Consider the group of covering transformations of the (based) covering map $p:(E,e_0)\to (B,b_0)$. On p. 488 Munkres writes that a covering transformation $h$ is uniquely determined by its value at $e_0$, but I don't see any explanation of this. Why is this so?
The covering transformation is determined by its value at the base point
algebraic-topologycovering-spacesgeneral-topology
Related Solutions
Here are some hints.
The order of the deck group is a divisor of the degree of the cover. (This statement assumes the deck group is a finite group and the covering is a finite sheeted cover, which is the case in these examples.)
If a covering $p:(E,e_0) \to (B,b_0)$ is regular (also called normal or Galois in some texts), then a based loop in $(B,b_0)$ that lifts to a based loop in $(E,e_0)$ will also lift to a based loop in $(E,e_1)$ for any choice of basepoint $e_1 \in p^{-1}(b_0)$. See if this is the case in your examples.
Now, it's important to understand why #1 and #2 are true. You can probably find the details in Munkres. The theorem (which is a challenging but worthwhile exercise to prove) that I find especially helpful (and will prove #1) is the following:
the deck group $D$ is canonically isomorphic to $N/K$, where $K = \pi_1(E,e_0)$ and $N$ is the normalizer in $G=\pi_(B,b_0)$ of the image of $K$ in $G$, i.e. of $p_*K$. Said in fancier language (and making the "canonical" statement precise), there is a short exact sequence
$$1 \to K \to N \to D \to 1,$$ where the map $N \to D$ is given by path lifting. By "given by path lifting," I mean that if $[\alpha] \in N$ is the homotopy class of an oriented based loop, then we can lift $\alpha$ at $e_0$ to a path with endpoint $e_1$. There is a unique lift $f:(E,e_0) \to (E,e_1)$ of $p$. (Existence of this lift uses the fact that $[\alpha]$ belongs to the normalizer.)
You can find a brisk but enlightening treatment of covering spaces using this short exact sequence as the principal tool for computation in the book Algebraic Topology: A First Course by Greenberg and Harper.
You have to show that for all $\gamma \in \pi_1(B,b_0)$ and all $\chi \in H_0$ one has $\gamma ^{-1} \chi \gamma \in H_0$ (not $\gamma \chi = \chi \gamma$ as you write).
With your notation this means $$[g]^{-1} * [p \circ f] * [g] \in H_0. $$
Let us lift the loop $g^{-1} * (p \circ f) * g$ in $B$ to $E$. Take the unique lift $\tilde g$ of $g$ such that $\tilde g(0) = e_0$. This is a path in $E$, but not necessarily a loop in $E$. Let $e_1 = \tilde g(1)$. The path $$F = \tilde g^{-1} * f * \tilde g$$ is a loop based at $e_1$.
Let $h : E \to E$ be a deck transformation such that $h(e_1) = e_0$. Then $h \circ F$ is a loop based at $e_0$, i.e. $[h \circ F] \in \pi_1(E,e_0)$. We get $$p_*([h \circ F]) = [p \circ h \circ F] = [p \circ F] = [(p \circ \tilde g^{-1}) * (p \circ f) * (p \circ \tilde g)] = [g^{-1} * (p \circ f) * g] \\= [g]^{-1} * [p \circ f] * [g]$$ which means that $$ [g]^{-1} * [p \circ f] * [g] \in H_0 .$$
Best Answer
It is not true for arbitrary covering maps (consider for example the projection $p : B \times F \to B$, where $F$ is discrete). However, if we assume that $E$ is path connected, then we can argue as follows:
In Lemma 54.1 Munkres establishes the unique path lifting property for covering maps. Here is a corollary:
Let $X$ be path connected and $f : X \to B$ be a map. If $\tilde{f}_0, \tilde{f}_1 : X \to E$ are lifts of $f$ such that $\tilde{f}_0(x) = \tilde{f}_1(x)$ for some $x \in X$, then $\tilde{f}_0 = \tilde{f}_1$.
Proof. Let $y \in X$ and $u : [0,1] \to X$ be a path from $x$ to $y$. Then $\tilde{f}_0u$ and $\tilde{f}_1u$ are lifts of the path $fu : [0,1] \to B$ which agree at $t = 0$. Hence they also agree at $t = 1$. This means $\tilde{f}_0(y) = (\tilde{f}_0u)(1) = (\tilde{f}_1u)(1) = \tilde{f}_1(y)$.
Now any covering transformation is a lift of $p : E \to B$. Hence if two covering transformations agree at $e_0$, then they are identical if $E$ is path connected.