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Let $f\colon X\to Y$ be a function. Prove that if for every $A\subseteq X$, we have $A=f^{-1}(f(A))$ then $f$ is one to one (injective)

My try: Let $x,x’\in X$ be such that $f(x)=f(x')$. We need to show that $x=x'$. So we have $$f^{-1}(f(x)) = f^{-1}(f(x'))$$According to the assumption we have $$x = x'$$I don't know whether my proof is correct and rigorous.

S.H.W
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1 Answers1

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It would be correct if $f$ had an inverse, but you are not assuming that.

You can say that, if $A=\{x\}$, then\begin{align}\{x\}&=A\\&=f^{-1}\bigl(f(A)\bigr)\\&=f^{-1}\bigl(\bigl\{f(x)\bigr\}\bigr)\\&\supset\{x,x'\}\end{align}and that therefore $x'=x$.

José Carlos Santos
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  • Thanks. Why my proof requires that $f^{-1}$ exists? – S.H.W Feb 05 '21 at 11:27
  • Because you wrote $f^{-1}\bigl(f(x)\bigr)$. This notation implies that $f^{-1}$ is the inverse of $f$. – José Carlos Santos Feb 05 '21 at 11:29
  • So my definition of $f^{-1}$ is different from yours? Also I couldn't understand how you got $f^{-1}\bigl(f(A)\bigr)=f^{-1}\bigl({x}\bigr)\supset{x,x'}$. – S.H.W Feb 05 '21 at 12:47
  • It was a typo; I meant $f^{-1}\bigl(\bigl{f(x)\bigr}\bigr)$. I've edited my answer. – José Carlos Santos Feb 05 '21 at 12:52
  • Thanks. I accepted your answer but there are some confusions still. What's the difference between my definition of $f^{-1}$ and yours? Why we have $f^{-1}\bigl(\bigl{f(x)\bigr}\bigr)\supset{x,x'}$? – S.H.W Feb 05 '21 at 13:10
  • My definition of $f^{-1}(A)$, with $A\subset Y$, is ${x\in X\mid f(x)\in A}$. So, since $f(x)=f(x')$, both $x$ and $x'$ belong to $f^{-1}\bigl(\bigl{f(x)\bigr}\bigr)$, and therefore ${x,x'}\subset f^{-1}\bigl(\bigl{f(x)\bigr}\bigr)$. How do you define $f^{-1}$? – José Carlos Santos Feb 05 '21 at 13:13
  • I see. So when we write $f^{-1}(a)$ where $a$ is an element not a set, we are assuming that inverse function exists, right? – S.H.W Feb 05 '21 at 14:08
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    Yes, although for some authors $f^{-1}(a)$ is a way of writing $f^{-1}\bigl({a}\bigr)$. – José Carlos Santos Feb 05 '21 at 14:35