Image of Intersection of sects not equal to intersection of images of sets [duplicate]

How to disprove, if $f$ is a function, $f(A \cap B) != f(A) \cap f(B)?$

marked as duplicate by user63181, Dan Rust, Davide Giraudo, Norbert, TZakrevskiyDec 20 '13 at 12:45

Counterexample: Let $f\colon\{1,2\}\rightarrow\{1\}$ be given by $f(1)=1,f(2)=1$ and let $A=\{1\},B=\{2\}$.
To see why this is a counter example, note that $A\cap B=\emptyset$ and so $f(A\cap B)=\emptyset$, but $f(A)\cap f(B)=\{1\}\cap\{1\}=\{1\}$ and so the LHS is not equal to the RHS.
• @user1063185 It's not an equation because equality does not hold, so it's not a well-posed question to ask when it fails. Note that this is more than 'an example'. It is precisely a proof that not for all $A$ and $B$ does $f(A\cap B)= f(A)\cap f(B)$. – Dan Rust Sep 3 '13 at 10:22