the co-domain N is the image of r – in the domain N . Thus, f is onto. Example Show that an onto function f : { , , } → { , , } is always one-one. Solution Suppose f is not one-one.
Then there exists two elements, say and in the domain whose image in the co-domain is same. Also, the image of under f can be only one element. Therefore, the range set can have at the most two elements of the co-domain { , , }, showing that f is not onto, a contradiction. Hence, f must be one-one.
Example Show that a one-one function f : { , , } → { , , } must be onto. Solution Since f is one-one, three elements of { , , } must be taken to different elements of the co-domain { , , } under f . Hence, f has to be onto. Remark The results mentioned in Examples and are also true for an arbitrary finite set X, i.e., a one-one function f : X → X is necessarily onto and an onto map f : X → X is necessarily one-one, for every finite set X.
In contrast to this, Examples and show that for an infinite set, this may not be true. In fact, this is a characteristic difference between a finite and an infinite set. EXERCISE . .
Show that the function f : R ∗∗∗∗∗ → R ∗∗∗∗∗ defined by f ( x ) = x is one-one and onto, where R ∗∗∗∗∗ is the set of all non-zero real numbers. Is the result true, if the domain R ∗∗∗∗∗ is replaced by N with co-domain being same as R ∗∗∗∗∗ ? . Check the injectivity and surjectivity of the following functions: (i) f : N → N given by f ( x ) = x (ii) f : Z → Z given by f ( x ) = x (iii) f : R → R given by f ( x ) =