An injection maps to unique elements. A surjection maps onto all elements. Math: f injection <==> (For all a_1, a_2 in A, f(a_1) == f(a_2) <==> a_1 == a_2) f surjection <==> (For all b in B, exists a in A st f(a) = b) Answer to question 1: yes, if A is not empty. Answer to question 2: yes. Proof of 1: A is not empty, by assumption. Let a0 be an element. Wish to construct a g(b) st for all a in A, exists b with g(b) = a Will need to make sure g(b) is defined for all b. Each element b, of B has zero or one a in A for which f(a) = b as f is an injection. (can't be two or more because if f(a_2) = f(a) a_2 = a) If the answer is zero, let g(b) = a0. If the answer is one, let g(b) = a Now g is defined for all b. Prove surjection: That says for all a in A exists b in B st g(b) = a Suppose false. Then exists a_z for which no g(b) = a_z. But g(f(a_z)) = a_z. Contradiction. Proof of 2: f surjection <==> (For all b in B, exists a in A st f(a) = b) To make an injection, need to construct a unique a_i for each b_i of B. Well we can construct at least one a from the defintion of surjection. Choose one of those a to be the unique a_i mapped to by b.