Is a covering space of a completely regular space also completely regular I'm trying to solve a problem in Munkres' Topology book. Let $p: E \rightarrow B$ be a covering map and suppose that $B$ is completely regular (for any closed subset $A$ and disjoint point $a$ there is a continuous function $f$ from $B$ to the closed unit interval $[0,1]$ such that $f(A) = \{0\}$ and $f(a) = 1$). Show that $E$ is also completely regular. 
I can't see how to do this. My main stumbling block seems to be that if $A$ is a closed subset of $E$ and $a$ is a point in $E - A$ it is quite possible for $A$ to intersect $p^{-1}(\{p(a)\})$ so I can't see how any continuous function from $B$ to the closed unit interval could be composed with the covering map $p$ to show complete regularity of $E$. 
Does anyone know if this result is definitely correct and how it can be proved? Thanks for your help.
 A: I know this is an old question, but still.
The problem is indeed how you describe it. However, it is more convenient to use the following definition of complete regularity: $X$ is completely regular iff for any point $x$ and its open neighborhood $U$ there is a function $f$ such that $f(x)=1$ and $f(X-U)=\{0\}$. (You can switch back and forth from the usual definition by taking $C=X-U$ where $C$ is closed.)
So, let $x\in E$ and $U$ its open neighborhood. $p$ being a covering map is open (see Munkres), so that $p(U)$ is an open neighborhood of $y=p(x)$. Let $V$ be another open neighborhood of $y$ evenly covered by $p$: $p^{-1}(V)=\cup_\alpha V_\alpha$, and $\beta$ is such that $x\in V_\beta$. Then, there is a function $f$ such that $f(y)=1$ and $f(B-V\cap p(U))=\{0\}$ (we take $W=V\cap p(U)$ as a neighborhood of $y$ so that it is both in $p(U)$ and evenly covered by $p$). Now, here comes the problem you have. If we simply take $f'=f\circ p$, then, indeed, $f'(x)=1$ and $f'$ is zero outside of $p^{-1}(W)$, but, for example, $f(e)=1$ for all $e\in p^{-1}(y)$ regardless of whether  $e$ is in $U$ or not.
To overcome this problem, we define $f'=f\circ p$ on $V_\beta\cap U$ only, and define it as $0$ otherwise. Now, we have a function that we need, $f(x)=1$ and $f$ equals $0$ outside of $U$, but we still need to show that it is continuous. This is easy.
In fact, $f'=f\circ p$ not only on $V_\beta\cap U$ but also outside of $p^{-1}(V)$ and on $V_\beta - U$, because both functions equal 0. So, they are equal and continuous on the closed set $E-\cup_{\alpha\neq\beta}V_\alpha$. Besides, $f'\equiv 0$ and is continuous on the closed set $E-V_\beta$. The two closed sets cover the whole space $E$, so that, $f'$ is continuous.
In few words, indeed, defining the function in a straightforward way leads to the problem you described, but instead you redefine it the way you need, so that it is still continuous.
