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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.

Krishna
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  • Your question is not clear. You don't understand why $f^{-1}(f(C))=C$ implies $f$ injective ? By the way, your statement is not clear. $f$ is injective if $C=f^{-1}(f(C))$ for all $C$ ! – Surb Mar 28 '20 at 13:10
  • Note: "doubt" in Indian English = "question" in US or UK English. – GEdgar Mar 28 '20 at 17:00

1 Answers1

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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?

Berci
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  • Oh thank you. I just forgot that $C$ could be any subset of $A$ and I was trying to pick a subset which particularly doesn't obey the condition. – Krishna Mar 28 '20 at 13:27