Abstract algebra: Take a group G, and a group H, with a group operation *. G is isomorphic to H if there exist a map f: G->H such that:
1: f is
injective
2: f is surjective
3: f(a*b)=f(a)*f(b) for all a,b in G
If f satisfies these three
properties, f is called an isomorphism.
The map f: Z -> E given by f(a)=
2a where Z is the integers and E is the even integers is an isomorphism.
Proof:
Showing injectivity
f(b)=f(a) => 2a=
2b (from the given function) a=b
Showing surjectivity
Suppose n is in E. n is an even
integer hence n=
2k for some integer k.
f(k)=2k=n, hence f is surjective.
Homomorphism:
f(a+b)=2(a+b)=2a+2b=f(a)+f(b)
Hence f is an isomorphism. Q. E. D.