[proofplan] The monodromy action of $\pi_1(X, x_0)$ on the fibre $p^{-1}(x_0)$ restricts to the orbit map $\ell([\gamma]) = \tilde{x}_0 \cdot [\gamma]$. By [Orbits and Stabilisers of the Monodromy Action](/theorems/1891), this orbit map descends to a bijection between the coset space $\pi_1(X, x_0) / \operatorname{Stab}(\tilde{x}_0)$ and the orbit of $\tilde{x}_0$. Two inputs from the hypothesis "$p$ is a universal cover" pin the picture down: simple connectivity of $\tilde{X}$ kills the stabiliser (by [Injectivity of the Covering Homomorphism](/theorems/1890)), and path-connectedness of $\tilde{X}$ makes the monodromy action transitive. Together these collapse the coset space to $\pi_1(X, x_0)$ itself and the orbit to the whole fibre, yielding the bijection $\ell: \pi_1(X, x_0) \to p^{-1}(x_0)$. [/proofplan] [step:Record the monodromy action and its orbit-stabiliser decomposition] Recall the **monodromy action** of $\pi_1(X, x_0)$ on the fibre $p^{-1}(x_0)$: given $\tilde{z} \in p^{-1}(x_0)$ and a class $[\gamma] \in \pi_1(X, x_0)$ with representative loop $\gamma: I \to X$, let $\tilde{\gamma}_{\tilde{z}}: I \to \tilde{X}$ be the unique lift of $\gamma$ with $\tilde{\gamma}_{\tilde{z}}(0) = \tilde{z}$ (this exists and is unique by the [Path Lifting Lemma](/theorems/1886) and [Uniqueness of Lifts](/theorems/1885)), and set \begin{align*} \tilde{z} \cdot [\gamma] := \tilde{\gamma}_{\tilde{z}}(1). \end{align*} Independence of the choice of representative $\gamma$ is the content of the [Lifted Path Homotopy](/theorems/1888) theorem, and associativity and neutrality of this assignment promote it to a right action of $\pi_1(X, x_0)$ on $p^{-1}(x_0)$. The orbit map at the distinguished fibre point $\tilde{x}_0$ is precisely \begin{align*} \ell: \pi_1(X, x_0) &\to p^{-1}(x_0) \\ [\gamma] &\mapsto \tilde{x}_0 \cdot [\gamma], \end{align*} as given in the theorem statement. [guided] Before we can speak of "the orbit map at $\tilde{x}_0$", we must fix exactly which action we are referring to. Let $[\gamma] \in \pi_1(X, x_0)$ and $\tilde{z} \in p^{-1}(x_0)$ be given, with $\gamma: I \to X$ a representative loop at $x_0$. Because $p: \tilde{X} \to X$ is a covering map and $\gamma(0) = x_0 = p(\tilde{z})$, the [Path Lifting Lemma](/theorems/1886) provides a lift $\tilde{\gamma}_{\tilde{z}}: I \to \tilde{X}$ with $\tilde{\gamma}_{\tilde{z}}(0) = \tilde{z}$ and $p \circ \tilde{\gamma}_{\tilde{z}} = \gamma$. By [Uniqueness of Lifts](/theorems/1885), applied to the connected space $I$, any two such lifts agreeing at one point agree everywhere — so $\tilde{\gamma}_{\tilde{z}}$ is unique. Projecting the endpoint gives $p(\tilde{\gamma}_{\tilde{z}}(1)) = \gamma(1) = x_0$, so the endpoint lies again in $p^{-1}(x_0)$. Define \begin{align*} \tilde{z} \cdot [\gamma] := \tilde{\gamma}_{\tilde{z}}(1) \in p^{-1}(x_0). \end{align*} Why is this well-defined on homotopy classes? If $\gamma \simeq \gamma'$ rel endpoints, the [Lifted Path Homotopy](/theorems/1888) theorem says their unique lifts starting at $\tilde{z}$ are themselves homotopic rel endpoints, so in particular share the same terminal point. The assignment respects concatenation of loops (lifting $\gamma \cdot \gamma'$ with initial point $\tilde{z}$ amounts to lifting $\gamma$, then $\gamma'$ from the new endpoint) and sends the constant loop to $\tilde{z}$ itself. Hence it is a right action of $\pi_1(X, x_0)$ on $p^{-1}(x_0)$. The map $\ell$ of the theorem statement is nothing other than the **orbit map** of this action at the basepoint $\tilde{x}_0$, sending a group element to the image of $\tilde{x}_0$ under the action. [/guided] [/step] [step:Collapse the stabiliser of $\tilde{x}_0$ using simple connectivity of $\tilde{X}$] The stabiliser of $\tilde{x}_0$ under the monodromy action is, by definition, the set of classes $[\gamma] \in \pi_1(X, x_0)$ whose unique lift $\tilde{\gamma}$ starting at $\tilde{x}_0$ is a **loop** at $\tilde{x}_0$. By [Injectivity of the Covering Homomorphism](/theorems/1890), applied to $p$ at the basepoint $\tilde{x}_0$, this stabiliser equals the image $p_*\pi_1(\tilde{X}, \tilde{x}_0)$: \begin{align*} \operatorname{Stab}(\tilde{x}_0) = p_* \pi_1(\tilde{X}, \tilde{x}_0). \end{align*} Since $p$ is a universal cover, $\tilde{X}$ is simply connected by definition, so $\pi_1(\tilde{X}, \tilde{x}_0) = \{e\}$. Hence \begin{align*} \operatorname{Stab}(\tilde{x}_0) = p_*\{e\} = \{e\} \subseteq \pi_1(X, x_0). \end{align*} [guided] We compute the stabiliser of $\tilde{x}_0$ under the action constructed in the previous step. By definition, \begin{align*} \operatorname{Stab}(\tilde{x}_0) = \{[\gamma] \in \pi_1(X, x_0) : \tilde{x}_0 \cdot [\gamma] = \tilde{x}_0\} = \{[\gamma] : \tilde{\gamma}(1) = \tilde{x}_0\}, \end{align*} where $\tilde{\gamma}$ denotes the lift of $\gamma$ starting at $\tilde{x}_0$. A lift $\tilde{\gamma}$ satisfies $\tilde{\gamma}(0) = \tilde{\gamma}(1) = \tilde{x}_0$ precisely when it is a loop in $\tilde{X}$ at $\tilde{x}_0$. So the stabiliser consists of those $[\gamma]$ whose lift from $\tilde{x}_0$ is a loop — which is exactly the image of $\pi_1(\tilde{X}, \tilde{x}_0)$ under $p_*$. This identification is the [Injectivity of the Covering Homomorphism](/theorems/1890), which states both that $p_*: \pi_1(\tilde{X}, \tilde{x}_0) \to \pi_1(X, x_0)$ is injective and that its image is characterised in exactly this way. We conclude \begin{align*} \operatorname{Stab}(\tilde{x}_0) = p_* \pi_1(\tilde{X}, \tilde{x}_0). \end{align*} Why does simple connectivity of $\tilde{X}$ help? By definition of a **universal cover**, the total space $\tilde{X}$ is simply connected, meaning $\pi_1(\tilde{X}, \tilde{x}_0) = \{e\}$ (see [Characterisation of Simply Connected Spaces](/theorems/1883)). Since $p_*$ is a group homomorphism it sends the identity to the identity, so $p_* \pi_1(\tilde{X}, \tilde{x}_0) = \{e\}$. Combining, \begin{align*} \operatorname{Stab}(\tilde{x}_0) = \{e\}, \end{align*} i.e. only the identity class of $\pi_1(X, x_0)$ fixes $\tilde{x}_0$. Geometrically: the *only* loops in $X$ whose lift closes up in $\tilde{X}$ are those already null-homotopic in $X$. [/guided] [/step] [step:Make the action transitive using path-connectedness of $\tilde{X}$] The orbit of $\tilde{x}_0$ under the monodromy action is the set $\{\tilde{x}_0 \cdot [\gamma] : [\gamma] \in \pi_1(X, x_0)\} = \operatorname{image}(\ell) \subseteq p^{-1}(x_0)$. We show it is all of $p^{-1}(x_0)$. Let $\tilde{z} \in p^{-1}(x_0)$ be arbitrary. Since $p$ is a universal cover, $\tilde{X}$ is path-connected, so there is a path $\tilde{\gamma}: I \to \tilde{X}$ with $\tilde{\gamma}(0) = \tilde{x}_0$ and $\tilde{\gamma}(1) = \tilde{z}$. Set $\gamma := p \circ \tilde{\gamma}: I \to X$. Then \begin{align*} \gamma(0) = p(\tilde{x}_0) = x_0, \qquad \gamma(1) = p(\tilde{z}) = x_0, \end{align*} so $\gamma$ is a loop in $X$ at $x_0$ and $[\gamma] \in \pi_1(X, x_0)$. By construction, $\tilde{\gamma}$ is a lift of $\gamma$ starting at $\tilde{x}_0$; by [Uniqueness of Lifts](/theorems/1885) it is **the** lift starting at $\tilde{x}_0$, so $\tilde{x}_0 \cdot [\gamma] = \tilde{\gamma}(1) = \tilde{z}$. Hence $\tilde{z} \in \operatorname{image}(\ell)$, and the action is transitive on the fibre. [guided] To show the orbit of $\tilde{x}_0$ is the entire fibre, we must produce, for each $\tilde{z} \in p^{-1}(x_0)$, a class $[\gamma] \in \pi_1(X, x_0)$ with $\tilde{x}_0 \cdot [\gamma] = \tilde{z}$. The universal-cover hypothesis is consumed here: one of the ingredients in "universal cover" is that $\tilde{X}$ is path-connected (not only connected). Fix $\tilde{z} \in p^{-1}(x_0)$. By path-connectedness of $\tilde{X}$, pick a continuous path \begin{align*} \tilde{\gamma}: I \to \tilde{X}, \quad \tilde{\gamma}(0) = \tilde{x}_0, \quad \tilde{\gamma}(1) = \tilde{z}. \end{align*} Project it: set $\gamma := p \circ \tilde{\gamma}: I \to X$. Since $\tilde{\gamma}(0), \tilde{\gamma}(1) \in p^{-1}(x_0)$, we have $\gamma(0) = \gamma(1) = x_0$, so $\gamma$ is a loop in $X$ based at $x_0$, giving $[\gamma] \in \pi_1(X, x_0)$. Now identify this candidate: $\tilde{\gamma}$ is *by construction* a lift of $\gamma$ to $\tilde{X}$ starting at $\tilde{x}_0$. By [Uniqueness of Lifts](/theorems/1885), applied to the connected space $I$, this is the unique such lift. So the monodromy action gives \begin{align*} \tilde{x}_0 \cdot [\gamma] = \tilde{\gamma}(1) = \tilde{z}, \end{align*} exhibiting $\tilde{z}$ as $\ell([\gamma])$. Because $\tilde{z}$ was arbitrary, $\ell$ is surjective. [/guided] [/step] [step:Assemble the orbit-stabiliser bijection to conclude] We combine the previous two steps with the orbit-stabiliser principle for group actions. By [Orbits and Stabilisers of the Monodromy Action](/theorems/1891), the orbit map $\ell$ factors through the coset space as a bijection \begin{align*} \pi_1(X, x_0) / \operatorname{Stab}(\tilde{x}_0) \xrightarrow{\sim} \operatorname{orbit}(\tilde{x}_0) \subseteq p^{-1}(x_0). \end{align*} From Step 2, $\operatorname{Stab}(\tilde{x}_0) = \{e\}$, so the quotient collapses: $\pi_1(X, x_0)/\{e\} = \pi_1(X, x_0)$. From Step 3, $\operatorname{orbit}(\tilde{x}_0) = p^{-1}(x_0)$. Substituting, the orbit map becomes \begin{align*} \ell: \pi_1(X, x_0) \xrightarrow{\sim} p^{-1}(x_0), \qquad [\gamma] \mapsto \tilde{x}_0 \cdot [\gamma], \end{align*} which is a bijection. This completes the proof. [/step]