Preimage of Cover is Cover

From ProofWiki
Jump to navigation Jump to search

Theorem

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

Let $\mathcal U$ be a cover of $T$.


Then the set:

$\left\{{\phi^{-1} \left({U}\right): U \in \mathcal U}\right\}$

is a cover of $S$.


Proof

Let $x \in S$.

Then $\phi \left({x}\right) \in T$.

Since $\mathcal U$ is a cover of $T$:

$\exists U \in \mathcal U: \phi \left({x}\right) \in U$

By definition of preimage, $x \in \phi^{-1} \left({U}\right)$.

So:

$\forall x \in S: \exists \phi^{-1} \left({U}\right) \in S: x \in \phi^{-1} \left({U}\right)$

That is, $\left\{{\phi^{-1} \left({U}\right): U \in \mathcal U}\right\}$ is a cover of $S$.

$\blacksquare$