s
Section 5.4 Functions
393
We have now seen that if f is a bijection, f: S S T, then there is a function
g: T S S with g + f = i S and f + g = i T . The converse is also true. To prove
the converse, suppose f: S S T and there exists g: T S S with g + f = i S and
f + g = i T . We can prove that f is a bijection. To show that f is onto, let t [ T.
Then t = i T (t) = ( f + g)(t) = f (g(t)). Because g: T S S, g(t) [ S, and g(t) is the
preimage under f of t. To show that f is one-to-one, suppose f (s 1 ) = f (s 2 ). Then
g( f (s 1 )) = g( f (s 2 )) and (g + f )(s 1 ) = (g + f )(s 2 ) implying i S (s 1 ) = i S (s 2 ), or
s 1 = s 2 . Thus, f is a bijection.
DefInItIon inveRSe Function
Let f be a function, f : S S T . If there exists a function g: T S S such that
g + f = i S and f + g = i T , then g is called the inverse function of f, denoted
by f
−1
.
We have proved the following theorem.
theoRem on BijectionS and inveRSe FunctionS
Let f: S S T . Then f is a bijection if and only if f
−1
exists.
Actually, we have been a bit sneaky in talking about the inverse function
of f. What we have shown is that if f is a bijection, this is equivalent to the existence of an inverse function. But it is easy to see that there is only one such
inverse function. When you want to prove that something is unique, the standard technique is to assume that there are two different such things and then
obtain a contradiction. Thus, suppose f has two inverse functions, f 1
−1
and f 2
−1
(existence of either means that f is a bijection). Both f 1
−1
and f 2
−1
are functions
from T to S; if they are not the same function, then they must act differently
somewhere. Assume that there is a t [ T such that f 1
−1
(t) ∙ f 2
−1
(t). Because f is
one-to-one, it follows that f ( f 1
−1
(t)) ∙ f ( f 2
−1
(t)), or ( f + f 1
−1
)(t) ∙ ( f + f 2
−1
)(t).
But both f + f 1
−1
and f + f 2
−1
are i T , so t ∙ t, which is a contradiction. We are
therefore justified in speaking of f
−1
as the inverse function of f. If f is a bijection, so that f
−1
exists, then f is the inverse function for f
−1
; therefore, f
−1
is
also a bijection.
PRaCtiCe 34
f : R S R given by f (x) = 3x + 4 is a bijection.
Describe f
−1
.
■
Section 5.4 Functions
393
We have now seen that if f is a bijection, f: S S T, then there is a function
g: T S S with g + f = i S and f + g = i T . The converse is also true. To prove
the converse, suppose f: S S T and there exists g: T S S with g + f = i S and
f + g = i T . We can prove that f is a bijection. To show that f is onto, let t [ T.
Then t = i T (t) = ( f + g)(t) = f (g(t)). Because g: T S S, g(t) [ S, and g(t) is the
preimage under f of t. To show that f is one-to-one, suppose f (s 1 ) = f (s 2 ). Then
g( f (s 1 )) = g( f (s 2 )) and (g + f )(s 1 ) = (g + f )(s 2 ) implying i S (s 1 ) = i S (s 2 ), or
s 1 = s 2 . Thus, f is a bijection.
DefInItIon inveRSe Function
Let f be a function, f : S S T . If there exists a function g: T S S such that
g + f = i S and f + g = i T , then g is called the inverse function of f, denoted
by f
−1
.
We have proved the following theorem.
theoRem on BijectionS and inveRSe FunctionS
Let f: S S T . Then f is a bijection if and only if f
−1
exists.
Actually, we have been a bit sneaky in talking about the inverse function
of f. What we have shown is that if f is a bijection, this is equivalent to the existence of an inverse function. But it is easy to see that there is only one such
inverse function. When you want to prove that something is unique, the standard technique is to assume that there are two different such things and then
obtain a contradiction. Thus, suppose f has two inverse functions, f 1
−1
and f 2
−1
(existence of either means that f is a bijection). Both f 1
−1
and f 2
−1
are functions
from T to S; if they are not the same function, then they must act differently
somewhere. Assume that there is a t [ T such that f 1
−1
(t) ∙ f 2
−1
(t). Because f is
one-to-one, it follows that f ( f 1
−1
(t)) ∙ f ( f 2
−1
(t)), or ( f + f 1
−1
)(t) ∙ ( f + f 2
−1
)(t).
But both f + f 1
−1
and f + f 2
−1
are i T , so t ∙ t, which is a contradiction. We are
therefore justified in speaking of f
−1
as the inverse function of f. If f is a bijection, so that f
−1
exists, then f is the inverse function for f
−1
; therefore, f
−1
is
also a bijection.
PRaCtiCe 34
f : R S R given by f (x) = 3x + 4 is a bijection.
Describe f
−1
.
■