The preimage of the union is the union of the preimages

up:: Preimage of Function

Let $f:X\to Y$ be a function and subsets $C,D\subset Y$.

Then let $x\in f^{-1}(C\cup D)$ a point in $f$‘s preimage. Then

$$x\in f^{-1}(C\cup D)⟺f(x)\in C\cup D⟺(f(x)\in C)\vee (f(x)\in f(D))$$

By the definition of the preimage of a function, we have that

$$y\in f^{-1}(C\cup D)⟺(x\in f^{-1}(C))\vee (x\in f^{-1}(D))⟺y\in f^{-1}(C)\cup f^{-1}(D)$$

Thus, $f^{-1}(C\cup D)\subseteq f^{-1}(C)\cup f^{-1}(D)$ and $f^{-1}(C\cup D)\supseteq f^{-1}(C)\cup f^{-1}(D)$.

Therefore, $f^{-1}(C\cup D)=f^{-1}(C)\cup f^{-1}(D)$.