How to disprove, if $f$ is a function, $f(A \cap B) != f(A) \cap f(B)?$
$\endgroup$ 01 Answer
$\begingroup$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.
$\endgroup$ 4
