Normal covering spaces - equivalent definitions for connected spaces A covering space $p: Y \to X$ is normal when for all $ x \in X$ and for all $x_1', x_2' \in p^{-1}(x)$ there is a deck transformation $\phi$ with $x_2' = \phi(x_1')$.
I am asked to show that when $X$ is connected, then this condition is equivalent to saying that there exists one $x_0 \in X$ so that for all $x_1', x_2' \in p^{-1}(x_0)$, there exists a deck transformation taking $x_1'$ to $x_2'$.
One direction of this is easy, the other I am having trouble with.
I have thought about defining the set $A$ to be the set of points in $X$ for which there exist points in its fibre that aren't mapped to teach other by any deck transformation.
If I can show that $A$ is both open and closed, then I will be done.
However I am not sure how to show this.
Is this the correct approach, and if so, how should I continue?
 A: Say that $x\in X$ has property $(\star)$ if whenever $y_1,y_2\in p^{-1}(x)$ there is a deck transformation $\phi:Y\rightarrow Y$ with $\phi(y_1)=y_2$.
Suppose $x_0\in X$ has $(\star)$. Then any point $x$ contained in a neighbourhood $U\subseteq X$ of $x_0$ over which $p$ is trivial also has $(\star)$. If $V\subset X$ is a second open subset of $X$ over which $p$ is trivial and $U\cap V\neq\emptyset$, then there is a point $x\in V\cap U\subseteq V$ with $(\star)$, so by the above all points of $V$ have $(\star)$.
Now suppose that $U_1,\dots, U_n\subseteq X$ is a finite chain of open subsets such that $1)$ $x_0\in U_1$, $2)$ $U_i\cap U_{i+1}\neq\emptyset$ for each $i=1,\dots,{n-1}$, $3)$ $p$ is trivialisable over each $U_i$. By inducting on the previous observation we see that each point of each $U_i$ has $(\star)$, and in particular each point of $U_n$ has $(\star)$.
The basic idea is apparent. To complete we need to show how any two points of $X$ can be joined by a finite chain of trivialising open sets when it is connected.
For the details let $\mathcal{U}$ be any open covering of $X$. For $V\in\mathcal{U}$ put
$$\mathcal{U}(V)=\{W\in\mathcal{U}\mid \exists\, U_1,\dots,U_n\in\mathcal{U},\, V\cap U_1\neq\emptyset,\;W\cap U_n\neq\emptyset,\;U_i\cap U_{i+1}\neq\emptyset,\;\forall i=1,\dots,n-1\}$$
and write $\widetilde V=\bigcup_{U\in\mathcal{U}(V)}U$. Notice that if $V_1,V_2\in\mathcal{U}$, then $\widetilde V_1\cap\widetilde V_2\neq\emptyset$ if and only if $\mathcal{U}(V_1)=\mathcal{U}(V_2)$ if and only if $\widetilde V_1=\widetilde V_2$. Thus $\{\widetilde V\mid V\in\mathcal{U}\}$ is a covering of $X$ by pairwise-disjoint clopen sets.
Finally assume that $X$ is connected. We take $\mathcal{U}$ to be any covering of $X$ by open sets which trivialise $p$. The argument above shows that $\{\widetilde V\mid V\in\mathcal{U}\}$ contains the single set $X$. Thus any two points of $X$ are connected by a finite chain of sets in $\mathcal{U}$. Returning to the open paragraphs we see that if any point $x_0\in X$ has property $(\star)$, then so does every other point.
A: Your approach is correct, but as far as I can see yo need further assumptions on $X$.
Call $x  \in X$ a normal point of $p$ if for all $y_1, y_2 \in p^{-1}(x)$ there is a deck transformation $\phi$ with $y_2 = \phi(y_1)$. Let us first prove the following
Lemma. Let $U$ be an evenly covered connected open subset of $X$. If some $\xi \in U$ is a normal point of $p$, then all $x \in U$ are normal points of $p$.
$p^{-1}(U)$ is the disjoint union of open $V_\alpha \subset Y$ which are mapped by $p$ homeomorphically onto $U$ ("sheet decomposition of $p^{-1}(U)$"). The $V_\alpha$ are the connected components of $p^{-1}(U)$. Let $x \in U$ and $y_i \in p^{-1}(x)$. There are unique $\alpha_i$ such that $y_i \in V_{\alpha_i}$. Let $\eta_i \in p^{-1}(\xi)$ be the unique point contained in $V_{\alpha_i}$. There exists a deck transformation $\phi$ such that $\eta_2 = \phi(\eta_1)$. The set $\phi(V_{\alpha_1})$ is a connected component of $p^{-1}(U)$ such that $\eta_2 = \phi(\eta_1) \in \phi(V_{\alpha_1})$. Thus $\phi(V_{\alpha_1}) = V_{\alpha_2}$. Therefore $y_2 = \phi(y_1)$.
Why do we need the connectedness of $U$? In the non-connected case the sheet decomposition of $p^{-1}(U)$ is not unique (see Covering projections: What are the sheets over an evenly covered set?), thus the sheet decomposition $\{\phi(V_\alpha) \}$ of $p^{-1}(U)$ may  differ from $\{V_\alpha \}$ and we cannot conclude that $\phi(V_{\alpha_1}) = V_{\alpha_2}$. Hence we cannot be sure that $y_2 = \phi(y_1)$. Of course there may exist a deck transformation $\phi'$ such that $y_2 = \phi'(y_1)$, but there is no general strategy to find it (and maybe it is different from $\phi$).
You might argue that $p^{-1}(U) \approx U \times F$ with a discrete $F$, thus certainly all $x \in U$ are normal points of the trivial covering $p_U : p^{-1}(U) \to U$. That is, for all $x \in U$ and all $y_i \in p^{-1}(x)$ there exists a deck tranformation $\phi_U$ for $p_U$ with $y_2 = \phi_U(y_1)$. But there is no reason to assume that $\phi_U$ extends to deck transformation for $p$.
Now let us assume that $X$ is locally connected.
Let $N$ denote the set of normal points of $p$. Since each $x \in X$ has an evenly covered connected open neigborhood, the above lemma shows that $N$ and $X \setminus N$ are open in $X$. Thus $N = X$.
