Let f ∈ F(A,B). Prove that if f is bijective, so is f-¹

Your notation looks a bit weird to me. Is that exactly from a question? What normal notation would be:

Let f:A⇒B. Prove that if f is bijective, so is f-¹

Well, we know that f is a function. Since f is a map:

For every element a∈A, there must exist a b∈B such that f(a) = b.

Applying an inverse to both sides, we get:

f-¹(f(a)) = f-¹(b)

Or:

a = f-¹(b)

For all a and some b. Thus, f-¹ must be onto.

Now let f-¹(b1) = f-¹(b2) for any b1, b2 in B. Since f is 1-1, f(f-¹(b1)) = f(f-¹(b2)). But f(f-¹(b1)) = b1 and f(f-¹(b2)) = b2. So b1 = b2. Thus, f-¹ is 1-1.

Let f ∈ F(A,B) and g ∈ F(B,C). prove that if f and g are invertible so is gf and that (gf)-¹ = f-¹g-¹

Notice how you can go from A->B and then from B->C. But remember, that f and g are invertible, so you can go from B->A and from C->B. Use this, and show that if an element is in C, then the element must correspond to some element in A.

Try this, and if you can get it, the thrid should become much easier.