Let $p:E\rightarrow B$ be a covering map. Assume that $B$ is connected and locally connected. Show that if $C$ is a component of $E$, then $p|_{C}:C\rightarrow B$ is a covering map.
My attempt
Without loss of generality, by using local connectedness of $B$, for every $x\in B$, we can choose a connected open neighborhood $U_{x}$ of $x$ such that $p^{-1}(U_{x})=\sqcup_{i\in I} V_{i}$ such that $p|_{V_{i}}$ is a homeomorphism for all $i\in I.$ Since $C$ is a component and $U_{x}$ are connected, the $V_{i}s$ will either be contained in $C$ or will have empty intersection with $C$. So, I chose $I'\subset I$ such that $V_{i'}\subset C, \forall i'\in I'.$
This can be done for every point in $B$. However, I do not understand how to obtain surjectivity of $p|_{C}$. I think we need connectedness of $B$ for the surjectivity argument but I am not sure how to proceed.
Best Answer
Let us show that $p(C)$ is open and closed in $B$; this implies $p(C) = B$.
$p(C)$ is open.
Let $x \in p(C)$. Take a connected open neigborhood $U_x$ as in your question. We have $p^{-1}(x) \cap C \ne \emptyset$, thus $p^{-1}(U_x) \cap C \ne \emptyset$. Hence some $V_i$ must intersect $C$. Therefore $V_i \subset C$ and $U_x = p(V_i) \subset p(C)$.
$p(C)$ is closed.
Let $x \notin p(C)$. Take again a connected open neigborhood $U_x$ as in your question. If some $V_i$ intersects $C$, we get $V_i \subset C$ and $x \in U_x = p(V_i) \subset p(C)$, a contradiction. Therefore no $V_i$ intersects $C$ and we conclude $p^{-1}(U_x) \cap C = \emptyset$. But then $U_x \cap p(C) = \emptyset$.