The Cantor-Schreuder-Berstien theorem


by gottfried
Tags: theorem
gottfried
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#1
Nov12-12, 11:11 AM
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The Cantor-Schreuder-Berstien theorem states that if there exists a one-to-one function from X to Y and the reverse then there exists a bijection between X and Y.
Does anybody know if this implies to Homorphisms. ie: If we can find an embedding between X and Y and the reverse does this imply that X and Y are isomorphic?
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lavinia
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Nov12-12, 12:12 PM
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Quote Quote by gottfried View Post
The Cantor-Schreuder-Berstien theorem states that if there exists a one-to-one function from X to Y and the reverse then there exists a bijection between X and Y.
Does anybody know if this implies to Homorphisms. ie: If we can find an embedding between X and Y and the reverse does this imply that X and Y are isomorphic?
it implies nothing except that there is a bijection between them. No structure needs to be preserved by the map.
micromass
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Nov12-12, 02:47 PM
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What do you mean with "homomorphism" in the first place?

gottfried
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Nov13-12, 02:27 AM
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The Cantor-Schreuder-Berstien theorem


What I mean by homomorphism is a function f:(G,.)->(H,*) where f(g.g')=f(g)*f(g')
micromass
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Nov13-12, 02:42 AM
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OK, so you're talking about group homomorphisms. Well, in that case, a version of Cantor-Shroder-Bernstein does not hold. A counterexample is given by free groups. Indeed, we can see [itex]F_3[/itex] (free group on three generators) as a subgroup of [itex]F_2[/itex] by considering the subset [itex]\{a^2,ab,b^2\}[/itex] as generators.


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