Solution
= Solution
\b[True.] For each $a\in A$, the definition of <preimage> gives
$$
a\in f^{-1}(Z\cap W)\iff f(a)\in Z\cap W\iff f(a)\in Z\text{ and }f(a)\in W\iff a\in f^{-1}(Z)\cap f^{-1}(W).
$$
Thus $\boxed{f^{-1}(Z\cap W)=f^{-1}(Z)\cap f^{-1}(W)}$. Unlike the corresponding statement for <images of a function>, this <intersection> identity requires no <injectivity>.