suppose that but that , that is, that one target in B has two different sources.

well, there's only so many points of A to go around. and if we've "doubled up" two of those to send to a single point in B, how can f be onto?

alternatively, begin with what you're given. f is ONTO.

this means that every point has a pre-image . note that every element of B must have a distinct pre-image

in A, since f is a FUNCTION. after we've sent every element b of B "back to (a) pre-image", are there any elements of A left over?