Yes.
How do we know that this map is well-defined?
Independent from the choices we can choose for ?
Well, let us assume there are two ways to expess an element in
and
Then the function maps them into,
, respectively.
But,
and because the conditon we had said that,
Thus, the map is well-defined.
The function is a group homomorphism,
=
The function is one-to-one,
Thus,
are both in same coset of and,
are both in same coset of .
The function is onto,
That result is trivial.
Any, can be obtained from mapping the element,