[proofplan] We reduce the unbased classification to the based [Galois correspondence](/theorems/1898) by studying how the associated subgroup $p_*\pi_1(\tilde{X}, \tilde{x}_0)$ depends on the choice of basepoint $\tilde{x}_0 \in p^{-1}(x_0)$. The key computation is that changing basepoint from $\tilde{x}_0$ to $\tilde{x}_0'$ conjugates the associated subgroup by a well-defined element of $\pi_1(X, x_0)$, so the conjugacy class is invariant. This defines an injective map from unbased isomorphism classes to conjugacy classes; surjectivity follows by realising any representative via the based correspondence, and injectivity follows from producing an unbased isomorphism between two based coverings whose subgroups are conjugate. [/proofplan]

[step:Show that changing basepoint conjugates the associated subgroup]Let $p: \tilde{X} \to X$ be a path-connected covering and let $\tilde{x}_0, \tilde{x}_0' \in p^{-1}(x_0)$ be two choices of basepoint. Since $\tilde{X}$ is path connected, choose a path $\tilde{\alpha}: [0, 1] \to \tilde{X}$ with $\tilde{\alpha}(0) = \tilde{x}_0$ and $\tilde{\alpha}(1) = \tilde{x}_0'$, and let $\alpha := p \circ \tilde{\alpha}: [0, 1] \to X$. Since $p(\tilde{\alpha}(0)) = p(\tilde{\alpha}(1)) = x_0$, the projection $\alpha$ is a loop in $X$ based at $x_0$, representing a class $[\alpha] \in \pi_1(X, x_0)$. Denote \begin{align*} H := p_*\pi_1(\tilde{X}, \tilde{x}_0), \qquad H' := p_*\pi_1(\tilde{X}, \tilde{x}_0'). \end{align*} We claim that $H' = [\alpha]^{-1} H [\alpha]$. Let $[\tilde{\gamma}'] \in \pi_1(\tilde{X}, \tilde{x}_0')$, so $\tilde{\gamma}'$ is a loop at $\tilde{x}_0'$. Then $\tilde{\alpha} \cdot \tilde{\gamma}' \cdot \tilde{\alpha}^{-1}$ is a loop at $\tilde{x}_0$, and the change-of-basepoint isomorphism [for the fundamental group](/theorems/???) \begin{align*} \Phi_{\tilde{\alpha}}: \pi_1(\tilde{X}, \tilde{x}_0') &\xrightarrow{\sim} \pi_1(\tilde{X}, \tilde{x}_0), & [\tilde{\gamma}'] &\mapsto [\tilde{\alpha} \cdot \tilde{\gamma}' \cdot \tilde{\alpha}^{-1}] \end{align*} is a group isomorphism. Applying $p_*$: \begin{align*} p_*[\tilde{\alpha} \cdot \tilde{\gamma}' \cdot \tilde{\alpha}^{-1}] = [\alpha] \cdot p_*[\tilde{\gamma}'] \cdot [\alpha]^{-1}. \end{align*} As $[\tilde{\gamma}']$ ranges over $\pi_1(\tilde{X}, \tilde{x}_0')$, the class $p_*[\tilde{\gamma}']$ ranges over $H'$, and the image on the left-hand side ranges over $H$ (by the identity $H = p_*\pi_1(\tilde{X}, \tilde{x}_0) = p_* \Phi_{\tilde{\alpha}}(\pi_1(\tilde{X}, \tilde{x}_0'))$). Therefore $H = [\alpha] H' [\alpha]^{-1}$, i.e., $H' = [\alpha]^{-1} H [\alpha]$, which shows $H$ and $H'$ are conjugate in $\pi_1(X, x_0)$.[/step]

[guided]The computation rests on pushing a change-of-basepoint isomorphism inside $\tilde{X}$ down through $p_*$ to $\pi_1(X, x_0)$. Concretely, the change-of-basepoint isomorphism $\Phi_{\tilde{\alpha}}$ in $\tilde{X}$ conjugates loops at $\tilde{x}_0'$ to loops at $\tilde{x}_0$ by prepending and appending $\tilde{\alpha}$. When we apply $p_*$ (which is a group homomorphism), this conjugation descends to conjugation in $\pi_1(X, x_0)$ by the class $[\alpha] = p_*[\tilde{\alpha}]$ — note that even though $\tilde{\alpha}$ is a path (not a loop) in $\tilde{X}$, its projection $\alpha = p \circ \tilde{\alpha}$ is a loop in $X$ because both endpoints lie in $p^{-1}(x_0)$. So the conjugacy class of the associated subgroup is an invariant of the unbased covering — any two basepoint choices give subgroups that differ by inner automorphism. The element $[\alpha]$ depends on the path $\tilde{\alpha}$ chosen, but different paths give elements differing by an element of $H$ (which is precisely what is invisible to the conjugacy class).[/guided]

[step:Define the unbased classification map into conjugacy classes] By the previous step, each path-connected unbased covering $p: \tilde{X} \to X$ determines a well-defined conjugacy class \begin{align*} [H]_p := \{g H g^{-1} : g \in \pi_1(X, x_0)\} \text{ where } H = p_*\pi_1(\tilde{X}, \tilde{x}_0) \text{ for any } \tilde{x}_0 \in p^{-1}(x_0). \end{align*} To see this is well-defined at the level of isomorphism classes: suppose $\psi: \tilde{X}_1 \to \tilde{X}_2$ is an unbased covering isomorphism ($p_2 \circ \psi = p_1$). Pick $\tilde{x}_1 \in p_1^{-1}(x_0)$ and let $\tilde{x}_2 := \psi(\tilde{x}_1) \in p_2^{-1}(x_0)$. Then $\psi$ becomes a based isomorphism $(\tilde{X}_1, \tilde{x}_1) \to (\tilde{X}_2, \tilde{x}_2)$, so \begin{align*} p_{1*}\pi_1(\tilde{X}_1, \tilde{x}_1) = (p_2 \circ \psi)_*\pi_1(\tilde{X}_1, \tilde{x}_1) = p_{2*}\psi_*\pi_1(\tilde{X}_1, \tilde{x}_1) = p_{2*}\pi_1(\tilde{X}_2, \tilde{x}_2), \end{align*} using that $\psi_*$ is an isomorphism onto $\pi_1(\tilde{X}_2, \tilde{x}_2)$. Hence the subgroups for the two coverings agree for some choice of basepoints and so have the same conjugacy class. This defines a map \begin{align*} \Psi: \{\text{unbased path-connected coverings of } X\}/{\cong} &\to \{\text{subgroups of } \pi_1(X, x_0)\}/\text{conj}, \end{align*} $p \mapsto [H]_p$. [/step]

[step:Prove $\Psi$ is surjective] Let $[H]$ be a conjugacy class of subgroups of $\pi_1(X, x_0)$, represented by some $H \leq \pi_1(X, x_0)$. By the [based Galois correspondence](/theorems/1898), there is a based path-connected covering $p_H: (\tilde{X}_H, \tilde{x}_0) \to (X, x_0)$ with $(p_H)_*\pi_1(\tilde{X}_H, \tilde{x}_0) = H$. Forgetting the basepoint, $p_H: \tilde{X}_H \to X$ is an unbased covering with $\Psi(p_H) = [H]$. Hence $\Psi$ is surjective. [/step]

[step:Prove $\Psi$ is injective via an unbased isomorphism between conjugate-subgroup coverings]Suppose $p_1: \tilde{X}_1 \to X$ and $p_2: \tilde{X}_2 \to X$ are path-connected coverings with $\Psi(p_1) = \Psi(p_2)$. Choose basepoints $\tilde{x}_1 \in p_1^{-1}(x_0)$ and $\tilde{x}_2' \in p_2^{-1}(x_0)$, and let \begin{align*} H_1 := p_{1*}\pi_1(\tilde{X}_1, \tilde{x}_1), \qquad H_2 := p_{2*}\pi_1(\tilde{X}_2, \tilde{x}_2'). \end{align*} By hypothesis, $H_1$ and $H_2$ are conjugate: there exists $g \in \pi_1(X, x_0)$ with $H_2 = g^{-1} H_1 g$. Choose a loop $\alpha$ at $x_0$ representing $g$, and let $\tilde{\alpha}_2: [0, 1] \to \tilde{X}_2$ be the unique lift of $\alpha$ starting at $\tilde{x}_2'$. Set $\tilde{x}_2 := \tilde{\alpha}_2(1) \in p_2^{-1}(x_0)$. By the computation in Step 1 applied to $p_2$ with path $\tilde{\alpha}_2$ connecting $\tilde{x}_2'$ to $\tilde{x}_2$: \begin{align*} p_{2*}\pi_1(\tilde{X}_2, \tilde{x}_2) = [\alpha] \cdot p_{2*}\pi_1(\tilde{X}_2, \tilde{x}_2') \cdot [\alpha]^{-1} = g H_2 g^{-1} = g (g^{-1} H_1 g) g^{-1} = H_1. \end{align*} Now $p_1: (\tilde{X}_1, \tilde{x}_1) \to (X, x_0)$ and $p_2: (\tilde{X}_2, \tilde{x}_2) \to (X, x_0)$ are based coverings with the same associated subgroup $H_1$. By the injectivity half of the [based Galois correspondence](/theorems/1898), there is a based covering isomorphism $\psi: (\tilde{X}_1, \tilde{x}_1) \to (\tilde{X}_2, \tilde{x}_2)$. Forgetting basepoints, $\psi$ is an unbased covering isomorphism, so $p_1$ and $p_2$ define the same unbased isomorphism class. Hence $\Psi$ is injective.[/step]

[guided]We are given that $H_1$ and $H_2$ are conjugate via some $g \in \pi_1(X, x_0)$, but they arise from our initial choices of basepoints $\tilde{x}_1, \tilde{x}_2'$. Conjugacy tells us we have freedom to move basepoints; we exploit this to move $\tilde{x}_2'$ to a new basepoint $\tilde{x}_2$ for which the subgroups literally coincide. The move is to replace $\tilde{x}_2'$ by the endpoint of the lift of a loop representing $g$. By Step 1, moving basepoint along a path $\tilde{\alpha}_2$ conjugates the associated subgroup by $[\alpha] = [p_2 \circ \tilde{\alpha}_2] = g$. Conjugating $H_2 = g^{-1}H_1 g$ by $g$ yields $H_1$, so at the new basepoint $\tilde{x}_2$ the subgroup is exactly $H_1$. Once the subgroups coincide, the based Galois correspondence hands us a based isomorphism. Forgetting basepoints gives an unbased isomorphism, which is what we wanted. Why does the based correspondence produce a homeomorphism that forgets to an unbased isomorphism? Because a based covering isomorphism is in particular a homeomorphism over $X$ (it is built as a homeomorphism satisfying $p_2 \circ \psi = p_1$); this property does not depend on the basepoint being tracked.[/guided]

[step:Conclude] The map $\Psi$ defined in Step 2 is surjective (Step 3) and injective (Step 4), so it is a bijection between isomorphism classes of path-connected unbased covering spaces of $X$ and conjugacy classes of subgroups of $\pi_1(X, x_0)$, as claimed. [/step]