Inverses and compositions (1) on the prep questions, you are proving that when f : a b and g : b c are bijections, g f : a c is also a bijection. Find a counterexample to show that this is not an if and only if claim. Here"s an example that satis es the second condition, as well, where neither f nor g is a bijection: Let a = {1} and b = {(cid:63), $} and c = {(cid:44)}. De ne f : a b by f (1) = $. De ne g : b c by g((cid:63)) = g($) = (cid:44). Which ones are injective? not (it fails to be injective). f1(x, y) = 2a + 3b f2(x, y) = x(y + 2) 2 f3(x, y) = 2 xy(x + y) (note: why is this well-de ned?) The function f1 is not surjective because a, b n tells us a, b 1 and so 2a + 3b 5.
