elementary set theory - how to prove $f^{-1}(B_1 cap B_2) = f^{-1}(B_1) cap f^{-1}(B_2)$

I am given this equation:
$f^{-1}(B_1 \cap B_2) = f^{-1}(B_1) \cap f^{-1}(B_2)$

I want to prove it: what i did is

I take any $a \in f^{-1}(B_1 \cap B_2)$, then there is $b \in (B_1 \cap B_2)$ so that $f(a)=b$. Because of $b \in (B_1 \cap B_2)$, it is true that $b \in B_1$ and $b \in B_2$, so $a \in f^{-1}(B_1)$ and $a \in f^{-1}(B_2)$.

this means $f^{-1}(B_1 \cap B_2) \subseteq f^{-1}(B_1) \cap f^{-1}(B_2)$.

is it ok?

Answer

Yeah...this can be actually written in this way;

$a\in f^{-1}(B_1\cap B_2)$, means $f(a)\in B_1\cap B_2$ and so $f(a)\in B_1$ and $f(a)\in B_2$. Hence, $a\in f^{-1}(B_1)$ and $a\in f^{-1}(B_2)$