f^-1(A) - f^-1(B)= f^-1 (A-B).

For :

Let .

Then

and ,

that is

and

so

or

.

Try the other inclusion the same way.

Just letIf f and g are one-ton-one, then f x g is one-to-one.

then

or

and

Since now f,g are both one to one, we get....

Let . We wish to find such that .If f maps onto A and g maps onto B then f x g maps on A x B

Now since and maps onto , there exists some such that .

Similarly, there must be some such that .

Now, check what is.

Ok, this is where latex officialy bores me

Let X={1} and Y={2,3}. Define f(1)=2; This f is not onto. But the inverse image of Y={2,3} through f is...If F^-1(Y) = X, then f maps onto Y

With X,Y as before, define now Z={4} and g(2)=g(3)=4. Then gof (composition) maps X onto Z, but...If g composed with f maps onto Z, then f maps onto Y