The following Theorem holds about the equivalence of coverings:
Let $B$ be locally path-connected, $X$ and $Y$ be path-connected and
let $p : X → B$ and $q : Y → B$ be coverings, $x_0 ∈ X$, $y_0 ∈ Y$,
$b_0 ∈ B$ and $p(x_0) = q(y_0) = b_0$. Then $p$ and $q$ are equivalent
coverings if and only if $p_∗(π_1(X, x_0)) = q_∗(π_1(Y, y_0))$.
Here we define $p : X → B$ and $q : Y → B$ with $p(x_0) = q(y_0) = b_0$ to be equivalent if there exists a homeomorphism $h: X \to Y$ with $q \circ h = p$ and $h(x_0) = y_0$.
I am looking for a counter-example showing that the theorem is false if we drop the assumption of $B$ being locally path-connected.
I want to use the Warsaw Circle as the base space $B$ that is path-connected and not locally path-connected.
Since the fundamental group of $B$ is trivial, all coverings of it would be equivalent if above theorem was true for non locally path-connected bases.
I am therefore trying to find two non-equivalent coverings of the Warsaw circle.
My initial idea is to use $X=B=$"Warsaw Circle", and to use the identity map for $p$.
For the second covering space $Y$, I constructed a space consisting of two Warsaw circles with opposite ends joined (see picture).
This construction is analogous to the covering $S^1 \to S^1$, $z \mapsto z^2$, where the circle covers itself twice. The map $q:Y \to B$ should then be the analogue of $z \mapsto z^2$, basically a projection down onto the Warsaw circle.
Then since $X$ and $Y$ are not homeomorphic, the coverings can not be equivalent, and hence this would be a counter-example.
However I just realized that the space $Y$ I constructed is not path-connected (for instance the points A and Z in the picture lie in different path components)
Is there any way to fix my example to produce a path-connected counter-example that shows that local path-connectedness is necessary?
Best Answer
Interesting example, despite not quite doing the job. It looks pretty easy to fix. Let's say, in your construction, that $A$ and $Z$ both cover the same point in $B$, the base point $p_0$.
Alter $B$, the Warsaw circle, by attaching a circle to $p_0$; call that $B_1$, which has fundamental group $\mathbb Z$.
Alter $Y$ by attaching two arcs each with one endpoint at $A$ and the other at $Z$, let $Y_1$ be the result. The fundmantal group of $Y_1$ is also infinite cyclic, and the covering map $Y_1 \to B_1$ induces a fundamental group isomorphism whose image in $\mathbb Z$ is $2\mathbb Z$.
Now let's construct the third covering space $Y_2$ by making a more severe alterion of $Y$. In your picture, you have a long arc from $A$ to the "lower" limit point that I'll denote $X_Z$, and a long arc from $Z$ to the "upper" limit point that I'll denote $X_A$. So, the first alteration of $Y$ is to remove those two long arcs, and glue in a long arc from $A$ to $X_A$ and another long arc from $Z$ to $X_Z$. And then, as before, attach two arcs with endpoints $A$ and $Z$. Again $Y_2$ has infinite cyclic fundamental group. And again there is a 2--1 covering map $Y_2 \mapsto B$, and induced homomorphism on fundamental groups is $2\mathbb Z$.
But, the two covering maps $Y_1 \mapsto B$ and $Y_2 \mapsto B$ are not equivalent. And $Y_1,Y_2$ are both path connected.