Let A be a set and f a function with f : A rightarrow A. Suppose f is one-to-one. Must f be onto? Suppose f is onto. Must f be one-to-one? Justify your answers. Solution a . since f is one to one , distinct elements have distinctimages. since A has n distinct elements , by the one one ness of f ,we follow that all the distinct elements of A are mapped to thedistinct elements of A. so, it follows that no element in the codomain is leftunmatched. f is onto. b. since f is onto, all the elements in thecodomain have preimages and since f is a function, all the elementsin the domain have images. further, the domain and codomain are one and the same , itfollows that the distinct elements of A are mapped all the distinctelements of A under f. in other words , distinct elements have distinct images underf. f is one to one..