Problem 5-12 John Lee's Smooth Manifolds. A smooth covering map restricted to a component of the boundary is a smooth covering map onto a component This is a problem I got stuck on while studying John Lee's Introduction to Smooth Manifolds.
Problem 5-12. Suppose $E$ and $M$ are connected smooth manifolds with boundary, and $\pi : E \to M$ is a smooth covering map, i.e. $\pi$ is smooth surjective and each point in $M$ has a neighborhood $U$ such that each component of $\pi^{-1}(U)$ is mapped diffeomorphically onto $U$ by $\pi$. Show that the restriction of $\pi $ to each connected component of $\partial E$ is a smooth covering map onto a component of $\partial M$. 
My attempt so far:
Let $F$ be a connected component of $\partial E$. Let $x \in \pi(F)$, and let $U$ be an evenly covered neighborhood of $x$. Take $e \in F$ such that $\pi(e)=x$. Let $\tilde U$ be the component of $\pi^{-1}(U)$ containing $e$. Then $\pi|_{\tilde U} : \tilde{U} \to U$ is a diffeomorphism by assumption. Since $\pi$ is a local diffeomorphism, $\pi$ maps boundary points to boundary points, so we have $\pi(\tilde{U} \cap F) = U \cap \partial M$. 
But this is not really getting me further. I need to show that first, the image $\pi(F)$ is a connected component of $\partial M$. And then for each $x \in \pi(F)$, there is an evenly covered neighborhood $U$ such that each component of $\pi^{-1}(U) \cap F$ is mapped diffeomorphically onto $U \cap \pi(F)$ by $\pi$. 
How could I prove this? This problem is in a chapter on submanifolds, but I cannot see how I could use theorems on embedded submanifolds  here.
 A: You correctly argue that $\pi$ maps boundary points to boundary points. Similarly $\pi$ maps interior points to interior points and we conclude that $\pi^{-1}(\partial M) = \partial E$. Hence the restriction  $\pi' : \partial E  \to \partial M$ of $\pi$ is a smooth covering map. Note that for any topological covering map $p : X \to Y$ and for any $A \subset Y$ the restriction $p' : p^{-1}(A)  \to A$ of $p$ is a covering map.
Now let $\{F_\iota\}_{\iota \in I}$ be the components of $\partial E$. Since $\partial E$ is a manifold, it is locally connected and hence the $F_\iota$ are open in $\partial E$. We conclude that the $G_\iota = \pi'(F_\iota)$ are connected open subsets of $\partial M$.
We claim that each $G_\iota$ is closed in $\partial M$, i.e. a clopen connected subset of $\partial M$. Such a set is obviously a component.
So let us assume that $G_\iota$ is not closed, i.e. there exists a point $x \in \overline G_\iota \setminus G_\iota$. Since $\partial M$ is locally connected, we find an open connected neigborhood $U$ of $x$ in $\partial M$ which is evenly covered by $\pi'$. Then $(\pi')^{-1}(U)$ is the disjoint union of open $V_\alpha \subset \partial E$ which are mapped by $\pi'$ homeomorphically onto $U$. We have $U \cap G_\iota \ne \emptyset$. Choose $y \in F_\iota$ such that $\pi'(y) \in U \cap G_\iota$. Then $y \in (\pi')^{-1}(U)$ so that $y \in V_\alpha$ for some $\alpha$. But $V_\alpha \approx U$ is connected, thus $F'_\iota = F_\iota \cup V_\alpha$ is connected. Since $F_\iota$ is a component, we infer that $F'_\iota = F_\iota$, i.e. $V_\alpha \subset F_\iota$. This shows $x \in U = \pi'(V_\alpha) \subset \pi'(F_\iota) = G_\iota$ which contradicts the choice of $x$.
Edited:
The proof that $\pi' : \partial E  \to \partial M$ maps components onto components works for any covering map $p : X \to Y$ with a locally connected $X$. Note that in that case also $Y$ must be locally connected since $p$ is a local homeomorphism.
Now let $F$ be a component of $X$ and $G = p(F)$ be the corresponding component of $Y$. Obviously $p_F : F \to G$ is a continuous surjection. To see that it is a covering map, consider $x \in G$. Since $Y$ is locally connected, $G$ is open and we can find a connected open neigborhood $U$ of $x$, $U \subset G$, which is evenly covered by $p$. Then $p^{-1}(U)$ is the disjoint union of open $V_\alpha \subset X$, $\alpha \in J$, which are mapped by $p$ homeomorphically onto $U$. The $V_\alpha$ are connected, thus we have either $V_\alpha \subset F$ or $V_\alpha \cap F = \emptyset$ since $F$ is a component of $X$. We see that
$$p_F^{-1}(U) = p^{-1}(U) \cap F$$
is the disjoint union of sets $V_\alpha$ with $\alpha  \in J'$ for a suitable $J' \subset J$. This shows that $p_F$ is a covering map.
