Proving properties of the images and inverse images of functions Let $f : X\to  Y; \;\;g : Y\to Z;\,$ and $\;g \circ f : X\to Z.$ 

Prove or disprove

a) For all subsets $\,A \subseteq X,\;\; f^{-1}(f(A)) = A$.

b) For all subsets $\,B \subseteq Y,\;\; f(f^{-1}(B)) = B$.

c) For all subsets $\,E \subseteq Z,\; \; (g \circ f)^{-1}(E) = f^{-1}(g^{-1}(E))$
a) False. Consider $f:\{0,1,2\} → \{0,1\}$ defined as $f(x)=0$   $\forall x$. Then $f^{-1}(f(A))=\{0,1,2\}$ for any subset $A⊆\{0,1,2\}$.

b)False. Consider f defined above. Then $f(f^{-1}(\{0,1\})=\{0\}.$

c)True. For $E⊆Z$, we have $x \in (g∘f)^{-1}(E) \iff (g∘f)(x) \in E \iff f(x)\in g^{-1}(E) \iff x \in f^{-1}(g^{-1}(E))$. Thus $(g∘f)^{-1}(E) = f^{-1}(g^{-1}(E))$. QED
Well for any element $x\in X$, $f^{-1}(f(x))=x$,and since $A\subseteq X$,$x \in A \implies x\in X$,
$y\in Y$$$f(f^{-1}(y))=y$$
Since $B\subseteq Y$,$y \in B \implies y\in Y,$
$e\in E$
$$(g^{-1}(f^{-1}(e))=(f^{-1}(g^{-1}(e))$$
Since $E\subseteq Z$,and $f^{-1}:Y\to X$ means that e might not be in $Y$ therefore it's false