Why $f^{-1}(f(A)) \not= A$ [duplicate]

Solution 1:

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 $f^{-1}(f(A))=[-2,-1]\cup[1,2]$.

Solution 2:

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 $$ f(A)=\{1\}\;\;\Longrightarrow \;\;f^{-1}(f(A))=f^{-1}(\{1\})=[0,\infty)\neq [0,1]=A. $$

Solution 3:

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 $f^{-1}(f(A))$.