$f: A \longrightarrow A$ be a function and let $C \subseteq A$. $f^{-1}(f(C)) \subseteq f\left(f^{-1}(C)\right)$ is false? $f: A \longrightarrow A$ be a function and let $C \subseteq A$.
True of false: $f^{-1}(f(C)) \subseteq f\left(f^{-1}(C)\right)$ ?
I know that $f\left(f^{-1}(C)\right) \subseteq f^{-1}(f(C))$ is true. Because $f\left(f^{-1}(C)\right)\subseteq C$ and $C \subseteq f^{-1}(f(C))$.
Therefore I believe this one should be false.
My question is what is the counterexample that could show it is false? And is my proof of
"$f\left(f^{-1}(C)\right) \subseteq f^{-1}(f(C))$ is true" correct?
 A: Your proof that $f(f^{-1}(C))\subseteq f^{-1}(f(C))$ is correct.
Suppose $a\in f^{-1}(f(C))$. By definition, you know that $f(a)\in f(C)$, so you have $f(a)=f(c)$, for some $c\in C$.
Now you wish to find some condition that guarantees $a\in f(f^{-1}(C))$. You should be able to find $b\in f^{-1}(C)$, that is, $f(b)\in C$, such that $a=f(b)$.
Well, take $a\notin f(A)$, that is, assume $f$ is not surjective. Take $C=\{a\}$. Then, obviously, $a\in f^{-1}(f(C))$, but $a\notin f(f^{-1}(C))\subseteq f(A)$.
A: egreg's answer shows that if $f : A \to A$ and $f$ is not surjective then there is a set $C \subseteq A$ such that $f^{-1}(f(C)) \nsubseteq f(f^{-1}(C))$.  It is also the case that if $f : A \to A$ and $f$ is not injective then there is a set $C \subseteq A$ such that $f^{-1}(f(C)) \nsubseteq f(f^{-1}(C))$.  Sketch of proof:  If $f$ is not injective then there are distinct $a_1$ and $a_2$ in $A$ such that $f(a_1) = f(a_2)$.  Let $C =\{a_1\}$.  Now verify that $a_2 \in f^{-1}(f(C))$ but $a_2 \notin f(f^{-1}(C))$.
Exercise:  Prove that if $f$ is both surjective and injective then for every $C \subseteq A$, $f^{-1}(f(C)) \subseteq f(f^{-1}(C))$.
