Theorem 3.7: Let

be a one to one correspondence, then

is also a one to one correspondence.

Proof: First we must show that it is a function. Let b be an element of B. Then since f is onto, there exists an element a in A such that

f (a) = b.

by Definition 3.6. By the definition of the inverse function, Definition 3.10,

f -1(b) = a,

so for every element of B, there is an element of A associated with it by f -1.

Next, we must show that it is well defined. Assume that

If b1 is an element of B, then since f is onto, then by Definition 3.6, there exists an element a1 in A such that

f (a1) = b1

If b2 is an element of B, then since f is onto, then by Definition 3.6, there exists an element a2 in A such that

f (a2) = b2

But if

b1 = b2,

then

f (a1) = f (a2).

Since f is one to one, we conclude that

a1 = a2.

by Definition 3.5. Then

f -1(b1) = a1 = a2 = f -1(b2)

and f-1 is well defined.

Next we show that f-1 is one to one. Assume that

f -1(b1) = f-1(b2)

Let

a = f-1(b1) = f-1(b2)

Then by the definition of f -1, Definition 3.10,

f(a) = b1 and f(a) = b2

Since f is a well defined function,

b1 = b2

by Definition 3.2.

Finally we show that f-1 is onto. Let a be an element of A. Then if we define

b = f(a),

then

f-1(b ) = a

and Definition 3.6 is satisfied.