- #1
DPMachine
- 26
- 0
I'm trying to show:
If f: S -> S' is an isomorphism of <S, *> with <S', *'>, then f^(-1) is homomorphic.
My take:
So I have to show that f^(-1)(x' *' y') = f^(-1)(x') * f^(-1)(y').
Since f is bijective (onto, more precisely) I know that f^(-1)(x') = x and f^(-1)(y') = y. So f^(-1)(x') * f^(-1)(y') = xy.
How can I simplify f^(-1)(x' *' y') though?
Thanks.
If f: S -> S' is an isomorphism of <S, *> with <S', *'>, then f^(-1) is homomorphic.
My take:
So I have to show that f^(-1)(x' *' y') = f^(-1)(x') * f^(-1)(y').
Since f is bijective (onto, more precisely) I know that f^(-1)(x') = x and f^(-1)(y') = y. So f^(-1)(x') * f^(-1)(y') = xy.
How can I simplify f^(-1)(x' *' y') though?
Thanks.