$X,Y$ 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 $f^{-1}(f(A))=A \quad \Longrightarrow f(A\cap B)=f(A)\cap B$?

  • 3
    $\begingroup$ Perhaps you mean $f(A \cap B) = f(A) \cap f(B)$. $\endgroup$ – Robert Israel Apr 24 at 19:42
  • 1
    $\begingroup$ Even with @RobertIsrael 's correction, I'm not convinced this is a true statement. Also, $X,Y$ are sets, not quantities. $\endgroup$ – Don Thousand Apr 24 at 19:44
  • $\begingroup$ 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 $\endgroup$ – Analysis Apr 24 at 19:45
  • $\begingroup$ Maybe this is worth reading: $f(A\cap B)=f(A)\cap f(B)$ $\iff$ $f$ is injective. $\endgroup$ – Minus One-Twelfth Apr 24 at 19:48
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    $\begingroup$ @MinusOne-Twelfth This is probably what OP meant. I would close as a duplicate of that problem $\endgroup$ – Don Thousand Apr 24 at 19:49

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.


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