we know that if f is injective then $f(A \cap B) = f(A) \cap f(B)$ for all $A$ and $B$ subsets of $X$ being the domain of $f$. I want to prove that the reciprocal is false, so ideally give an example of a non injective function $f$ for which the equality is held for all $A$ and $B$ subsets of domain $X$ of $f$. Can you give such an example?
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I thought about the absolute value of x on IR, quite hard for the case a=-b – user1478012 Nov 16 '24 at 16:45
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Such an example is not possible. If $f$ is not injective there exists two different elements $a,b\in X$ with $f(a)=f(b)$, but then $$ f(\{a\}\cap\{b\})=f(\emptyset)=\emptyset. $$ While $f(\{a\})\cap f(\{b\})$ is clearly non empty.
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Your reasoning is nice, however I'm looking for f non injective for which the statement is true FOR all A, B subsets. – user1478012 Nov 16 '24 at 17:05
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@zkutch You're right, I think I was confused because for the empty function we always have $f(A\cap B)=f(A)\cap f(B)$, but that doesn't really matters. I'll edit the answer. – Ignacio Henríquez Nov 16 '24 at 17:32
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@user1478012 The point of my answer is to show that what you're looking for does not exist, that is, if $f(A\cap B)=f(A)\cap f(B)$ for every pair of $A,B\subset X$ then $f$ is injective. – Ignacio Henríquez Nov 16 '24 at 17:35
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