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If we have a function $f:X\rightarrow Y$ where $A\subset X$, is it true to say that $f^{-1}(f(A))=A$?

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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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No. Not in general. Note that if you take the constant map $x\mapsto 1$ mapping $\mathbb{R}\to\mathbb{R}$ then $f^{-1}(f(\{0\}))=\mathbb{R}$. In fact, the equality you wrote holds true for all subsets of $X$ precisely when $f$ is injective.

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No. This need not be true.

For example, $X=Y=\{0,1\}$, $f(x)=0$ and $A=\{1\}$.

$f(A) = \{0\}$ and $f^{-1}(\{0\})=X$, so $f^{-1}(f(A))=X\neq A$.

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  • $\begingroup$ I think you have some typos here, but I don't want to edit your work. $\endgroup$ – Squirtle Sep 10 '12 at 15:12
  • $\begingroup$ @dustanalysis: Why do you think I have typos here? $\endgroup$ – Asaf Karagila Sep 10 '12 at 15:26
  • $\begingroup$ Why do you use f(x), do you mean f(X)? $\endgroup$ – Squirtle Sep 10 '12 at 15:28
  • $\begingroup$ @dustanalysis: No, but it would be the same thing. When I write that $f(x)=0$ it is a defining property, it means that for every $x\in\operatorname{dom}(f)$ we have that $f(x)=0$. $\endgroup$ – Asaf Karagila Sep 10 '12 at 15:31

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