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Math Help - Show there is an isomorphism...

  1. #1
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    Show there is an isomorphism...

    If there is a bijection f: X -> Y, prove there is an isomorphism o: S(subx) -> S(suby)

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  2. #2
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    Re: Show there is an isomorphism...

    if X and Y are finite, this is pretty easy, because in that case, both S_X,S_Y are isomorphic to S_n, where n = |X| = |Y|.

    what's not so clear is that this is also true even if X are Y are infinite sets like \mathbb{Z} and \mathbb{Q}.

    well, your isomorphism o is going to have to use f somehow, because it's the only mapping X-->Y we're able to get our hands on.

    what does an element of S_X look like? well, it in turn, is a bjiection φ: X--->X, x-->φ(x).

    so how can we create a mapping ψ:Y-->Y using f? how about defining ψ(f(x)) = f(φ(x)), which takes f(x)-->f(φ(x)).

    f is onto, so f(x) runs through all values of y, and the x in f(x) is uniquely defined, because f is injective, so ψ is well-defined.

    now, i've done the hard part, all you have to do is show that o:φ -->ψ (that is if φ:x-->φ(x), then o(φ):f(x)-->f(φ(x))

    is, in fact, an isomorphism. (you might want to let \phi_1,\phi_2 be two arbitrary elements of S_X).
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