0
$\begingroup$

I need help with this exercise, I prove the injectivity, but i'm stuck in the surjectivity of the function.

$f:\;\mathbb{Z\rightarrow Z}_{5} $ $f(x)=[{x}]$

Injective: Be $x_{1},x_{2}\in\mathbb{Z}$ Suppose $f(x_{1})=f(x_{2})$ Then $[x_{1}]=\left\{ x\in\mathbb{Z}:\;x\sim x_{1}\right\} =\left\{ x\in\mathbb{Z}:\;x\sim x_{2}\right\} =[x_{2}] \rightarrow x_{1}=x_{2}$

Surjective: Let $[y]\in\mathbb{Z}_{5}$, does there exists an $x\in\mathbb{Z}$ such that $f(x)=[y]$?

  • 0
    It seems you haven't understood what is a equivalence class. The implication $[x_1]=[x_2]\Longrightarrow x_1=x_2$ is not true. Furthermore, the reason $f$ is not a bijection is that it is not injective. It is surjective though since $f(\{0,1,2,3,4\})=\mathbb{Z}_{5}$. – 2017-01-20

1 Answers 1

2

Actually, $f(0)=f(5)=[0]$, but $0\neq 5$ as integers. Hence $f$ is not injective. It is clearly surjective.