Throughout, $\bar f$ stands for the direct image of a function, $f$, under some subset of its domain. $\hat f$ stands for the inverse image of $f$ under some subset of its co-domain.
Problem statement as written:
Suppose $f\colon A\rightarrow B$; let $C\subseteq A$.
a) Prove that $\bar f[\hat f[\bar f(C)]] = \bar f(C)$.
b) Use the result of (a) to prove that $\bar f \circ \hat f \circ \bar f = \bar f$.
I've proven (a), which is straighforward and only requires repeated application of the definitions of $\bar f$ and $\hat f$. I just want to check with the community that (b) requires no extra work, except perhaps mention that function composition obeys associativity, or citing the familiar lemma that if $f\colon A\rightarrow B$ and $g\colon A\rightarrow B$, then $f=g$ if and only if $\forall x\in A, f(x) = g(x)$. Both of these seem too marginal to warrant a separate section of the problem- I'd just like reassurance that I'm not majorly misunderstanding anything, please.