If we have a function $f:X\rightarrow Y$ where $A\subset X$, is it true to say that $f^{-1}(f(A))=A$?
Answer
As noted, the asserted equality is not true.
In general, one inclusion always holds: $$A\subseteq f^{-1}(f(A)).$$
How to see that? Remember that $x\in f^{-1}(B)$ if and only if $f(x)\in B$.
Now, to show $A$ is contained in $f^{-1}\Bigl( f(A)\Bigr)$, let $a\in A$; we need to show that $a\in f^{-1}\Bigl( f(A)\Bigr)$. But this holds if and only if $f(a)\in f(A)$, which holds since $a\in A$ and $f(A) = \{f(x)\mid x\in A\}$.
The other inclusion does not hold in general, but you do have the following:
Proposition. Let $f\colon X\to Y$ be a function. Then $f$ is one to one (injective) if and only if for every $A\subseteq X$, we have $A=f^{-1}(f(A))$.
Proof. Assume first that $f$ is injective, and let $A\subseteq X$. We already know that $A\subseteq f^{-1}(f(A))$, so we only need to show that $f^{-1}(f(A))\subseteq A$. Let $x\in f^{-1}(f(A))$; we want to prove that $x\in A$. That means that $f(x)\in f(A)$, so there exists $a\in A$ such that $f(x)=f(a)$. But since $f$ is one-to-one, this implies $x=a$, so $x\in A$, as desired.
Conversely, assume that for every $A\subseteq X$, $A=f^{-1}(f(A))$. Let $x,x’\in X$ be such that $f(x)=f(x')$. We need to show that $x=x'$. Let $A=\{x\}$; then $f(x')\in f(A)$, so $x'\in f^{-1}(f(A))$. By assumption, $f^{-1}(f(A))=A=\{x\}$, so we can conclude that $x'\in \{x\}$; but this means $x'=x$, which is what we needed to prove. $\Box$
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