Be $f:\;\mathbb{Z\rightarrow Z}_{5} $ Prove $f$ is not bijective.

Question

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 $f(x)=[y]$?

Answer 1

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