Question

f:A->B is a one to one isomorphism, where A,B are sets
I need to prove the inverse is a isomorphism.

Answer 1

I need to prove that the inverse is a bijection and I need to use the fact that f is well defined to prove f^-1 is 1-1, and the fact that f is defined everywhere to prove that f^-1 is onto

Answer 2

before I even get to the homomorphism

Answer 3

saying 'a one to one isomorphism' is overkill since an isomorphism is a \(\underline{\text{bijective}}\) homomorphism

or are you using a different definition.

Answer 4

nah, I dont know why I wrote it like that

Answer 5

what have you tried for 1-1?

Answer 6

does this look right @Zarkon ?

Answer 7

\[f:A\rightarrow B\\let \ b_1, b_2 \in B\\since \ f^{-1} \ is \ a function \ we \ know \ f^{-1}(b) \ \in A \ for \ all \ b \ \in B \\Suppose \ f^{-1}(b_1)=f^{-1}(b_2)\\since \ f \ is \ a \ function, \ it \ is \ well \ defined.\\Thus \ f(f^{-1}(b_1)) = f(f^{-1}(b_2)\\So \ b_1=b_2 \ by \ the \ definition \ of \ an \ inverse\]

Answer 8

it has what you need .

Answer 9

I was more a little confused on the onto part

Answer 10

let \(a\in A\)
then \(f(a)=b\) and therefore \(f^{-1}(b)=a\)

so for all \(a\in A\) there exists a \(b\in B\) such that \(f^{-1}(b)=a\)

Answer 11

so the fact that you can even take f(a) is because f is defined everywhere.
ie
let a be in A
the there exists b in B st f(a) = b (because f is everywhere defined).
....
qed