I have the following problem:
Suppose $p:Y\rightarrow X$ and $p':Y'\rightarrow X$ are covering maps and $\phi:Y\rightarrow Y'$ is a continuous map s.t. $p'\circ \phi=p$. Suppose that $X,Y,Y'$ are connected and $X$ is also locally connected. Show that $\phi$ is a covering map.
Definition of a covering map A continuous map $p:Y\rightarrow X$ is a covering map if forall $x\in X$ there exists an open neighbourhood $N$ of $x$ such that $p^{-1}(N)=\bigcup_\alpha N_\alpha$ where
- $N_\alpha$ are open in $Y$
- $N_\alpha\cap N_\beta=\emptyset$ if $\alpha \neq \beta$
- $p|_{N_\alpha}:N_\alpha \rightarrow N$ is a homeomorphism
Proof
To check that $\phi$ is a covering map I need to show the following two points. First I need to check that $\phi$ is continuous, but this is trivially correct by the exercise. Now I need also to check the statements in the above definition. Therefore let $y'\in Y'$, then $p'(y')\in X$. Now since $p$ is a covering there exists an open neighbourhood $U\in \mathfrak{U}\left (p'(y')\right)$ such that $$p^{-1}(U)=\dot\bigcup_\alpha U_\alpha$$ where $U_\alpha$ are open in $Y$ and $p|_{U_\alpha}:U_\alpha \rightarrow U$ is a homeomorphism. Then clearly $$p'(y')\in U=\dot\bigcup_\alpha p\left(U_\alpha\right)\,\,\,\,\Leftrightarrow \,\,\,\,\,y'\in p'^{-1}\left(\dot\bigcup_\alpha p(U_\alpha)\right)=:N$$ where $N\in \mathfrak{U}(y')$ is open in $Y'$. Furthermore $$\phi^{-1}(N)=p^{-1}\left(p'(N)\right)=\dot\bigcup_\alpha U_\alpha$$ (thus point $1.,2.$ from the above definiton are satisfied). Now I only need to show that $$\phi|_{U_\alpha}:U_\alpha \rightarrow N$$ is a homeomorphism.
So my question is the following. Is the above part correct so? If yes could you give me a hint how to prove the homeomorphism part?
Thanks for your help.
