Preimage of Cover is Cover

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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 $S$.

$\blacksquare$