Classification of covering spaces for spaces that are not locally path connected: counterexamples?

The standard theory treats the case where the base space $$B$$ is path connected, locally path connected, and semi-locally simply connected. While being path connected and semi-locally simply connected is necessary to have a universal covering (which by definitions is just a covering with simply connected total space), the condition local path connectedness is not so natural. So I'd like to see counterexamples for the following:

1. Let $$B$$ be path connected and semi-locally simply connected. Then $$B$$ not necessarily has a universal covering.
2. Let $$B$$ be path connected and semi-locally simply connected and have a universal covering. Then $$B$$ is not necessarily locally path connected.
3. Let $$B$$ be path connected and semi-locally simply connected and have a universal covering $$p:E\to B$$. Do we still have the usual theory that connected coverings of $$B$$ correspond to subgroups of $$\pi_1(B)$$? In particular, is the group of deck transformations of $$E$$ isomorphic to the group $$\pi_1(B)$$?

• There is an alternative (and in my opiniuon better) definition of a universal covering. Consider all connected coverings $p: X' \to X$ of a connected X (connected covering means that $X'$ is connected). A map from $p_1$ to $p_2$ is a map $f : X'_1 \to X'_2$ such that $p_2 \circ f = p_1$. Then call $p_u$ a universal covering if for each $p$ there exists a map $f : p_u \to p$. Then it is a theorem that a simply connected covering of a connected locally path connected $X$ is a universal covering. – Paul Frost Jan 6 at 9:21

1. Let $$X$$ be any path connected simply connected space which is not locally connected (which implies that it is not locally path connected). As an example take the Warsaw circle (see https://de.wikipedia.org/wiki/Datei:Warsaw_Circle.png, https://en.wikipedia.org/wiki/Shape_theory_(mathematics), How to show Warsaw circle is non-contractible?). Then $$id : X \to X$$ is a universal covering.
• Interesting question, but I do not know the answer although I guess there are no non-trivial coverings of $X$. Perhaps you should ask an additional question. – Paul Frost Jan 7 at 9:40
• By the way, $id$ is a universal covering in the sense of my comment to your question if and only if $X$ does not have nontrivial coverings. To see this, assume that $id$ is universal and let $p$ any covering of $X$. Then the map $f : id \to p$ is a section of $p$. Now see math.stackexchange.com/q/256951. – Paul Frost Jan 8 at 0:59
1. As in the answer to 2., let $$W$$ be the Warsaw circle which is path connected simply connected. It has a universal covering, $$id : W \to W$$. However, it has infintely many distinct connected coverings, and these cannot be classified by subgroups of $$\pi_1(W) = 0$$. These coverings are obtained by pasting together $$n$$ copies of the closed toplogist's sine curve $$S$$ into a "circular" pattern and mapping this space in the obvious way to $$W$$ by wrapping it $$n$$-times around $$W$$. Another covering is obtained by pasting together infinitely copies of $$S$$ into a "linear" pattern and mapping this space in the obvious way to $$W$$. This is in complete analogy to the coverings $$z^n : S^1 \to S^1$$ and $$e^{2\pi it} : \mathbb{R} \to S^1$$.
Note that all these coverings are not path connected. It therefore potentially makes a difference whether we work with connected coverings or with path connected coverings. For a locally path connected base space $$X$$ this is the same.