4
$\begingroup$

I am reading Lee's Introduction to Topological Manifolds. I got stuck on the problem 11-7 on pages 303. The below is the problem.

Prove : If $q: E \rightarrow X$ is a covering map and $A \subseteq X$ is a locally path-connected subset, then the restriction of $q$ to each component of $q^{-1}(A)$ is a covering map onto its image.

I proved that $q^{-1}(A)$ is locally path-connected and so are its components. The hardest thing is to show it is evenly covered. But I can't find a clue. I'd like to know how to construct an evenly covered neiborhood and hints for the proof.

$\endgroup$
  • 4
    $\begingroup$ If $a\in A$, take an evenly covered neighborhood of $a$ for $q$, and intersect it with $A$. That should work! $\endgroup$ – Mariano Suárez-Álvarez Jul 25 '15 at 8:42
2
$\begingroup$

Let $C$ be a component of $q^{-1}(A)$ and let $p:C\to A$ be the restriction of $q$. Now if $x\in A$, intersect the evenly covered neighborhood of $x$ with $A$, and choose a smaller path connected neighborhood $U$ of $x$ relative to $A$. Then $U$ is evenly covered. The path components of $q^{-1}(U)$ are mapped homeomorphically onto $U$. Since $C$ is a path component of $q^{-1}(A)$, the path components of $p^{-1}(U)$ are path components of $q^{-1}(U)$ and are mapped homeomorphically to $U$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.