Suppose that X and Y are finite sets, with m and n elements respectively. Suppose
further that the function f : X → Y is one-to-one and the function g : X → Y is
onto.
(i) Use the function f to show that m ≤ n.
(ii) Use the function g to show that m ≥ n.
(iii) Is f : X → Y onto? Justify your assertion.
(iv) Is the function g : X → Y one-to-one? Justify your assertion.
Solution:
(i) f : X → Y is one-to-one, so every element in X must have a distinct image in Y. For this to be
true there can't be more elements in Y than there are in X else you'd have one or more elements in
X mapping to the same element in Y. So m ≤ n.
your answer is partly right. Here is the Right answer:
F : X Y is one-to-one, so every element in X must have a distinct image in Y. For this to be true
we can have more elements in Y than X. but that extra element will not have pre-image in X. So m
≤ n.
Example:
f: XY is an one-one function. f is not onto since 25 belongs to Y does not have a pre-image in X
2. f: X → Y is onto, so each element of Y is the image of some element in X. There must be at least
as many elements in X as there are in Y for this to be true otherwise g: X → Y does not conform to
a definition of a function i.e. every x in X must belong to a unique ordered pair in (x,y) in f. So m ≥
n.
Your answer is right.
One more thing A function f : X Y is said to be an onto function if its range is equal to its codomain Y.
(iii) You have no way of knowing whether f : X → Y is onto. For example, if Y had one more element
than X then one element of Y under f would not map to an element in X.
Your answer is absolutely right.
(iv) Again, I don't think there's any way of knowing, is there?
Here is the right answer:
Here given g is given onto function and we are asked to find g :X Y is onto or not.
Case:1
It can be on-to. if its range is equal to its co-domain Y.
Example:
Case :2
It is one-one function but not onto, when its range is not equal to its co-domain Y.