Why $f^{-1}(f(A)) \not= A$ Let $A$ be a subset of the domain of a function $f$.
Why $f^{-1}(f(A)) \not= A$.
I was not able to find a function $f$ which satisfies the above equation.
Can you give an example or hint.
I was asking for an example function which is not addressed 
here
 A: Let $f:\mathbb{R}\to\mathbb{R}$ be given by $f(x)=1$ for $x\geq 0$ and $f(x)=0$ for $x<0$ and $A=[0,1]$, then
$$
f(A)=\{1\}\;\;\Longrightarrow \;\;f^{-1}(f(A))=f^{-1}(\{1\})=[0,\infty)\neq [0,1]=A.
$$
A: First, you mean, for a function $f:X\to Y$ and a subset $A\subset X$, why is it that $f^{-1}(f(A))\ne A$. Note that in order to describe a function you must be clear about its domain and codomain. 
Now, try this: $X=Y=\mathbb N$ and $f:X\to Y$ given by $f(n)=1$. Then for the set $A=\{1\}$ compute $f^{-1}(f(A))$. 
A: Any noninjective function provides a counterexample. To be more specific, let $X$ be any set with at least two elements, $Y$ any nonempty set, $u$ in $X$, $v$ in $Y$, and $f:X\to Y$ defined by $f(x)=v$ for every $x$ in $X$. Then $A=\{u\}\subset X$ is such that $f(A)=\{v\}$ hence $f^{-1}(f(A))=X\ne A$.
In general, for $A\subset X$, $A\subset f^{-1}(f(A))$ but the other inclusion may fail except when $f$ is injective.
Another example: define $f:\mathbb R\to\mathbb R$ by $f(x)=x^2$ for every $x$. Then, $f^{-1}(f(A))=A\cup(-A)$ for every $A\subset\mathbb R$. For example, $A=[1,2]$ yields $f^{-1}(f(A))=[-2,-1]\cup[1,2]$. 
A: If $f:X\longrightarrow Y$ is not injective and $A\subset X$. Then take $x \in X-A$ such that $f(x)=f(a)$ for some $a\in A$ then we have that $x\in f^{-1}(f(A))$ although $x\not \in A$. 
