The restriction fo covering to a component is a covering map onto its image.

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.

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

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 $U$.