Functions and sets. Show that if A is a subset of X, then $f^{-1}(f(A))\supseteq A$. Find an example where $f^{-1}(f(A))\neq A$. I have proved that if $A$ is subset of $X$, then $f^{-1}(f(A))\supseteq A$ for any function $f:X\rightarrow Y$. However, I can't find an example where $f^{-1}(f(A))\neq A$.
 A: You can find such an example by taking a non injective function, e.g. $f : \mathbb R \to \mathbb R^+ : x \mapsto x^2$. If $A = [0, 1]$, then $f(A) = [0, 1]$. But $f^{-1}(A) = [-1, 1]$. And $f^{-1}(f(A)) = f^{-1}([0, 1]) = [-1, 1] \neq [0, 1]$
A: You can take any non-injective function.
A trivial example is $f:\{1,2\}\to \{1,2\}$ defined as $f (1)=f (2)=1$. Then take $A=\{1\} $.
A: As you have proven $A \subset f^{-1}(f(A))$ in order for $A \ne f^{-1}(f(A))$ there must be and $x\in f^{-1}(f(A))$ so that $x \not \in A$.  
But $x \in f^{-1}(f(A))$ means $f(x) \in f(A)$ which means there is an $y \in A$ so that $f(y) = f(x)$.  But we know $y\ne x$ because $x \not \in A$.  That's not unusual.  That just means $f$ isn't injective.
If we take any old non-injective function $f$, say $f(x) = x^2$ and any pair of $x,y$ so that $f(x) = f(y); x \ne y$, say $x = -1$ and $y=1$.... then we just take any set $A$ so that $y\in A$ and $x \not \in A$.... say $A= \{1\}$. 
The $A= \{1\}$, $f(A) = \{f(w)|w\in A\} = \{f(1)\} = 1$.  And $f^{-1}(f(A)) = \{y| f(y) \in f(A)\} =\{y|f(y) \in \{1\}\} = \{1, -1\}$.
And that's a valid example..
Or $A = \{-3,3, 1, 7\}$.  Or $B = \mathbb R^+$ or $C = \mathbb R\setminus \{-1\}$.
Then $f(A) = \{9,1,49\}; f^{-1}(f(A)) = \{\pm 3, \pm 1, \pm 49\}$.
$f(B) = \mathbb R^+; f^{-1}(f(B)) = \mathbb R$, $f(C) = \mathbb R^+; f^{-1}(f(C)) = \mathbb R^+$. 
Etc.
