben.

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