Fiber of covering map is orbit of deck transformation group Some definitions: Let $\lambda: \mathbb H \to \mathbb C \setminus \{-1, +1 \}$ be the covering map, i.e. $\lambda$ is surjective and for every $z \neq \pm 1$ there exists a neighborhood $V_z \subseteq \mathbb C$ such that $\lambda : U_\alpha \to V_z$ is a biholomorphism for each connected component $U_\alpha$ of $\lambda^{-1} (V_z)$. Let $G$ be the decking transformations, i.e. automorphisms of the upper half plane $T: \mathbb H \to \mathbb H$ satisfying $\lambda(Tz) = \lambda(z)$. Clearly this forms a group under composition. 
I want to show that, for every $z \in \mathbb H$,
\begin{equation}
\lambda^{-1}(\lambda(z)) = \{ Tz : T \in G \}.
\end{equation}
The right hand side is clearly contained in the left, so it remains to show that for every $w \in \mathbb H$ satisfying $\lambda(z) = \lambda(w)$, we have $w = Tz$ for some $T \in G$. 
I was able to prove that any $T \in G$ which is not the identity cannot have a fixed point on the fiber, but this isn't good enough to obtain the result. By the Lindelof covering property of the complex plane, the fiber is at most countable. If it were finite, then maybe some pigeonhole-type argument could be used, but this seems to be a dead end avenue. 
Constructing the map explicitly seems a bit difficult since I'm not too sure how $\lambda$ behaves (I'm following Marshall's Complex Analysis; he constructs the map via an inductive application of the reflection principle), though it seems it's the only avenue left. 
 A: My written up solution for posterity, inspired by the ever so helpful comments by Conrad in the comments: Recall that automorphisms of the upper half plane are Mobius transformations. Let $G$ be the set of automorphisms $T: \mathbb H \to \mathbb H$ satisfying $\lambda(T(z)) = \lambda(z)$, and suppose $T, S \in G$. Then 
        \begin{equation}
 \lambda((T \circ S)(z)) = \lambda(S(z)) = \lambda(z), \qquad \lambda(z) = \lambda((T \circ T^{-1}) (z)) = \lambda(T^{-1} (z)), \end{equation}
    for all $z \in \mathbb H$, i.e. $T \circ S \in G$ and $T^{-1} \in G$. Clearly $G$ contains the identity, so we conclude it is a group under composition. Fix $z \in \mathbb H$, by construction of $G$, we have the inclusion
        \begin{equation} \{ T(z) : T \in G \} \subseteq \lambda^{-1} (\lambda(z)). \end{equation}
    To show equality, we need to show that if $w \in \mathbb H$ satisfies $\lambda(w) = \lambda(z)$, then $T(z) = w$ for some $T \in G$. Since $\lambda$ is a covering map, every $p \in \mathbb C \setminus \{-1, 1\}$ admits a neighborhood $V_p \subseteq \mathbb C$ such that 
        \begin{equation} \lambda: U_i \to V_p \end{equation}
    is a biholomorphism on each connected component $U_i$ of $\lambda^{-1} (V_p)$. For $p = \lambda(z)$, suppose $z \in U_0$ and $w \in U_1$. Denoting $g_0 : V_{\lambda(z)} \to U_1$ the local inverse of $\lambda$, we have
        \begin{equation} \lambda \circ g_0 \circ \lambda = \lambda, \qquad (g_0 \circ \lambda)(z) = w. \end{equation}
    We want to analytically continue $g_0 \circ \lambda: U_0 \to U_1$ to an automorphism on $\mathbb H$. Fix $\zeta \in \mathbb H$ and let $\gamma: [0, 1] \to \mathbb H$ be a continuous curve with initial point $z$ and terminal point $\zeta$. The image $(\lambda \circ \gamma)([0, 1])$ is compact, so the open cover
        \begin{equation} (\lambda \circ \gamma)([0, 1]) \subseteq \bigcup_{t \in [0, 1]} V_{(\lambda \circ \gamma) (t)}\end{equation}
    admits a Lebesgue covering number $\epsilon > 0$. Since $\lambda \circ \gamma$ is uniformly continuous, we can choose a partition of the unit interval $0 = t_0 < \cdots < t_n = 1$ with mesh sufficiently small such that 
        \begin{equation} (\lambda \circ \gamma)([t_i, t_{i + 1}]) \subseteq V_{(\lambda \circ \gamma)(t_i)}. \end{equation}
    For brevity, denote $V_i = V_{(\lambda \circ \gamma)(t_i)}$ and $w_i = (\lambda \circ \gamma)(t_i)$; in particular, $V_0 = V_{\lambda(z)}$. Proceeding inductively, given function element $(g_i \circ \lambda, \lambda^{-1} (V_i))$, we have $w_{i + 1} \in V_i \cap V_{i + 1}$, so $g_i (w_{i + 1}) \in U_{i + 2}$ for some connected component of $\lambda^{-1} (V_{i + 1})$. Denote $g_{i + 1} : V_{i + 1} \to U_{i + 2}$ a local inverse of $\lambda$, then
\begin{equation} (g_{i} \circ \lambda, \lambda^{-1} (V_i)), \qquad (g_{i + 1} \circ \lambda, \lambda^{-1} (V_{i + 1}))  \end{equation}
    the latter is a direct analytic continuation of the former. This proves $g_0 \circ \lambda$ admits an analytic continuation along $\gamma$. Since $\mathbb H$ is simply connected (in fact, contractible), the monodromy theorem tells us that the terminal analytic continuation does not depend on $\gamma$, so by construction there exists $T: \mathbb H\to \mathbb H$ satisfying 
        \begin{equation} \lambda \circ T = \lambda, \qquad T(z) = w. \end{equation}
    It remains to show that $T$ is an automorphism. Arguing similarly to our construction of $T$, we can construct an analytic map $S: \mathbb H \to \mathbb H$ such that $S = (g_0 \circ \lambda)^{-1}$ on $U_1$. In particular, $T \circ S$ is the identity on an open set $U_0$, so by the uniqueness theorem and connectedness of the upper half plane, $(T \circ S)(z) = z$ for all $z \in \mathbb H$. Hence $T \in G$, completing the proof. 
