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Mappings

Definition 5.1.1   Let $ A$ and $ B$ be two sets. A mapping from $ A$ to $ B$ is a binary relation, whose graph $ \mathcal{G}
\subset A \times B$ verify the following condition:

$\displaystyle \forall a \in A, \; \exists ! \; b \in B \; \vert \; (a,b) \in \mathcal{G}.$    

i.e. a mapping establishes a correspondence between the elements of $ A$ and the elements of $ B$ such that every element of $ A$ appears in a single pair in $ \mathcal{G}$.

As for general binary relations, $ A$ is called the domain of the mapping and $ B$ is the range of the mapping. Usually, mappings are denoetd in another way: the mapping is denoted by a single letter $ f=(A,B,\mathcal{G})$, and the triple is displayed as in the following diagram:
$ f:$ $ A$ $ \longrightarrow$ $ B$
     $ a$ $ \mapsto$ $ b$
The element $ B$ of $ B$ is denoted by $ f(a)$ and is called the image of $ a$ by the mapping $ f$. If $ b=f(a)$, the element $ a$ in $ A$ is called a pre-image of $ b$ by $ f$. For example, let $ f: \mathbb{R}\longrightarrow \mathbb{R}$ such that for any $ x \in \mathbb{R}$, $ f(x)=x^2$. Then $ 4$ is the image of $ 2$ by $ f$, $ 9$ has two pre-images by $ f$, namely $ -3$ and $ 3$, and $ -2$ has no pre-image by $ f$. Figure 1 displays diagrams of binary relations. The first diagram determines a mapping, as it fulfills the requirements of Definition 1.1. The second one does not determine a mapping, as there exists at least one element in the domain without an image; the third one does not determine a mapping, as there exist a domain element related to two different elements in the range.

Figure 1: Mapping or not.
\begin{figure}
\centering
\mbox{\subfigure[a mapping]{\epsfig{file=Mapping-01...
...ot a mapping]{\epsfig{file=NotMapping-02.eps, height=3.5cm}}
}
\end{figure}

Example 5.1.2       

Definition 5.1.3 (Composition of mappings)   Let $ A$, $ B$ and $ C$ be three sets and let $ f: \; A \rightarrow B$ and $ g: \; B \rightarrow C$ be two mappings. The composition of $ f$ by $ g$ is the mapping denoted by $ gof: A \rightarrow C$ such that

$\displaystyle \forall x \in A, \; (gof)(x)=g(f(x)).$    

Figure 2: Composition of two mappings.
\begin{figure}
\centering
\mbox{\epsfig{file=Composition.eps, width=4cm}}
\end{figure}

Example 5.1.4   Let $ f$ and $ g$ be two mappings from $ \mathbb{R}$ to $ \mathbb{R}$ given respectiveley by $ f(x)=2x-1$ and $ g(x)=3x+5$. Then we have:

$\displaystyle \forall x \in \mathbb{R}, \; (gof)(x)=g(f(x))=3f(x)+5=3(2x-1)+5=6x+2.$    

In general, we cannot compute both compositions $ gof$ and $ fog$ (if the given sets are all different), but even when both compositions are defined, they are generally different. For example, let $ f: \; x \mapsto 3x-1$ and $ g: \; x \mapsto -2x+7$ be two mappings from $ \mathbb{R}$ to itself. Then we have:

$\displaystyle \forall x \in \mathbb{R}, \;$ $\displaystyle (gof)(x)= g(f(x))= -2f(x)+7= -2(3x-1)+7=-6X+9$    
$\displaystyle \quad$ $\displaystyle (fog)(x)=f(g(x)=3(-2x+7)-1=-6x+20.$    

The mappings $ gof$ and $ fog$ have the same domain and the same range, but their graphs are different (i.e. the correspondences they define are different). For example, we have $ gof(1)=3$ and $ (fog)(1)=14$, hence $ (gof)(1) \neq (fog)(1)$.
next up previous contents
Next: Injections Up: Mappings Previous: Mappings   Contents
root 2002-06-10