If there are injections both ways, then there is a bijection of $A$ with $B$ and the sets have the same cardinality (Schroeder-Bernstein theorem, proved in exercises of Section 7). Otherwise, there can be an injection one way only, and we need to show that there is one. If there is a surjection from $A$ onto $B$ , then there is an injection $B\rightarrow A$ (which is a right inverse for the surjection, which exists due to the axiom of choice, which is equivalent to the well-ordering theorem, see Exercise 5(a) of §9). If there is no surjection $A\rightarrow B$ , then we can well-order both sets and use Exercise 10 to show that there is an injective function from $A$ into $B$ (the fact that it is injective follows immediately from the definition $(\ast)$ ).