Let \\(f(A,*) \\implies (B, \\cdot)\\) be homomorphic. Prove that if \\(f^{-1}\\

Question

Let (f(A,*) implies (B, cdot)) be homomorphic. Prove that if (f^{-1})is a function then it is homomorphic
Inorder for (f^{-1}) to be a function then f must be a bijection

Explanation / Answer

This is quite easy. So we have f:(A,*) -> (B,.) is a homomorphic that means that f(x*y) = f(x).f(y) [see that the operations change] so now we are given that f^-1 (I will call it g for convenience of typing) is a function. Now g is a function means that fog is identity (def. of inverse) That also means that f is a bijection. [if its not : ie f(x) = f(y) then g(f(x)) cannot be defined and also if one of the elements in B is not mapped to any x then we do not have a g for that, hence f is bijection] to prove that g is a homomorphism we need to prove g(u.v) = g(u)*g(v) consider f(x) = u and f(y) = v they exist as f is a bijection. now g(u) = g(f(x)) = x : f, g are inverses g(v) = y Hence we need to prove that g(u.v) = x*y iff g(f(x).f(y)) = x*y but f(x*y) = f(x).f(y) hence g( f(x).f(y)) =g( f(x*y) ) = x*y Hence g is also a homomorphism. Infact both f and g are isomorphisms now. message me if you have any doubts

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