Solution to Problem 6 on Baby Exam
M366
Spring 2012
Problem 6 Prove that if X and Y are finite sets with the same number of
elements and
f :X→Y
is a function from X to Y , then f is injective if and only if it is surjective.
Proof: we are given that X and Y are finite sets with the same number of
elements. Without loss of generality, throughout the proof we assume:
|X| = |Y | = n
for some positive integer n and we denote the elements of X by x1 , . . . , xn and
the elements of Y by y1 , . . . , yn .
f injective implies f surjective: we assume that f is an injective function. This
means that if xi 6= xj then f (xi ) 6= f (xj ).
Therefore the set R = {f (x1 ), f (x2 ), . . . , f (xn )} consists of n distinct elements of Y . R is a subset of Y and it has the same number of elements as Y :
this means that R = Y .
Therefore for every y ∈ Y , y ∈ R, i.e. y = f (xi ) for some i. But this is the
definition of f being surjective.
f surjective implies f injective: now it is part of our hypotheses that f is a
surjective function.
We proceed to make a proof by contradiction. Let us assume that f is not
injective and derive a contradiction to our hypotheses.
If f is not injective there must be two distinct elements of X, that we
can name x1 and x2 , such that f (x1 ) = f (x2 ). Then the set R = {f (x1 ) =
f (x2 ), f (x3 ), . . . , f (xn )} has at most n − 1 elements and is therefore a proper
subset of Y. This means that there is an element of Y that does not belong to
R, i.e. that is not the image of any element of X via f . But this contradicts
the assumption that f is a surjective function.