[proofplan] The three claims are logically linked. For (1) we produce a correspondence between orbits of the monodromy action and path components of $\tilde{X}$: given $\tilde{x}, \tilde{x}' \in p^{-1}(x_0)$, the two points lie in the same orbit iff a loop $\gamma$ at $x_0$ lifts from $\tilde{x}$ to $\tilde{x}'$, which by projecting and reversing the path between them in $\tilde{X}$ happens iff $\tilde{x}, \tilde{x}'$ lie in the same path component of $\tilde{X}$. Transitivity of the action then follows from path connectedness of $\tilde{X}$. For (2) we unpack the stabiliser condition: $[\gamma]$ fixes $\tilde{x}_0$ iff the lift of $\gamma$ starting at $\tilde{x}_0$ is a loop in $\tilde{X}$, which is exactly the condition that $[\gamma] \in p_* \pi_1(\tilde{X}, \tilde{x}_0)$. For (3) we combine (1) and (2) via the orbit-stabiliser theorem for right actions. The only subtlety is verifying that the monodromy action is indeed a well-defined right action — this uses the [Lifted Path Homotopy theorem](/theorems/1888) to check that the action respects path homotopy classes. [/proofplan] [step:Record the monodromy action and verify it is well-defined] The monodromy action of $\pi_1(X, x_0)$ on $p^{-1}(x_0)$ is defined by \begin{align*} \Phi: p^{-1}(x_0) \times \pi_1(X, x_0) &\to p^{-1}(x_0), \\ (\tilde{x}, [\gamma]) &\mapsto \tilde{\gamma}_{\tilde{x}}(1), \end{align*} where $\tilde{\gamma}_{\tilde{x}}: I \to \tilde{X}$ is the unique lift of $\gamma$ starting at $\tilde{x}$, produced by the [Homotopy Lifting Lemma](/theorems/1887). We write $\tilde{x} \cdot [\gamma] := \Phi(\tilde{x}, [\gamma])$. *Well-definedness (independence of representative).* Suppose $\gamma \simeq \gamma'$ as loops at $x_0$, with lifts $\tilde{\gamma}_{\tilde{x}}$ and $\tilde{\gamma}'_{\tilde{x}}$ both starting at $\tilde{x}$. By the [Lifted Path Homotopy theorem](/theorems/1888), the two lifts are path-homotopic in $\tilde{X}$; in particular, $\tilde{\gamma}_{\tilde{x}}(1) = \tilde{\gamma}'_{\tilde{x}}(1)$. Hence $\tilde{x} \cdot [\gamma]$ depends only on $[\gamma]$. *Image lies in $p^{-1}(x_0)$.* $p(\tilde{\gamma}_{\tilde{x}}(1)) = \gamma(1) = x_0$. *Right-action axioms.* Identity: $\tilde{x} \cdot [c_{x_0}] = c_{\tilde{x}}(1) = \tilde{x}$, where $c_{\tilde{x}}$ is the constant loop at $\tilde{x}$ (which is the unique lift of $c_{x_0}$ starting at $\tilde{x}$, by uniqueness of lifts). Composition: for loops $\alpha, \beta$ at $x_0$, the lift $\widetilde{\alpha * \beta}_{\tilde{x}}$ is obtained by concatenating the lift $\tilde{\alpha}_{\tilde{x}}$ (starting at $\tilde{x}$, ending at $\tilde{x} \cdot [\alpha]$) with the lift $\tilde{\beta}_{\tilde{x} \cdot [\alpha]}$ (starting at $\tilde{x} \cdot [\alpha]$); the endpoint of the concatenated lift is $(\tilde{x} \cdot [\alpha]) \cdot [\beta]$. Therefore $\tilde{x} \cdot ([\alpha][\beta]) = \tilde{x} \cdot [\alpha * \beta] = (\tilde{x} \cdot [\alpha]) \cdot [\beta]$. [/step] [step:Set up the orbit vs path-component correspondence for part (1)] We prove (1) by showing: \begin{align*} \tilde{x}, \tilde{x}' \in p^{-1}(x_0) \text{ lie in the same orbit} \iff \tilde{x}, \tilde{x}' \text{ lie in the same path component of } \tilde{X}. \end{align*} [claim:Two fibre points are in the same orbit iff they are joined by a path in $\tilde{X}$] For $\tilde{x}, \tilde{x}' \in p^{-1}(x_0)$, there exists $[\gamma] \in \pi_1(X, x_0)$ with $\tilde{x} \cdot [\gamma] = \tilde{x}'$ if and only if there is a continuous path $\tilde{\alpha}: I \to \tilde{X}$ with $\tilde{\alpha}(0) = \tilde{x}$ and $\tilde{\alpha}(1) = \tilde{x}'$. [/claim] [proof] *($\Rightarrow$).* Suppose $\tilde{x} \cdot [\gamma] = \tilde{x}'$ for some $[\gamma] \in \pi_1(X, x_0)$. Let $\tilde{\gamma}_{\tilde{x}}: I \to \tilde{X}$ be the unique lift of $\gamma$ starting at $\tilde{x}$. Then $\tilde{\gamma}_{\tilde{x}}$ is a continuous path in $\tilde{X}$ with $\tilde{\gamma}_{\tilde{x}}(0) = \tilde{x}$ and $\tilde{\gamma}_{\tilde{x}}(1) = \tilde{x} \cdot [\gamma] = \tilde{x}'$. *($\Leftarrow$).* Suppose $\tilde{\alpha}: I \to \tilde{X}$ is a continuous path with $\tilde{\alpha}(0) = \tilde{x}$, $\tilde{\alpha}(1) = \tilde{x}'$. Define $\gamma := p \circ \tilde{\alpha}: I \to X$. Then $\gamma$ is continuous, $\gamma(0) = p(\tilde{x}) = x_0$, and $\gamma(1) = p(\tilde{x}') = x_0$ (since both $\tilde{x}, \tilde{x}' \in p^{-1}(x_0)$). So $\gamma$ is a loop at $x_0$ and $[\gamma] \in \pi_1(X, x_0)$. Moreover $\tilde{\alpha}$ is itself a continuous lift of $\gamma$ with $\tilde{\alpha}(0) = \tilde{x}$, so by uniqueness of path lifts $\tilde{\alpha} = \tilde{\gamma}_{\tilde{x}}$, giving \begin{align*} \tilde{x} \cdot [\gamma] = \tilde{\gamma}_{\tilde{x}}(1) = \tilde{\alpha}(1) = \tilde{x}'. \end{align*} [/proof] [guided] The orbit through $\tilde{x}$ consists of all endpoints of lifted loops starting at $\tilde{x}$. A lifted loop is a continuous path in $\tilde{X}$ starting at $\tilde{x}$ and ending somewhere in the fibre $p^{-1}(x_0)$. Conversely, any continuous path in $\tilde{X}$ between two points of $p^{-1}(x_0)$ projects under $p$ to a loop at $x_0$ in $X$, and then — by uniqueness of path lifts — is itself the lift of that loop starting at the first endpoint. So the two notions coincide exactly: "reachable by lifted loop" $=$ "reachable by continuous path in $\tilde{X}$". The subtle check is that the lifted-path operation does not have to be performed from scratch — an existing path $\tilde{\alpha}$ in $\tilde{X}$ is already the lift of $p \circ \tilde{\alpha}$ starting at $\tilde{\alpha}(0)$, by uniqueness of path lifts starting at a prescribed point. [/guided] [/step] [step:Deduce part (1) — orbits correspond to path components] The claim in Step 2 shows that the orbit of $\tilde{x} \in p^{-1}(x_0)$ under $\pi_1(X, x_0)$ equals \begin{align*} \{\tilde{x}' \in p^{-1}(x_0) : \tilde{x}' \text{ is in the same path component of } \tilde{X} \text{ as } \tilde{x}\} = p^{-1}(x_0) \cap \tilde{P}(\tilde{x}), \end{align*} where $\tilde{P}(\tilde{x})$ denotes the path component of $\tilde{X}$ containing $\tilde{x}$. *Orbits and path components correspond bijectively.* Let $\{\tilde{P}_\alpha\}_{\alpha \in A}$ be the path components of $\tilde{X}$. Each $\tilde{P}_\alpha$ either meets $p^{-1}(x_0)$ or does not; let $A' \subseteq A$ index the components that do meet it. For each $\alpha \in A'$, the intersection $\tilde{P}_\alpha \cap p^{-1}(x_0)$ is a single orbit by the claim of Step 2. Conversely, every orbit is of this form. So the map \begin{align*} A' &\to \{\text{orbits of } \pi_1(X, x_0) \text{ on } p^{-1}(x_0)\}, \\ \alpha &\mapsto \tilde{P}_\alpha \cap p^{-1}(x_0) \end{align*} is a bijection. *Every path component of $\tilde{X}$ meets $p^{-1}(x_0)$.* Given $\tilde{P}_\alpha$, pick $\tilde{z} \in \tilde{P}_\alpha$. Since $X$ is path connected, there is a path $\beta: I \to X$ from $p(\tilde{z})$ to $x_0$. Lifting $\beta$ at $\tilde{z}$ gives a continuous path $\tilde{\beta}: I \to \tilde{X}$ from $\tilde{z}$ to some point $\tilde{z}' \in p^{-1}(x_0)$. Since $\tilde{\beta}$ is continuous, $\tilde{z}$ and $\tilde{z}'$ lie in the same path component, so $\tilde{z}' \in \tilde{P}_\alpha \cap p^{-1}(x_0)$. Hence $A' = A$, and the bijection above is between all path components of $\tilde{X}$ and the orbits. *Transitivity.* The action is transitive iff there is exactly one orbit, iff $A = A'$ consists of a single index, iff $\tilde{X}$ is path connected. [guided] The hard direction is the claim that every path component of $\tilde{X}$ meets the fibre $p^{-1}(x_0)$. This is where path connectedness of $X$ enters: in an arbitrary covering space $\tilde{X}$ one can have path components that project into $X$ but miss the base point $x_0$. Path connectedness of $X$ rules this out — for any point $\tilde{z} \in \tilde{X}$, the path from $p(\tilde{z})$ to $x_0$ in $X$ lifts to a path in $\tilde{X}$ landing in the fibre $p^{-1}(x_0)$, proving that every path component touches the fibre. Without path connectedness of $X$, the correspondence between orbits and path components of $\tilde{X}$ would still hold at the level of path components that meet the fibre, but there could be extra path components in $\tilde{X}$ that project to other path components of $X$ and are invisible to the monodromy action at $x_0$. Path connectedness of $X$ exactly forces $A' = A$. [/guided] [/step] [step:Prove part (2) — the stabiliser is the image subgroup $p_* \pi_1(\tilde{X}, \tilde{x}_0)$] Fix $\tilde{x}_0 \in p^{-1}(x_0)$. We show \begin{align*} \operatorname{Stab}(\tilde{x}_0) = p_* \pi_1(\tilde{X}, \tilde{x}_0) \le \pi_1(X, x_0). \end{align*} *($\supseteq$).* Let $[\tilde{\beta}] \in \pi_1(\tilde{X}, \tilde{x}_0)$, represented by a continuous loop $\tilde{\beta}: I \to \tilde{X}$ at $\tilde{x}_0$. Set $\beta := p \circ \tilde{\beta}$; then $[\beta] = p_*([\tilde{\beta}])$, and $\beta$ is a loop at $x_0$. The path $\tilde{\beta}$ is a continuous lift of $\beta$ starting at $\tilde{x}_0$, so by uniqueness of path lifts $\tilde{\beta}_{\tilde{x}_0} = \tilde{\beta}$, and \begin{align*} \tilde{x}_0 \cdot [\beta] = \tilde{\beta}_{\tilde{x}_0}(1) = \tilde{\beta}(1) = \tilde{x}_0. \end{align*} Hence $[\beta] \in \operatorname{Stab}(\tilde{x}_0)$, so $p_* \pi_1(\tilde{X}, \tilde{x}_0) \subseteq \operatorname{Stab}(\tilde{x}_0)$. *($\subseteq$).* Let $[\gamma] \in \operatorname{Stab}(\tilde{x}_0)$, so $\tilde{x}_0 \cdot [\gamma] = \tilde{x}_0$. Let $\tilde{\gamma}_{\tilde{x}_0}: I \to \tilde{X}$ be the unique lift of $\gamma$ starting at $\tilde{x}_0$. Then $\tilde{\gamma}_{\tilde{x}_0}(0) = \tilde{x}_0$ and $\tilde{\gamma}_{\tilde{x}_0}(1) = \tilde{x}_0 \cdot [\gamma] = \tilde{x}_0$, so $\tilde{\gamma}_{\tilde{x}_0}$ is a loop in $\tilde{X}$ based at $\tilde{x}_0$. Define $[\tilde{\gamma}] := [\tilde{\gamma}_{\tilde{x}_0}] \in \pi_1(\tilde{X}, \tilde{x}_0)$. Then $p_*([\tilde{\gamma}]) = [p \circ \tilde{\gamma}_{\tilde{x}_0}] = [\gamma]$, so $[\gamma] \in p_* \pi_1(\tilde{X}, \tilde{x}_0)$. [guided] The stabiliser condition $\tilde{x}_0 \cdot [\gamma] = \tilde{x}_0$ says: when you lift the loop $\gamma$ starting at $\tilde{x}_0$, you return to $\tilde{x}_0$ at time $1$. Equivalently: the lift $\tilde{\gamma}_{\tilde{x}_0}$ is already a loop in $\tilde{X}$ based at $\tilde{x}_0$, not just a path ending in the fibre. This is exactly the condition to produce an element of $\pi_1(\tilde{X}, \tilde{x}_0)$ that $p_*$ sends to $[\gamma]$. In the other direction, any homotopy class in $\pi_1(\tilde{X}, \tilde{x}_0)$ is represented by some loop $\tilde{\beta}$ at $\tilde{x}_0$, which projects to a loop $\beta$ at $x_0$; the lift of $\beta$ at $\tilde{x}_0$ is $\tilde{\beta}$ itself (by uniqueness of lifts), so $\tilde{x}_0 \cdot [\beta] = \tilde{\beta}(1) = \tilde{x}_0$, showing $[\beta]$ stabilises. The entire content of part (2) is that "lifts of loops at $x_0$ starting at $\tilde{x}_0$ that are also loops at $\tilde{x}_0$" is the same thing as "loops in $\tilde{X}$ at $\tilde{x}_0$, projected to $X$" — and uniqueness of lifts is what identifies these two descriptions. [/guided] [/step] [step:Apply the orbit-stabiliser theorem to prove part (3)] Assume now that $\tilde{X}$ is path connected. By part (1) the action of $\pi_1(X, x_0)$ on $p^{-1}(x_0)$ is transitive. By part (2), the stabiliser of $\tilde{x}_0$ is $H := p_* \pi_1(\tilde{X}, \tilde{x}_0)$, a subgroup of $G := \pi_1(X, x_0)$. The orbit-stabiliser theorem for a transitive right action of a group $G$ on a set $S$ with a chosen basepoint $s_0 \in S$ gives a bijection \begin{align*} H \backslash G &\to S, \\ H g &\mapsto s_0 \cdot g, \end{align*} where $H = \operatorname{Stab}_G(s_0)$ and $H \backslash G$ denotes the set of right cosets. Well-definedness: if $Hg_1 = Hg_2$ then $g_2 = h g_1$ for some $h \in H$, so $s_0 \cdot g_2 = s_0 \cdot h g_1 = (s_0 \cdot h) \cdot g_1 = s_0 \cdot g_1$ since $h$ stabilises $s_0$. Injectivity: if $s_0 \cdot g_1 = s_0 \cdot g_2$ then $s_0 \cdot (g_2 g_1^{-1}) = s_0$ (using $(s_0 \cdot g_2) \cdot g_1^{-1} = s_0 \cdot g_1 \cdot g_1^{-1} = s_0$), so $g_2 g_1^{-1} \in H$ and $H g_1 = H g_2$. Surjectivity: transitivity. Applying this with $G = \pi_1(X, x_0)$, $H = p_* \pi_1(\tilde{X}, \tilde{x}_0)$, $S = p^{-1}(x_0)$, and $s_0 = \tilde{x}_0$, we obtain the bijection \begin{align*} p_* \pi_1(\tilde{X}, \tilde{x}_0) \backslash \pi_1(X, x_0) &\to p^{-1}(x_0), \\ p_* \pi_1(\tilde{X}, \tilde{x}_0) \cdot [\gamma] &\mapsto \tilde{x}_0 \cdot [\gamma]. \end{align*} This proves part (3) and completes the proof of the theorem. [/step]