Section Inverse Functions
Now that we have studied composite functions, we will move on to consider another important idea: the inverse of a function. In previous mathematics courses, you probably learned that the exponential function (with base \(e\)) and the natural logarithm functions are inverses of each other. You may have seen this relationship expressed as follows: For each \(x \in \R\) with \(x \gt 0\) and for each \(y \in \R\text{,}\)
\(y = \ln(x)\) if and only if \(x = e^y\text{.}\) Notice that \(x\) is the input and \(y\) is the output for the natural logarithm function if and only if \(y\) is the input and \(x\) is the output for the exponential function. In essence, the inverse function (in this case, the exponential function) reverses the action of the original function (in this case, the natural logarithm function). In terms of ordered pairs (input-output pairs), this means that if \(( {x, y} )\) is an ordered pair for a function, then \(( {y, x} )\) is an ordered pair for its inverse. The idea of reversing the roles of the first and second coordinates is the basis for our definition of the inverse of a function.
Definition 2.6.
Let \(f : A \to B\) be a function. The inverse of \(f\text{,}\) denoted by \(f^{ - 1}\text{,}\) is the set of ordered pairs
\begin{equation*}
f^{ - 1} = \left\{ { {( {b, a} ) \in B \times A} \mid ( {a, b} ) \in f} \right\}\text{.}
\end{equation*}
Notice that this definition does not state that
\(f^{-1}\) is a function. Rather,
\(f^{-1}\) is simply a subset of
\(B \times A\text{.}\) In
Activity 2.5, we will explore the conditions under which the inverse of a function
\(f: A \to B\) is itself a function from
\(B\) to
\(A\text{.}\)
Activity 2.5.
Let \(A = \left\{ {a, b, c} \right\}\text{,}\) \(B = \left\{ {a,b,c,d} \right\}\text{,}\) and \(C = \left\{ {p, q, r} \right\}\text{.}\) Define
\begin{align*}
f: A \amp \to C \text{ by} \amp g: A \amp \to C \text{ by} \amp h: B \amp \to C \text{ by}\\
f( a ) \amp = r \amp g( a ) \amp = p \amp h( a ) \amp = p\\
f( b ) \amp = p \amp g( b ) \amp = q \amp h( b ) \amp = q\\
f( c ) \amp = q \amp g( c ) \amp = p \amp h( c ) \amp = r\\
\amp \amp \amp \amp h ( d ) \amp = q
\end{align*}
(a)
Determine the inverse of each function as a set of ordered pairs.
(b)
(i)
Is \(f^{ - 1}\) a function from \(C\) to \(A\text{?}\) Explain.
(ii)
Is \(g^{ - 1}\) a function from \(C\) to \(A\text{?}\) Explain.
(iii)
Is \(h^{ - 1}\) a function from \(C\) to \(B\text{?}\) Explain.
(c)
Make a conjecture about what conditions on a function \(F: S \to T\) will ensure that its inverse is a function from \(T\) to \(S\text{.}\)
The result of the
Activity 2.5 should have been the following theorem.
Theorem 2.7.
Let \(A\) and \(B\) be nonempty sets, and let \(f: A \to B\text{.}\) The inverse of \(f\) is a function from \(B\) to \(A\) if and only if \(f\) is a bijection.
The proof of
Theorem 2.7 is outlined in the following activity.
Activity 2.6.
Theorem 2.7 is a biconditional statement, so we need to prove both directions. Let
\(A\) and
\(B\) be nonempty sets, and let
\(f: A \to B\text{.}\)
(a)
Assume that
\(f\) is a bijection. We will prove that
\(f^{-1}\) is a function, that is that
\(f^{-1}\) satisfies the conditions of
Definition 2.1.
(i)
Let \(b \in B\text{.}\) What property does \(f\) have that ensures that \((b,a) \in f^{-1}\) for some \(a \in A\text{?}\) What conclusion can we draw about \(f^{-1}\text{?}\)
(ii)
Now let \(b \in B\text{,}\) \(a_1 , a_2 \in A\) and assume that
\begin{equation*}
( {b, a_1 } ) \in f^{ - 1} \text{ and } ( {b, a_2 } ) \in f^{-1}\text{.}
\end{equation*}
What does this tell us about elements that must be in \(f\text{?}\) What property of \(f\) ensures that \(a_1=a_2\text{?}\) What conclusion can we draw about \(f^{-1}\text{?}\)
(b)
Now assume that \(f^{-1}\) is a function from \(B\) to \(A\text{.}\) We will prove that \(f\) is a bijection.
(i)
What does it take to prove that \(f\) is an injection? Use the fact that \(f^{-1}\) is a function to prove that \(f\) is an injection.
(ii)
What does it take to prove that \(f\) is a surjection? Use the fact that \(f^{-1}\) is a function to prove that \(f\) is a surjection.
In the situation where
\(f: A \to B\) is a bijection and
\(f^{-1}\) is a function from
\(B\) to
\(A\text{,}\) we can write
\(f^{-1} : B \to A\text{.}\) In this case, we frequently say that
\(f\) is an
invertible function, and we usually do not use the ordered pair representation for either
\(f\) or
\(f^{-1}\text{.}\) Instead of writing
\(( {a, b} ) \in f\text{,}\) we write
\(f( a ) = b\text{,}\) and instead of writing
\(( {b, a} ) \in f^{-1}\text{,}\) we write
\(f^{-1} ( b ) = a\text{.}\) Using the fact that
\(( {a, b} ) \in f\) if and only if
\(( {b, a} ) \in f^{-1}\text{,}\) we can now write
\(f( a ) = b\) if and only if
\(f^{-1} ( b ) = a\text{.}\) Theorem 2.8 formalizes this observation.
Theorem 2.8.
Let \(A\) and \(B\) be nonempty sets, and let \(f: A \to B\) be a bijection. Then \(f^{ - 1} : B \to A\) is a function, and for every \(a \in A\) and \(b \in B\text{,}\)
\begin{equation*}
f( a ) = b \text{ if and only if } f^{ - 1} ( b ) = a\text{.}
\end{equation*}
The next result provide useful information about inverse functions. The proofs are left for
Exercise 8.
Corollary 2.9.
Let \(A\) and \(B\) be nonempty sets, and let \(f: A \to B\) be a bijection. Then
For every \(x\) in \(A\text{,}\) \(\left( f^{ - 1} \circ f \right)(x) = x\text{.}\)
For every \(y\) in \(B\text{,}\) \(\left( f \circ f^{ - 1}\right)(y) = y\text{.}\)
The next question to address is what we can say about a composition of bijections. In particular, if \(f: A \to B\) and \(g: B \to C\) are both bijections, then \(f^{ - 1} : B \to A\) and \(g^{ - 1} : C \to B\) are both functions. Must it be the case that \(g \circ f\) is invertible and, if so, what is \((g \circ f)^{-1}\text{?}\)
Activity 2.7.
Let \(f: A \to B\) and \(g: B \to C\) both be bijections.
(a)
Why do we know that \(g \circ f\) is invertible?
(b)
Now we determine the inverse of \(g \circ f\text{.}\) We might be tempted to think that \((g \circ f)^{-1}\) is \(g^{-1} \circ f^{-1}\text{,}\) but this composite is not defined because \(g^{-1}\) maps \(B\) to \(C\) and \(f^{-1}\) maps \(B\) to \(A\text{.}\) However, \(f^{_1} \circ g^{-1}\) is defined. To prove that \((g \circ f)^{-1} = f^{-1} \circ g^{-1}\text{,}\) we need to prove that two functions are equal. How do we prove that two functions are equal?
(c)
Suppose \(c \in C\text{.}\)
(i)
What tells us that there is a \(b \in B\) so that \(g(b) = c\text{?}\)
(ii)
What tells us that there is an \(a \in A\) so that \(f(a) = b\text{?}\)
(iii)
What element is \((g \circ f)^{-1}(c)\text{?}\) Why?
(iv)
What element is \(f^{-1}(b)\text{?}\) Why? What element is \(g^{-1}(c)\text{?}\) Why?
(v)
What element is \((f^{-1} \circ g^{-1})(c)\text{?}\) Why? What can we conclude about \((g \circ f)^{-1}\) and \(f^{-1} \circ g^{-1}\text{?}\) Explain.
The result of
Activity 2.7 is contained in the next theorem.
Theorem 2.10.
Let \(f: A \to B\) and \(g: B \to C\) be bijections. Then \(g \circ f\) is a bijection and \(( {g \circ f} )^{ - 1} = f^{ - 1} \circ g^{ - 1}\text{.}\)
