Then f- 1 = {

Because f is surjective, for every b∈B, there is

(This means that the f- 1 domain has searched the whole set B)

F is injective, so for every b∈B, there is exactly one a∈A that makes < A, B > ∈ F, so there is only one a∈A for every B, which makes.

(This means that f- 1 is a single-valued mapping. )

So f- 1 satisfies two necessary conditions of a function, so it is a function.

And because ran(f- 1)=dom(f)=A, f- 1 is surjective.

It is proved that f- 1 is injective, thus it is denied. Assuming that f-1(b1) = f-1(b2) holds when b 1≠b2, then we might as well assume that

F-1(b1) = a1,F- 1 (B2) = A2, and a 1=a2, then F (a1) = B/kloc-.