Why the assumption of locally-connected, while defining a monodromy group?

Question

The Wikipedia page on monodromy groups start with an assumption that the base space $X$ of the covering map $p: E \longrightarrow X $ is connected and locally-connected. However, I am unable to understand the reason for the assumption of locally-connectedness here.

The only reason I could suspect, is that it somehow becomes important while proving that the action of the fundamental group on the fibre is well-defined. However, I have failed to come up with a proper argument to prove the well-definedness either. Specifically, I understand that the homotopy lifting does give us a homotopy between the liftings of two homotopic loops, but I do not see why the endpoints of the lifts are the same, in that case.

So can someone please help me with any of these questions?

1. Why the assumption of locally-connected?
2. How to prove the well-definedness?

Moreover, if anyone could point to references for definition of monodromy group and its properties (some of the books I checked seemed to explain it away in a sort of hand-wavy style, and preferred to focus on its applications), I would really appreciate it!

Edit: I think I found an answer to the second question myself. I might update that later as an answer, probably. However, if someone could point me on why local-connectedness is relevant, that would be great!

Answer 1

There is no reason to assume that the base space is connected and locally connected. For any covering $p : E \to X$ we get the monodromy action of $\pi_1(X,x)$ on the fiber $F_x = p^{-1}(x)$.

In fact, for each $y \in F_x$ and each loop $u : I \to X$ based at $x$ and we can lift $u$ to a path $u_y : I \to E$ such that $u_y(0) = y$ and define $y \cdot u = u_y(1)$. It is an easy exercise to show that $y \cdot u = y \cdot v$ if $[u] = [v]$ in $\pi_1(X,x)$. Thus $y \cdot [u] = u_y(1)$ is well-defined. This construction gives a map $$\mu : F_x \times \pi_1(X,x) \to F_x$$ which is easily seen to be a (right) group action. This means that $y \cdot e = y$ and $y \cdot (g \cdot h) = (y \cdot g) \cdot h$.

Hence for each $g \in \pi_1(X,x)$ we get a bijection $$\mu_g : F_X \to F_x, \mu_g(y) = y \cdot g .$$ Thus we get a map $$\bar \mu : \pi_1(X,x) \to \operatorname{Aut}(F_x), \bar \mu(g)(y) = \mu_g(y) = y \cdot g .$$ Here $\operatorname{Aut}(F_x)$ is the set of bijections ("automorphisms") on $F_x$. It becomes a group via composition of functions $\phi \circ \psi$.

Unfortunately $\bar \mu$ is in general no group homomorphism. In fact, we have $\bar \mu (g \cdot h) = \bar \mu (h) \circ \bar \mu (g) $. The reason for this phenomenon is this: The composition $u \cdot v$ of paths $u, v$ is defined by first taking $u$ and then $v$, but in the composition of functions $\phi \circ \psi$ we first take $\psi$ and then $\phi$.

One can resolve this by endowing $\pi_1(X,x)$ with the group multiplication $g * h = h \cdot g$. Alternatively one could work with the left group action $$\pi_1(X,x) \times F_x \to F_x, (g,y) \mapsto y \cdot g^{-1} .$$

Anyway, the image of $\bar \mu$ is a subgroup of $\operatorname{Aut}(F_x)$ which is called the monodromy group.

I recommend to have a look at Recovering a covering space via monodromy representation. Here you can see that under certain conditions the monodromy action allows to reconstruct the covering. For general coverings without any niceness conditions this is impossible.