Let $f:X\rightarrow Y$ be a function from one set $X$ to another set $Y$, let $S$ be a subset of $X$, and let $U$ be a subset of $Y$. What, in general, can one say about $f^{-1}(f(S))$ and $S$? What about $f(f^{-1}(U))$ and $U$?
MY ATTEMPT (EDIT)
One can say that $S\subseteq f^{-1}(f(S))$. Indeed, let us take $s\in S$. Then $f(s)\in f(S)$. Consequently, $s\in f^{-1}(f(S))$, since we have \begin{align*} x\in f^{-1}(f(S)) \Longleftrightarrow f(x) \in f(S) \end{align*}
However the inclusion $S\supseteq f^{-1}(f(S))$ is false in general. Let us consider, for instance, that $f:\{0,1\}\rightarrow\{0\}$ is given by $f(0) = f(1) = 0$. If we choose $S = \{0\}$, one has that $f(S) = \{0\}$. Consequently, $f^{-1}(f(S)) = \{0,1\}$, which is different from $S$.
On the other hand, one has that $U\supseteq f(f^{-1}(U))$. Indeed, if $u\in f(f^{-1}(U))$, then one may claim \begin{align*} u\in f(f^{-1}(U)) & \Longrightarrow u = f(x)\,\,\text{for some}\,\,x\in f^{-1}(U)\\\\ & \Longrightarrow (\exists x)(u = f(x))\wedge(f(x)\in U) \Longrightarrow u\in U \end{align*}
However the inclusion $U\subseteq f(f^{-1}(U))$ is not true in general. Consider, for instance, the function $f:\{0\}\rightarrow\{0,1\}$ such that $f(0) = 0$. Consequently, for $U = \{0,1\}$, one has that $f^{-1}(U) = \{0\}$ and, consequently, $f(f^{-1}(U)) = \{0\}$, which is different from $U$.
The correct inclusions are $S \subseteq f^{-1}(f(S))$ (the unit, if you think of the direct and inverse image as adjoint functors) and $f(f^{-1}(U)) \subseteq U$ (the counit, again if you think of the direct and inverse image as adjoint functors). The former will be an equality if $f$ is injective, while the latter will be an equality if $f$ is surjective, or more generally, if $U \subseteq f(X)$.