Exercise 53.3

Proof. Since $p$ is a covering map, for any $b$ with $|p^{-1}(b)|=j$ there exists a neighborhood $U\ni b$ such that $p^{-1}(U)$ is the disjoint union of $j$ open neighborhoods $V_{\alpha }$ homeomorphic to $U$, for each $V_{\alpha }$ contains a unique preimage of $b$, which must be one point since $U,V_{\alpha }$ are homeomorphic. By the same argument, all $x\in U$ are such that $|p^{-1}(x)|=j$. Thus, we can partition $B$ into disjoint open sets $A_j$ where each $x\in A_j$ are such that $p^{-1}(x)=j$ (note that $j$ can be any cardinal since the argument above does not depend on $j$ being finite), and $A_k\ni b_0$ is one of the $A_j$; however, since $B$ is connected, $B=A_k$, for otherwise the $A_j$ will be a separation of $B$. □