Injections.

Definition 2.2.1   A function $ f: \; A \longrightarrow B$ is an injection (or is injective) if every element of $ B$ has at most one pre-image in $ A$ .

$ f:A \longrightarrow B$ is an injection if, and only if:

$\displaystyle \forall x_1 \in A, \forall x_2 \in A, \quad x_1 \neq x_2 \Longrightarrow f(x_1)_ \neq f(x_2).$    

or equivalently:

$\displaystyle \forall x_1 \in A, \forall x_2 \in A, f(x_1) = f(x_2) \Longrightarrow \quad x_1 = x_2.$    

Example 2.2.2   Consider

$\displaystyle f:$ $\displaystyle \mathbb{R}$ $\displaystyle \longrightarrow$ $\displaystyle \quad \mathbb{R}$    
$\displaystyle \quad$ $\displaystyle x$ $\displaystyle \mapsto$ $\displaystyle \quad 3x+5$    

We prove that $ f$ is an injection: let $ x_1$ and $ x_2$ be two real numbers such that $ f(x_1)=f(x_2)$ . Then we have:
$\displaystyle f(x_1)$ $\displaystyle =$ $\displaystyle f(x_2)$  
$\displaystyle 3x_1+5$ $\displaystyle =$ $\displaystyle 3x_2+5$  
$\displaystyle 3x_1$ $\displaystyle =$ $\displaystyle 3x_2$  
$\displaystyle x_1$ $\displaystyle =$ $\displaystyle x_2$  

Example 2.2.3   Consider

$\displaystyle f:$ $\displaystyle \mathbb{R}$ $\displaystyle \longrightarrow$ $\displaystyle \quad \mathbb{R}$    
$\displaystyle \quad$ $\displaystyle x$ $\displaystyle \mapsto$ $\displaystyle \quad x^2$    

We have $ f(-2)=f(2)=4$ , therefore $ f$ is not an injection.

Remark 2.2.4       

Proposition 2.2.5   Let $ f:A \longrightarrow B$ and $ g: B \longrightarrow C$ be two applications. If $ f$ and $ g$ are injective, then $ g o f : A \longrightarrow C$ is an injection.

Proposition 2.2.6   Let $ f:A \longrightarrow B$ and $ g: B \longrightarrow C$ be two applications. If $ g o f : A \longrightarrow C$ is injective, then $ f$ is injective.

Noah Dana-Picard 2007-12-28