Direct Image of a Set
Before we define what an inverse function is necessarily, let's first define some important terms leading us there.
| Definition: Suppose $f : A \to B$ is a function where $A = D(f)$ and $R(f) \subseteq B$. If $E \subseteq A$, then we say the direct image of $E$ under the function $f$ is the subset of $B$ such that $f(E) = \{ f(x) : x \in E \}$. |
For example, consider the function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = x + 1$, and suppose that $E \subset A = D(f)$ where $E = \{ x : 1 ≤ x ≤ 2 \}$. The direct image of $E$ under $f$ would be $f(E) = \{ y : 2 ≤ y ≤ 3 \}$ since any $x$ where $1 ≤ x ≤ 2$ maps to a value $y$ where $2 ≤ y ≤ 3$ under $f$.
Example 1
Show that if $f: A \to B$ is a function and $G \subseteq B$ and $H \subseteq B$, then $f^{-1} (G \cup H) = f^{-1} (G) \cup f^{-1} (H)$.
Suppose that $a \in f^{-1} (G \cup H)$. Then $f(a) \in (G \cup H)$ $\implies$ $f(a) \in G$ or $f(a) \in H$ or $f(a) \in G \cap H$ so then $a \in f^{-1} (G)$ or $a \in f^{-1} (H)$ or $a \in f^{-1} (G \cap H)$ so $a \in f^{-1} (G) \cup f^{-1} (H)$ so $f^{-1} (G \cup H) \subseteq f^{-1} (G) \cup f^{-1} (H)$.
Now suppose that $a \in f^{-1} (G) \cup f^{-1} (H)$. Then $a \in f^{-1} (G)$ or $a \in f^{-1} (H)$ or $a \in f^{-1} (H) \cap f^{-1} (G)$. Therefore $f(a) \in G$ or $f(a) \in H$ or $f(a) \in G \cap H$ so then $f(a) \in G \cup H$. Therefore $a \in f^{-1} (G \cup H)$ and thus $f^{-1} (G) \cup f^{-1} (H) \subseteq f^{-1} (G \cup H)$.
Since $f^{-1} (G \cup H) \subseteq f^{-1} (G) \cup f^{-1} (H)$ and $f^{-1} (G) \cup f^{-1} (H) \subseteq f^{-1} (G \cup H)$ it follows that $f^{-1} (G \cup H) = f^{-1} (G) \cup f^{-1} (H)$.
Inverse Image of a Set
| Definition: If $f : A \to B$ is a function where $H \subseteq B$ then the inverse image of $H$ under the function $f$ is defined to be the set $f^{-1}(H) = \{ x \in A : f(x) \in H \}$ containing elements $x \in A$ such that $f(x) \in H$. |
For example, consider the function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = x^3$ and suppose that $H \subset B$ where $H = \{ y : y ≥ 8 \}$. The inverse image of $H$ under $f$ would be $f^{-1} (H) = \{ x : x ≥ 2$. That is, all elements $x \in f^{-1} (H)$ map to an element in $H$.
Inverse Functions
| Definition: Suppose $f: A \to B$ is a bijection of the set $A$ onto $B$, the denote the inverse of $f$ as $f^{-1} = \{ (b, a) \in B \times A : (a, b) \in f \}$ is the subset of ordered pairs $(b, a)$ from the cartesian product $B \times A$ such that the ordered pair $(a, b) \in f$. |
For example, consider the function $f: \mathbb{R} \to \mathbb{R}$ where $f(x) = x - 1$. This function is surjective since $R(f) = B$ and this function is injective since whenever $a ≠ b$, $f(a) ≠ f(b)$ (passes the horizontal line test). Therefore, $f$ is bijective, so an inverse function exists, namely $f^{-1} (x) = y - 1$. We note that if $(a, b) \in f$, then $(b, a) \in f^{-1}$. For example, $(1, 0) \in f$ since $f(1) = 0$. Furthermore, $(0, 1) \in f^{-1}$ since $f^{-1} (1) = 0$.