suppose S is the smallest set (fewest members) for which there exists a 1-1 function f that is not onto. Call the elements of S x1 x2 … xn. By assumption there is an i such that no member of S is mapped onto xi. yet xi uniquely maps onto xj (1-1). Remove xi from S, and eliminate xi,xj from the function. We have constructed a smaller set, possessing a 1-1 function that is not onto, since nothing maps onto xj. This contradicts the "smallest" hypothesis.
Similarly, use induction to prove the converse. Suppose S is the smallest set possessing an onto function f that is not 1-1. By assumption, there exists i j and k such that xi and xj both map to xk. Remove xi from S, and eliminate xi,xk from the function. If f use to map xm onto xi, replace this with xm,xk. There is at least one such entry, f being onto. We have constructed a smaller set possessing an onto function that is not 1-1, since xm and xj both map to xk. This contradicts the "smallest" hypothesis.
When S is infinite, 1-1 and onto are independent properties of f. Let S be the positive integers. f = 2*x is 1-1, but not onto. f = (x+1)/2 (integer divide) is onto, but not 1-1.
If you want to be rigorous, we can't really prove this theorem yet, nor the next one, because we don't know what a finite set is, nor have we verified the method of induction. Finite and transfinite induction depend on ordinals, which are developed as part of ZF set theory. That's the first subtopic beneath this one.