Preimage of Cover is Cover

Theorem
Let $\phi: S \to T$ be a mapping between the sets $S$ and $T$.

Let $\UU$ be a cover of $T$.

Then the set:
 * $\set {\map {\phi^{-1} } U: U \in \UU}$

is a cover of $S$.

Proof
Let $x \in S$.

Then $\map \phi x \in T$.

Since $\UU$ is a cover of $T$:
 * $\exists U \in \UU: \map \phi x \in U$

By definition of preimage, $x \in \map {\phi^{-1} } U$.

So:
 * $\forall x \in S: \exists \map {\phi^{-1} } U \in S: x \in \map {\phi^{-1} } U$

That is, $\set {\map {\phi^{-1} } U: U \in \UU}$ is a cover of $S$.