# Is $f^{-1}(f(A))=A$ always true?

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

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. $$\Box$$

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.

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 $f^{-1}(f(A))=X\neq A$.

• I think you have some typos here, but I don't want to edit your work. – Squirtle Sep 10 '12 at 15:12
• @dustanalysis: Why do you think I have typos here? – Asaf Karagila Sep 10 '12 at 15:26
• Why do you use f(x), do you mean f(X)? – Squirtle Sep 10 '12 at 15:28
• @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 $f(x)=0$. – Asaf Karagila Sep 10 '12 at 15:31