For any subset $B$, we have $f^{-1}(f(f^{-1}(B)))=f^{-1}(B)$ Let $E$ and $F$ be sets and $f:E\to F$ be a map. Let $B\subset F$. I want to show that $f^{-1}(f(f^{-1}(B)))=f^{-1}(B)$.
We already have that $f^{-1}(B)\subset f^{-1}(f(f^{-1}(B)))$. Indeed,  for each subset $Y \subset F$ we have that $Y\subset f^{-1}(f(Y))$.
It remains the other inclusion. Take $x\in f^{-1}(f(f^{-1}(B)))$, then $f(x)\in f(f^{-1}(B))$ so $\exists x'\in f^{-1}(B)$ such that $f(x)=f(x')$. So $f(x')\in B$ thus $f(x)\in B$ which means that $x\in f^{-1}(B)$. 
Is my reasoning correct ? thanks for your help!
 A: I will try to give you a general viewpoint of this phenomenon. Please take the time to read each paragraph carefully if anything is too new to you. There are some words and comments you shouldn't care about now, but might be useful to others, or for future reference. 
Consider a function $f:A\to B$. This functions defines two functions $f_\#:\wp(A)\to \wp(B)$ and $f^\#:\wp(B)\to\wp(A)$ that send a subset $S$ of $A$ to $f(S)$, and a subset $T$ of $B$ to $f^{-1}(T)$. Let us denote set inclusion by $\leqslant$, that is, $S\subseteq S'$ will be written $S\leqslant S'$. 
Note that if $S\leqslant S'$ in $A$, then $f_\#(S)\leqslant f_\#(S')$, and if $T\subseteq T'$ in $B$, then $f^\#(T)\leqslant f^\#(T')$ in $A$, that is, both $f_\#$ and $f^\#$ are order preserving. 
There is an additional property: $f(T) \subseteq S$ if and only if $T\subseteq f^{-1}(S)$: indeed, saying that $f(T)$ lies in $S$ means that every element of $T$ is mapped into $S$ by $f$. Saying that $T$ is in $f^{-1}(S)$ is the same as saying that every element of $T$ is sent into $S$ by $f$. 
A partially ordered set or poset is a set $X$ endowed with a relation $\leqslant$ from $X$ to $X$ that is reflexive, transitive and antisymmetric: first, we always have $x\leqslant x$, second, $x\leqslant y$ and $y\leqslant z$ implies $x\leqslant z$, finally, if $x\leqslant y$ and $y\leqslant x$; then forcefully $x=y$. An example of a poset is the powerset $\wp(X)$ of any set $X$ under the relation $A\leqslant B$ if $A\subseteq B$. Note that there might exist elements in a poset that are not comparable. For example, in $\wp(\{1,2\})$, $\{1\}$ and $\{2\}$ are not related. This explains the term partial order. 
Thus we have shown that $f_\#(T)\leqslant S$ if and only if $T\leqslant f^\#(S)$. This shows that any function $f:A\to B$ defines a Galois connection between the posets $\wp(A)$ and $\wp(B)$ of subsets of $A$ and $B$ respectively, ordered by inclusion: a pair of order preserving maps $\Phi:P\to Q$ and $\Psi:Q\to P$ such that $\Phi(p)\leqslant q$ if and only if $p\leqslant \Psi(q)$. Call this last property the adjunction property. 
In more fancy terms, by viewing a poset as a category, $(f_\#,f^\#)$ is an adjoint pair (and this is perhaps the most elementary example of such thing), and is perhaps the first example one encounters! 
Returning to your question, I claim that given any Galois connection $(P,Q,\Phi,\Psi)$ between posets $P,Q$, it is always the case that $\Psi\Phi\Psi(q)=\Psi(q)$ and $\Phi\Psi\Phi(p)=\Phi(p)$ for any $p\in P,q\in Q$. Indeed, since $\Phi(p)\leqslant \Phi(p)$, the adjunction property implies that $p\leqslant \Psi\Phi(p)$,and so applying $\Phi$ we get $\Phi(p)\leqslant \Phi\Psi\Phi(p)$. The same reasoning shows  $\Phi\Psi(q)\leqslant q$ always holds and letting $q=\Phi(p)$ gives the other inequality. The same argument shows that $\Psi\Phi\Psi(q)=\Psi(q)$. 
In conclusion, we have proven that in particular $f_\#f^\#f_\#=f_\#$ and that $f^\#f_\# f^\# =f^\#$, as desired. 
