next up previous contents
Next: The characteristic function of Up: Mappings Previous: Surjections   Contents

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$.

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

Figure 5: A bijection or not.
\begin{figure}
\centering
\mbox{\subfigure[a bijection]{\epsfig{file=Bijectio...
... bijection]{\epsfig{file=NotBijection-02.eps, height=3.5cm}}
}
\end{figure}

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.
\begin{figure}
\centering
\mbox{\epsfig{file=InvertibleMap-01.eps,height=3cm}}
\end{figure}

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:

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

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:

$\displaystyle fog=Id_{\mathbb{R}} \Longleftrightarrow \forall x \in \mathbb{R},$ $\displaystyle f(g(x))=x$    
$\displaystyle \quad$ $\displaystyle 2 g(x)-1=x$    
$\displaystyle \quad$ $\displaystyle 2 g(x) = x+1$    
$\displaystyle \quad$ $\displaystyle g(x) = \frac 12 (x+1).$    

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

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

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


next up previous contents
Next: The characteristic function of Up: Mappings Previous: Surjections   Contents
root 2002-06-10