Topology: An Inquiry-Based Approach

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 $x\in \mathbb{R}$ with $x>0$ and for each $y\in \mathbb{R}\text{,}$

$y=\ln (x)$ 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.

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 $A\text{.}$

The result of the Activity 2.5 should have been the following theorem.

The proof of Theorem 2.7 is outlined in the following activity.

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.

The next result provide useful information about inverse functions. The proofs are left for Exercise 8.

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 $(g\circ f{)}^{-1}\text{?}$

The result of Activity 2.7 is contained in the next theorem.