$X,Y$ are Quantites and $f:X\rightarrow Y$ a function that is injective.
i have already proven that $f^{-1}(f(A))=A$ when $f$ is injective.
How to prove $f^{-1}(f(A))=A \quad \Longrightarrow f(A\cap B)=f(A)\cap B$?
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3Perhaps you mean $f(A \cap B) = f(A) \cap f(B)$. – Robert Israel Apr 24 '19 at 19:42
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1Even with @RobertIsrael 's correction, I'm not convinced this is a true statement. Also, $X,Y$ are sets, not quantities. – Rushabh Mehta Apr 24 '19 at 19:44
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Hello Don Thousand, i have to show the equivalence of five statements. And I thought that i'd proof the equivalence like this: a => b => c => d => e => a – Apr 24 '19 at 19:45
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Maybe this is worth reading: $f(A\cap B)=f(A)\cap f(B)$ $\iff$ $f$ is injective. – Minus One-Twelfth Apr 24 '19 at 19:48
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1@MinusOne-Twelfth This is probably what OP meant. I would close as a duplicate of that problem – Rushabh Mehta Apr 24 '19 at 19:49
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https://math.stackexchange.com/questions/3198722/showing-equivalences-functions-injective – Apr 24 '19 at 19:49
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Do you have that $f^{-1}(A\cap B)=f^{-1}(A) \cap f^{-1}(B)$ ? – Fareed Abi Farraj Apr 24 '19 at 19:52
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No. I have not. – Apr 24 '19 at 19:55
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Yes, $f^{-1}(f(A))=A$
when f is injective. So what? Prove that
if for all A, $f^{-1}(f(A))=A,$
then f is injective.
Now use the injectiveness of f to show the desired conclusion for all A and B.
The use of variables in your question was disastrous for not being properly quantified.
William Elliot
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