Showing a function $\Bbb Z/5\to\Bbb Z/5$ is surjective

Question

How can I prove that $f: ℤ/5 → ℤ/5$ given by $f(x) = x \cdot [2]_5$ is surjective?

Answer 1

for injectiv, jaust look where 1,2,3,4,0 go to. for the inverse multiply the Equation with the inverse of 2 trula

Answer 2

Omitting the $[x]_5$ notation, i.e $x=[x]_5$

$$ 0 \rightarrow 0 \\ 1 \rightarrow 2 \\ 2 \rightarrow 4 \\ 3 \rightarrow 6=1 \\ 4 \rightarrow 8=3 \\ 5 \rightarrow 10=5 $$ Thus surjective as every value gets mapped to.

Answer 3

Speaking more abstractly than others: First note that $Z_5$ is a field, hence at least an integral domain - i.e there are no zero-divisors in the system. Hence if $2x = 2y$ in the system, then $2 (x - y) = 0$, and so since there are no zero-divisors, necessarily $x - y = 0$, or $x = y$. Hence the map $x \rightarrow 2x$ is one-to-one, hence onto by finiteness. Note that this argument generalizes beyond the specific values 5 and 2.