A related question was asked some years ago. Here is Lee's definition :
Suppose $E$ and $X$ are topological spaces. A map $\pi :E \to X$ is called a covering
map if $E$ and $X$ are connected and locally path-connected, $\pi$ is surjective and continuous, and each point $p \in X$ has an open neighborhood $U$ that is evenly covered by $\pi$, meaning that each component of $\pi^{-1}(U)$ is mapped homeomorphically onto $U$ by $\pi$. In this case, $X$ is called the base of the covering, and $E$ is called a covering space of $X$. If $U$ is an evenly covered subset of $X$, the components of $\pi^{-1}(U)$ are called the sheets of the covering over $U$.
The standard definition is this:
A covering map is a continuous surjective map $\pi : E \to X$ such that each point of $X$ has an evenly covered open neighborhood $U$ which means that $\pi^{-1}(U)$ is a disjoint union of open subsets of $E$, each of which is mapped homeomorphically by $\pi$ onto $U$.
Question:
Does it make sense to use Lee's definition for general topological spaces without assuming that they are connected and locally path-connected? Or more generally, for which type of spaces is it equivalent to the standard definition?
Best Answer
Lee's definition is inadequate for general topological spaces. Its generalization would be
Note that each evenly covered $U$ in the sense of Lee must be connected because it is the continuous image of a connected set.
Example 1 (due to ronno). The identity on $\mathbb Q$ is not a covering map in the sense of Lee. More generally, the identity map on any space $X$ having a point $q$ which does not have a connected open neighborhood is not a covering map in the sense of Lee.
Example 2. The projection $$\pi : \mathbb R \times \mathbb Q \to \mathbb R$$ is a covering map in the sense of Lee. This does not make much sense because the fibers $\pi^{-1}(x) = \{ x\} \times \mathbb Q$ are not discrete.
The problem with the above definition is that the components of $\pi^{-1}(U)$ are not necessarily open in $\pi^{-1}(U)$ which prevents $\pi$ from being a covering map in the standard sense.
Lemma. Let $\pi : E \to X$ be a map with a locally connected $E$. Then the open subsets of $X$ which are evenly covered in the sense of Lee agree with the connected open subsets of $X$ which are evenly covered in the standard sense.
Proof. Let $U \subset X$ be open and evenly covered in the sense of Lee. We know that $U$ must be connected. It is moreover well-known that components of open sets in a locally connected space $E$ are open in $E$. See here. Hence the components $V_\alpha$ of $\pi^{-1}(U)$ are open in $E$. This implies that $U$ is the disjoint union of open subsets $V_\alpha$ of $E$, each of which is mapped homeomorphically by $\pi$ onto $U$. Therefore $U$ is a connected open subset of $X$ which is evenly covered in the standard sense.
Conversely, let $U \subset X$ be connected, open and evenly covered in the standard sense. We know that $\pi^{-1}(U)$ is a disjoint union of open subsets $V_\alpha$ of $E$, each of which is mapped homeomorphically by $\pi$ onto $U$. Each $V_\alpha$ is connected, thus contained in a component $C(V_\alpha)$ of $\pi^{-1}(U)$. We claim that $V_\alpha = C(V_\alpha)$ which shows that $U$ is evenly covered in the sense of Lee. Let $V^*_\alpha = \bigcup_{\alpha' \ne \alpha} V_{\alpha'}$. This is an open subset of $E$ and we have $C(V_\alpha) = (C(V_\alpha)\cap V_\alpha) \cup (C(V_\alpha) \cap V^*_\alpha)$. The sets on the RHS are disjoint open subsets of $C(V_\alpha)$. Since $C(V_\alpha)\cap V_\alpha = V_\alpha \ne \emptyset$, we conclude that $C(V_\alpha) \cap V^*_\alpha = \emptyset$. This proves $V_\alpha = C(V_\alpha)\cap V_\alpha = C(V_\alpha)$, i.e. our claim is true.
Corollary. Let $\pi : E \to X$ be a map with a locally connected $E$. Then $\pi$ is a covering map in the sense of Lee if and only if it is a covering map in the standard sense.
Proof. By the above Lemma a covering map in the sense of Lee trivially is a covering map in the standard sense.
Conversely, let $\pi : E \to X$ be a covering map in the standard sense. By the above Lemma it suffices to show that each $q \in X$ has a connected open neighborhood in $X$ which is evenly covered in the standard sense. We know that $q$ has an open neighborhood $U'$ in $X$ which is evenly covered in the standard sense. Since $\pi$ is a local homeomorphism, $X$ is locally connected. Thus $U'$ contains a connected open neighborhood $U$ of in $X$. As an open subset of $U'$, the set $U$ is trivially evenly covered in the standard sense.
Remark 1. Covering maps $\pi : E \to X$ in the standard sense are local homeomorphisms, hence $E$ is locally connected if and only $X$ is locally connected. For covering maps in the sense of Lee this is not true. The above Corollary shows that the local connectedness of $E$ implies the local connectedness of $X$, but the above example $\pi : \mathbb R \times \mathbb Q \to \mathbb R$ means that the converse is false.
Anyway, as long as all spaces under consideration are locally connected, covering maps in the sense of Lee agree with covering maps in the standard sense. Lee's focus is on manifolds, thus everything works nicely.
Remark 2. For a covering map $\pi : E \to X$ between locally connected spaces each open neighborhood $W$ of a point $p \in X$ contains an evenly covered connected open neighborhood $U$ of $p$. In fact, take any evenly covered open neighborhood $U'$ of $p$. There exists a connected open neighborhood $U$ of $p$ which is contained in $W \cap U'$. As an open subset of $U'$, the set $U$ is trivially evenly covered.
Another variant of defining covering maps could be to consider only connected evenly covered neighborhoods:
The problem with this definition is that for non-locally connected base $X$ it may be impossible to find connected open neighborhoods. Thus there exist less covering maps in the above sense than standard covering maps. Moreover, even if connected open neighborhoods exist, it is not guaranteed that points have arbitrarily small evenly covered open neighborhoods (which is frequently needed in the theory of covering maps).
Note that also covering maps in the above sense are local homeomorphisms, hence again $E$ is locally connected if and only $X$ is locally connected.
Therefore, as long as all spaces under consideration are locally connected, covering maps in the above sense agree with covering maps in the standard sense.