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Discrete Mathematical Structures

Dan Yasaki, Sebastian Pauli

Subsection 2.3.6 Bijective functions

Definition 2.3.6.1.

Let \(f \colon A \to B\text{.}\) The function \(f\) is bijective if \(f\) is injective and surjective (or one-to-one and onto). A bijective function is called a bijection or one-to-one correspondence.
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Example 2.3.6.2. Identity function is bijective.

The identity function \(f \colon \RR \to \RR\) defined by \(f(x) = x\) is bijective.
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Remark 2.3.6.4.

If \(A\) and \(B\) are finite sets of the same size, a function \(f \colon A \to B\) is surjective if and only if \(f\) is injective. Furthermore, if \(\size{A} \lt \size{B}\) then \(f\) cannot be surjective. If \(\size{B} \lt \size{A}\) then \(f\) cannot be injective.
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Example 2.3.6.5. Area of a circle.

The area of a circle is a function of its radius. This function is injective, because if the area of a circle of radius \(r\) is equal to the area of a circle of radius \(r'\text{,}\) we must have \(r = r'\text{.}\) This function is not surjective, when viewed as a function from \(\RR_{> 0}\) to \(\RR\text{,}\) since there is no radius that would give rise to area \(-1\text{.}\) When we restrict the codomain to \(\RR_{>0}\text{,}\) this function is surjective since we can construct a circle for any given area.
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Definition 2.3.6.7.

Let \(f \colon A \to B\) be a bijection. The inverse of \(f\text{,}\) denoted \(f^{-1}\) , is a function \(f^{-1} \colon B \to A\) that assigns to an element \(b \in B\text{,}\) the unique element \(a \in A\) such that \(f(a) = b\text{.}\)
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From Definition 2.3.6.7, it immediately follows that \(f\) is an invertible function if and only if
\begin{align*} f(f^{-1}(b)) \amp = b \quad \text{for all }b \in B; \text{ and} \\ f^{-1}(f(a)) \amp = a \quad \text{for all }a \in A. \end{align*}
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Two sets A and B are represented each by an oval. The ovals do not overlap. An arrow labeled f goes from A to B. An arrow labeled f to the -1 goes from B to A.
Figure 2.3.6.8. A function \(f\) from \(A\) to \(B\) and its inverse from \(B\) to \(A\text{.}\)
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Remark 2.3.6.9.

  1. The definition of inverse makes sense since \(f\) is a
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  2. \(f^{-1}\) does not mean \(\frac{1}{f}\text{.}\)
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  3. Non-bijective functions do not have an inverse.
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Example 2.3.6.10. Inverse of a function.

Consider the function \(f(x) = 3x + 2\text{.}\) Find the inverse of \(f\text{.}\)
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Solution.
Let \(y = f(x)\text{.}\) Then \(f^{-1}(y) = x\text{.}\) That means we can just take \(y = 3 x + 2\text{,}\) and solve for \(x\text{.}\) Doing so, we see that \(x = \frac{y - 2}{3}\text{.}\) That means \(f^{-1}(y) = \frac{y - 2}{3}\text{.}\) Convention often dictates the input variable be named \(x\text{,}\) so we write it as
\begin{equation*} f^{-1}(x) = \frac{x - 2}{3}. \end{equation*}
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Finding the inverse of a function is more difficult than checking if given functions are inverses of each other.
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Example 2.3.6.11. Check the inverse.

Let \(f\) and \(g\) be functions from \(\RR\) to \(\RR\) given by \(f(x) = x^3 - 4\) and \(g(x) = \sqrt[3]{x + 4}\text{.}\) Determine whether \(g\) is the inverse of \(f\text{.}\)
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Solution.
We need to check that and both hold. We compute each one separately. For all \(x \in \RR\text{,}\) we have
\begin{align*} f(g(x)) \amp = f(\sqrt[3]{x + 4}) \\ \amp = (\sqrt[3]{x + 4})^3 - 4 \\ \amp = (x + 4) - 4 \\ \amp = x; \end{align*}
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\begin{align*} g(f(x)) \amp =g(x^3 - 4) \\ \amp = \sqrt[3]{(x^3 - 4) + 4} \\ \amp = \sqrt[3]{x^3 } \\ \amp = x. \end{align*}
Thus \(g\) is the inverse of \(f\text{.}\)
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Note: This also shows that \(f\) is the inverse of \(g\text{.}\)
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In Example 2.3.6.12 observe how changing the plot of the graph of a function changes its graph and the properties of the function.
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Example 2.3.6.12. Interactive plot of a function with invertibility.

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