How come $f^{-1}(Q \cap R)$ = $f^{-1}(Q) \cap f^{-1}(R)$ is true? $f(S\cap T) \neq f(S) \cap f(T)$
but 
$f^{-1}(Q \cap R)=f^{-1}(Q) \cap f^{-1}(R)$
Can you explain it in simple terms, so I understand why and develop the intuition to see if a statement is true or false just by looking at it?

Suppose $x\in f^{-1}(Q\cap R)$. Then $f(x)\in Q\cap R$, and in particular, $f(x)\in Q$ and $f(x)\in R$. Thus $x\in f^{-1}(Q)$ and $x\in f^{-1}(R)$, and we have $x\in f^{-1}(Q)\cap f^{-1}(R)$. This gives us that $f^{-1}(Q\cap R)\subset f^{-1}(Q)\cap f^{-1}(R)$.
On the other hand, suppose $x\in f^{-1}(Q)\cap f^{-1}(R)$. Then $x\in f^{-1}(Q)$ and $x\in f^{-1}(R)$, so $f(x)\in Q$ and $f(x)\in R$, hence $f(x)\in Q\cap R$. This tells us $x\in f^{-1}(Q\cap R)$, therefore $f^{-1}(Q)\cap f^{-1}(R)\subset f^{-1}(Q\cap R)$.
With these two arguments, we have $f^{-1}(Q)\cap f^{-1}(R)= f^{-1}(Q\cap R)$.
