Define $f : Z/3Z → Z/3Z$ by $f ([a]) = [2a + 1]$

Question

Just finished proving this is well-defined, how do I prove it's surjective and injective?

I know that injective means that if $x1 \neq x2$, then $f(x_1) \neq f(x2)$, i.e. each value in the domain is uniquely mapped to an element in the codomain, with no duplicates.

Not sure how to even begin to prove this though.

Answer 1

$[2a+1]=[2b+1]$

$\implies 2a+1-(2b+1)\equiv 0 $mod 3

$\implies 2(a-b) \equiv 0$ mod 3

$\implies 3$ divides $2(a-b)$ and since 3 is prime 3 divides $a-b$

$\implies [a]=[b]$

Since the set is finite injective implies surjective