providing a function with sets

Question

So I'm trying to show $F(S)\cap F(T)⊆F(S\cap T)$ is false.

Let $Y\in F(S)\cap F(T) $ then $ X\in S$ and $X\in T$ with $F(x) =Y$.

So $X\in S\cap T$

But after this I can seem to only prove true and not false.

Answer 1

To show it is false, it suffices to exhibit a counterexample. Let $S=\{1,2\}$ and $T=\{1,3\}$ and consider the function $F\colon \{1,2,3\}\to \{1,2\}$ given by $F(1)=1$ and $F(2)=F(3)=2$.

Then $F(S)=F(T)=\{1,2\}$, whereas $F(S\cap T)=\{1\}$. So $F(S)\cap F(T)=\{1,2\}$ which is not a subset of $\{1\}$. (In fact, the reverse inclusion holds.)