Consider the mapping given by:
.
Is this a bijection?
OK, for one thing the INVERSE of f (if it exists, which you have NOT shown) would be a function .
so you would need to show TWO things:
1)
2)
but you can't simply say "g exists" you have to DEFINE it.
Perhaps a more straight-forward method is to prove:
A) (that is, f is injective)
B) for EVERY , there exists SOME with: (that is, f is surjective).