Let $f:A\to B$ be a function. Show that $f^{-1}(f(A))=A$

Asked Aug 05 '22 at 01:23

Active Aug 05 '22 at 01:46

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I am taking a basic course on proof and trying some questions on relations and functions I am trying from the very basic please understand my poor proving skills :( I believe I have to use the definition of the image and the inverse image of f to prove

Proof) I want to show $f^{-1}(f(A))=A$

($\Rightarrow$) I want to show $f^{-1}(f(A))\subseteq A$. By definition, $f(A)=\{f(x):x\in A\}$. By definition, $f^{-1}(f(A))=\{x\in A:f(x)\in f(A)\}\subseteq A$

($\Leftarrow$) I want to show $A\subseteq f^{-1}(f(A))$. Pick an arbitrary $x\in A$, I want to show $x\in f^{-1}(f(A))$. $\forall x\in A,\exists!f(x)\in D\subseteq B$ call $D=f(A)$. $f(x)\in f(A)$ by definition, $f^{-1}(f(A))=\{x\in A:f(x)\in f(A)\}$ therefore $x\in f^{-1}(f(A))$ and $A\subseteq f^{-1}(f(A))$

edited Aug 05 '22 at 01:46

Brian Moehring

21,088

asked Aug 05 '22 at 01:23

slowpoke

1

What you want to show is not true. Consider $f : [-1, 1] \to \mathbb{R}$ defined by $f(x) = x^2$. We have $f([0,1]) = [0,1]$. But $f^{-1}(f([0,1])) = f^{-1}([0,1]) = [-1, 1]$. – Charles Hudgins Aug 05 '22 at 01:33
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@CharlesHudgins To match the problem, $A$ would need to be the domain of $f$, so you'd have to have $A = [-1,1]$. – Brian Moehring Aug 05 '22 at 01:35
@BrianMoehring I missed that $A$ had to be the domain. You're right. In that case the statement is true. – Charles Hudgins Aug 05 '22 at 01:37
Thank you so much for editing my question It is my first time using latex I will try harder to make it neat :) – slowpoke Aug 05 '22 at 01:49
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It looks good. Proving that the preimage of the image is the domain is more or less a definition check. There isn't much to it. – CyclotomicField Aug 05 '22 at 01:51
Thank you so much guys! :) – slowpoke Aug 05 '22 at 01:55

Let $f:A\to B$ be a function. Show that $f^{-1}(f(A))=A$

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