Bijections

Definition 5.4.1   The function $f: \; A \mapsto B$ is bijective if it is both injective and surjective, i.e. if every element of the range $B$ has a unique pre-image by $f$ in $A$.

$$\forall y \in B, \; \exists ! \; x \in A , \; y=f(x).$$

Figure 5: A bijection or not.

The mapping shown in Example 3.2 is a bijection from $\mathbb{R}$ to $\mathbb{R}$: when we looked for a pre-image of an element in $\mathbb{R}$, we actually proved that every real number has a unique pre-image by $f$. The following proposition is a direct consequence of Prop. 2.4 and Prop. 3.4.

Proposition 5.4.2   Let $A$,$B$,$C$ be three sets and let $f:A \rightarrow B$ and $g: A \rightarrow C$ be two mappings. If $f$ and $g$ are bijective, then $gof$ is a bijection.

The following proposition is a direct consequence of Prop. 2.5 and Prop. 3.5.

Proposition 5.4.3   Let $A$,$B$,$C$ be three sets and let $f:A \rightarrow B$ and $g: A \rightarrow C$ be two mappings. If $gof$ is a bijection, then $f$ is injective and $g$ is surjective.

Definition 5.4.4   Let $E$ be a set. The identity mapping of $A$ is the mapping $Id_A: \; A \rightarrow A$ such that for any $x \in A$, $Id_A (x)=x$.

Definition 5.4.5   Let $A$ and $B$ be two sets and let $f$ be a mapping from $A$ to $B$. The mapping $f$ is invertible if there exists a mapping $g: \; B \rightarrow A$ such that $gof=Id_A$ and $fog=Id_B$ (see Figure 6). If these conditions hold, we say that $g$ is the inverse mapping of $f$ and we denote it by $f^{-1}$.

Figure 6: Invertible mappings.

Example 5.4.6   Let $f : \; \mathbb{R}\longrightarrow \mathbb{R}^+$ such that $f(x)=x^2$ and $g: \; \mathbb{R}^+ \longrightarrow \mathbb{R}$ such that $f(x)=\sqrt{x}$. We have:

$$\forall x \in \mathbb{R}^+, \; (fog)(x)= (\sqrt{x})^2=x=Id_{\mathbb{R}^+}(x)$$

$$\forall x \in \mathbb{R}, \; (gof)(x)=\sqrt{x^2}=\vert x\vert.$$

Thus $gof \neq Id_{\mathbb{R}}$. We have two comments here:

1. The mapping $g$ is not an inverse mapping for $f$.
2. This does not mean that $f$ is not invertible! Perhaps there exists another mapping for which the conditions hold. Actually, there is none, but we still did not prove it!

Example 5.4.7   Let $f: \; \mathbb{R}\longrightarrow \mathbb{R}$ such that $f(x)=2x-1$. We look for a mapping $g : \; \mathbb{R}\longrightarrow \mathbb{R}$ such that the conditions of Definition 4.5 hold. First we have:

$$fog=Id_{\mathbb{R}} \Longleftrightarrow \forall x \in \mathbb{R}, f(g(x))=x$$

$$\quad 2 g(x)-1=x$$

$$\quad 2 g(x) = x+1$$

$$\quad g(x) = \frac 12 (x+1).$$

Now we must check whether the mapping $g$ verifies the second condition:

$$\forall x \in \mathbb{R}, \; (gof)(x)=g(f(x))=\frac 12 (f(x)+1) =\frac 12 (2x-1 +1)=x.$$

Thus $gof=Id_{\mathbb{R}}$. we proved that $f$ is invertible and that $g=f^{-1}$.