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

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- Oct 22nd 2011, 08:19 PMjzelltShow there is an isomorphism...
If there is a bijection f: X -> Y, prove there is an isomorphism o: S(subx) -> S(suby)

Thanks - Oct 22nd 2011, 08:55 PMDevenoRe: Show there is an isomorphism...
if X and Y are finite, this is pretty easy, because in that case, both $\displaystyle S_X,S_Y$ are isomorphic to $\displaystyle S_n$, where n = |X| = |Y|.

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

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 $\displaystyle S_X$ 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 $\displaystyle \phi_1,\phi_2$ be two arbitrary elements of $\displaystyle S_X$).