User:Dunloskinbeg/sandbox

Alternate proof
Statement    If there is a one-to-one mapping f from A into B, and a one-to-one mapping g from B into A, then there is a one-to-one mapping h from A onto B.

Proof    We construct h by taking f for some of the points in A, and g -1 for the others.

We may suppose there are points in A and not in g(B), else g is onto.

Take a point a in A – g(B) and form the sequence a, f (a), g(f (a)), f (g(f (a))),. . .of points belonging alternately to A and B. Call these points ‘sequence points’.

Form similar sequences beginning with all points in A – g(B).

Define h = f for all sequence points in A.

The remaining points of A are in g(B) so for them we can define h = g -1.

To verify that h is a 1-1 mapping we need to check that the image of a sequence point cannot coincide with the image of a non-sequence point.

Suppose therefore that as is  a sequence point and an is a non-sequence point and that f (as) = g -1(an).

It follows that  g(f (as)) = g(g -1(an)). But the LHS is a sequence point in A while the RHS is just the non-sequence point an  — a contradiction.

Next we show that h is onto. Consider any b in B then :

Case 1 if b is a sequence point, then f-1(b) exists and is a sequence point. Hence h = f maps f-1(b) to b.

Case 2 if b is not a sequence point then neither is g(b).

Hence h = g -1 maps g(b) to b. $$\ \blacksquare$$