>>11185003well, you've got the right map F. now you need to show it's a bijection. remember bijection means surjection + injection (onto + one-to-one).
take an arbitrary function g in C. what should you put into F in order to get g out of F? this is how you prove F is surjective.
now, just follow the definition of injectivity to find that F is injective - in other words, if F(a1, a2) = F(b1, b2), show that a1 = b1 and that a2 = b2.
carefully write down why!
then you'll have shown F is a bijection, and you'll be done.
alternatively, you could write down an inverse for F. this is not hard to do in this case. then you just need to show it's an inverse by showing G(F(a1, a2)) = (a1, a2) and F(G(f)) = f where G is your inverse to F.
