A doubt on the question $C = f^{-1}(f(C)) \iff f$ is injective and the similar surjective version

Question

Prove that $C = f^{-1}(f(C)) \iff f$ is injective and $f(f^{-1}(D)) = D \iff f$ is surjective I have a doubt in the question asked above.

In this statement,

$C = f^{-1}(f(C)) \iff f$ is injective

I can't get why the following is not true.

$f$ is injective is not a necessary condition as if we choose $C$ contains $a,b\in C$ , $a\neq b$ and $f(a) = f(b)$. Which means $f$ can have a many to one relation and holds the equality.

In the same manner I'm thinking about the second part of the original question. Can anyone make it clear.

Edit: A similar question published earlier

$A\subset f^{-1}(f(A))$ with equality if and only $f$ is injective.

Answer 1

Note that $C$ [respectively $D$] is (implicitly) bound by a universal quantifier, so that the condition needs to hold for all subsets $C$.

In the given circumstances $C=\{a\}$ would be a better choice to exhibit a counterexample. Can you see why?