[proofplan] We prove that the assignment $(p: (\tilde{X}, \tilde{x}_0) \to (X, x_0)) \longmapsto p_*\pi_1(\tilde{X}, \tilde{x}_0)$ is a bijection by establishing injectivity and surjectivity separately. Injectivity follows from the [lifting criterion](/theorems/???) applied symmetrically to two coverings with the same associated subgroup: the mutual lifts compose to the identity by uniqueness of lifts, yielding a based homeomorphism. Surjectivity is constructive: given $H \leq \pi_1(X, x_0)$, we take the quotient of the universal cover $\bar{X}$ by the deck action of $H$, verify that the induced map $q: \bar{X}/H \to X$ is a covering, and compute $q_*\pi_1(\bar{X}/H) = H$. The auxiliary statements (sheet count, universal cover correspondence, quotient description) follow directly from this construction. [/proofplan] [step:Establish the lifting setup relating two coverings with the same associated subgroup] Suppose $p_1: (\tilde{X}_1, \tilde{x}_1) \to (X, x_0)$ and $p_2: (\tilde{X}_2, \tilde{x}_2) \to (X, x_0)$ are path-connected based coverings with \begin{align*} p_{1*}\pi_1(\tilde{X}_1, \tilde{x}_1) = p_{2*}\pi_1(\tilde{X}_2, \tilde{x}_2) \leq \pi_1(X, x_0). \end{align*} To apply the [lifting criterion for covering spaces](/theorems/???) to lift $p_1: (\tilde{X}_1, \tilde{x}_1) \to (X, x_0)$ through $p_2: (\tilde{X}_2, \tilde{x}_2) \to (X, x_0)$, we must verify the criterion's hypotheses: - $\tilde{X}_1$ is path connected: by hypothesis. - $\tilde{X}_1$ is locally path connected: since $X$ is locally path connected and $p_1$ is a covering map, local path-connectedness lifts through covering maps (local homeomorphisms preserve local topology). - The $\pi_1$ condition: $p_{1*}\pi_1(\tilde{X}_1, \tilde{x}_1) \subseteq p_{2*}\pi_1(\tilde{X}_2, \tilde{x}_2)$. This holds with equality by hypothesis. All hypotheses are verified, so the lifting criterion yields a unique continuous map \begin{align*} h: (\tilde{X}_1, \tilde{x}_1) &\to (\tilde{X}_2, \tilde{x}_2) \end{align*} satisfying $p_2 \circ h = p_1$. [guided] The lifting criterion is the main engine of injectivity, so we must be careful about its hypotheses. The criterion states: given a covering $p: (\tilde{X}, \tilde{x}_0) \to (X, x_0)$ and a map $f: (Y, y_0) \to (X, x_0)$ with $Y$ path connected and locally path connected, $f$ lifts to $\tilde{f}: (Y, y_0) \to (\tilde{X}, \tilde{x}_0)$ if and only if $f_*\pi_1(Y, y_0) \subseteq p_*\pi_1(\tilde{X}, \tilde{x}_0)$. We are applying this with $Y = \tilde{X}_1$, $y_0 = \tilde{x}_1$, $f = p_1$, and the target covering $p_2$. The $\pi_1$ condition becomes $(p_1)_*\pi_1(\tilde{X}_1, \tilde{x}_1) \subseteq (p_2)_*\pi_1(\tilde{X}_2, \tilde{x}_2)$, which follows from the equality hypothesis. Why does $\tilde{X}_1$ inherit local path-connectedness from $X$? Given $\tilde{y} \in \tilde{X}_1$, there is an evenly covered neighbourhood $V \ni p_1(\tilde{y})$ in $X$, and $p_1$ restricts to a homeomorphism on the sheet through $\tilde{y}$ onto $V$. A path-connected neighbourhood basis around $p_1(\tilde{y})$ (which exists because $X$ is locally path connected) pulls back through this homeomorphism to a path-connected neighbourhood basis around $\tilde{y}$. [/guided] [/step] [step:Construct the inverse lift and conclude $h$ is a homeomorphism] By symmetry — swapping the roles of $(\tilde{X}_1, \tilde{x}_1, p_1)$ and $(\tilde{X}_2, \tilde{x}_2, p_2)$ in the previous step — the lifting criterion also yields a unique continuous map \begin{align*} k: (\tilde{X}_2, \tilde{x}_2) &\to (\tilde{X}_1, \tilde{x}_1) \end{align*} with $p_1 \circ k = p_2$. Consider the composition $k \circ h: (\tilde{X}_1, \tilde{x}_1) \to (\tilde{X}_1, \tilde{x}_1)$. Then \begin{align*} p_1 \circ (k \circ h) = (p_1 \circ k) \circ h = p_2 \circ h = p_1. \end{align*} So $k \circ h$ is a lift of $p_1$ to $\tilde{X}_1$ fixing $\tilde{x}_1$. But $\operatorname{id}_{\tilde{X}_1}$ is also such a lift. By the [uniqueness of path lifts](/theorems/???) (equivalently, the uniqueness clause of the lifting criterion), these two lifts coincide: $k \circ h = \operatorname{id}_{\tilde{X}_1}$. The identical argument shows $h \circ k = \operatorname{id}_{\tilde{X}_2}$. Therefore $h$ is a homeomorphism with inverse $k$, and since $p_2 \circ h = p_1$, it is a based covering isomorphism. This proves injectivity of the assignment. [guided] Having both $h$ and $k$ is not enough — we need them to be mutually inverse. The trick is to use the uniqueness half of the lifting criterion: any two lifts of the same map that agree at a single basepoint must agree everywhere on a path-connected space. We compute $k \circ h$: applying $p_1$ gives $p_1 \circ k \circ h = p_2 \circ h = p_1$, so $k \circ h$ covers $p_1$. It also sends $\tilde{x}_1 \mapsto h(\tilde{x}_1) = \tilde{x}_2 \mapsto k(\tilde{x}_2) = \tilde{x}_1$. So $k \circ h$ and $\operatorname{id}_{\tilde{X}_1}$ are both lifts of $p_1$ passing through $\tilde{x}_1$, hence equal. Why must they be equal? The uniqueness principle for covering lifts says: if two continuous maps $\tilde{f}_1, \tilde{f}_2: Y \to \tilde{X}$ satisfy $p \circ \tilde{f}_1 = p \circ \tilde{f}_2$ and agree at one point of a connected $Y$, then they agree everywhere. The set $\{y : \tilde{f}_1(y) = \tilde{f}_2(y)\}$ is both open and closed (using local triviality of the covering), hence all of $Y$. [/guided] [/step] [step:Construct a covering from a subgroup $H \leq \pi_1(X, x_0)$ via the deck action on the universal cover] For surjectivity, fix $H \leq \pi_1(X, x_0)$. Let $\bar{p}: \bar{X} \to X$ be a [universal covering](/theorems/???) of $X$, whose existence is guaranteed by the hypothesis that $X$ is path connected, locally path connected, and semi-locally simply connected. By the [deck transformation theorem for the universal cover](/theorems/???), the group of deck transformations of $\bar{p}$ is canonically isomorphic to $\pi_1(X, x_0)$: \begin{align*} \Phi: \pi_1(X, x_0) &\xrightarrow{\sim} \operatorname{Deck}(\bar{p}), \end{align*} and the action of $\operatorname{Deck}(\bar{p})$ on $\bar{X}$ is free and properly discontinuous. Restricting the action to the subgroup $\Phi(H) \leq \operatorname{Deck}(\bar{p})$ gives an action of $H$ on $\bar{X}$ that remains free and properly discontinuous. Form the orbit space $\tilde{X} := \bar{X}/H$ with the quotient topology, and let $\pi: \bar{X} \to \tilde{X}$ be the quotient map. Since $\bar{p}: \bar{X} \to X$ is $H$-invariant ($\bar{p} \circ \varphi = \bar{p}$ for every deck transformation $\varphi \in \Phi(H)$), $\bar{p}$ factors uniquely through $\pi$: \begin{align*} q: \tilde{X} &\to X, & q \circ \pi &= \bar{p}. \end{align*} Pick a basepoint $\tilde{x}_0 := \pi(\bar{x}_0) \in \tilde{X}$, where $\bar{x}_0 \in \bar{p}^{-1}(x_0)$ is the chosen basepoint of $\bar{X}$. [guided] The construction takes the universal cover $\bar{X}$ — which is "too big" (it realises the trivial subgroup) — and quotients just enough to realise $H$. The quotient is by the subgroup of deck transformations corresponding to $H$ under the canonical isomorphism $\Phi: \pi_1(X, x_0) \xrightarrow{\sim} \operatorname{Deck}(\bar{p})$. Why does the quotient $\bar{X}/H$ make sense topologically? Because the $H$-action (via $\Phi(H)$) is free (no non-identity element has fixed points) and properly discontinuous (every $\bar{x}$ has a neighbourhood $\bar{U}$ such that $\varphi(\bar{U}) \cap \bar{U} = \varnothing$ for $\varphi \neq \operatorname{id}$). These two properties imply that $\pi: \bar{X} \to \bar{X}/H$ is itself a covering map, and in particular the quotient is Hausdorff and inherits the local topology. Why does $\bar{p}$ descend to $\tilde{X}$? Because $\bar{p}$ is constant on $H$-orbits: if $\varphi \in \Phi(H)$, then $\bar{p}(\varphi(\bar{x})) = \bar{p}(\bar{x})$, since every deck transformation commutes with the projection. By the universal property of the quotient topology, the orbit-invariant continuous map $\bar{p}$ factors uniquely through $\pi$ to give $q: \tilde{X} \to X$. [/guided] [/step] [step:Verify that $q: \tilde{X} \to X$ is a covering map] Let $x \in X$ and let $U \ni x$ be an evenly covered open neighbourhood for $\bar{p}$: $\bar{p}^{-1}(U) = \bigsqcup_{i \in I} \bar{U}_i$ where each $\bar{p}|_{\bar{U}_i}: \bar{U}_i \to U$ is a homeomorphism. The deck action of $\pi_1(X, x_0)$ permutes the sheets $\{\bar{U}_i\}$ freely and transitively. The $H$-orbits among the sheets $\{\bar{U}_i\}$ partition the index set $I$ into $H$-orbits. Write this partition as $I = \bigsqcup_{j \in J} I_j$, where each $I_j$ is a single $H$-orbit. Then \begin{align*} q^{-1}(U) = \pi(\bar{p}^{-1}(U)) = \bigsqcup_{j \in J} \pi\Bigl(\bigsqcup_{i \in I_j} \bar{U}_i\Bigr). \end{align*} For each $j$, the set $\pi(\bigsqcup_{i \in I_j} \bar{U}_i)$ is a single open subset $\tilde{U}_j \subseteq \tilde{X}$: the quotient map identifies each $\bar{U}_i$ ($i \in I_j$) homeomorphically onto $\tilde{U}_j$, because the $H$-action permutes these sheets freely and there is exactly one sheet per $\tilde{U}_j$-point. Concretely, picking any $i_0 \in I_j$, the composition $q \circ (\pi|_{\bar{U}_{i_0}})^{-1}: \tilde{U}_j \to U$ coincides with $\bar{p} \circ (\pi|_{\bar{U}_{i_0}})^{-1}|_{\tilde{U}_j}$, which is a homeomorphism onto $U$. So $q|_{\tilde{U}_j}$ is a homeomorphism onto $U$. For distinct $j \ne j'$, the sets $\tilde{U}_j$ and $\tilde{U}_{j'}$ are disjoint, because sheets in different $H$-orbits map to distinct $H$-orbits in $\tilde{X}$. Hence $q^{-1}(U) = \bigsqcup_{j \in J} \tilde{U}_j$ is a disjoint union of open sets each mapped homeomorphically to $U$ by $q$. Therefore $U$ is evenly covered by $q$, and since $x$ was arbitrary, $q$ is a covering map. [guided] The strategy is to take an evenly covered neighbourhood $U$ for the universal covering $\bar{p}$, and show that $U$ is also evenly covered for the induced covering $q$. The sheets of $q$ over $U$ are the $H$-orbits of the sheets of $\bar{p}$ over $U$. Picture this concretely: $\bar{p}^{-1}(U)$ is a disjoint union of copies of $U$ indexed by the deck group $\pi_1(X, x_0)$, and $\pi_1(X, x_0)$ permutes these copies freely. The $H$-action restricts to a sub-permutation, partitioning the copies into $H$-orbits. After quotienting by $H$, each orbit collapses to a single copy of $U$ in $\tilde{X}$. Different orbits stay separate, so $q^{-1}(U)$ is still a disjoint union — just with one summand per $H$-orbit rather than one per element of $\pi_1(X, x_0)$. The homeomorphism property $q|_{\tilde{U}_j}: \tilde{U}_j \to U$ works because within a single $H$-orbit, the quotient map $\pi$ maps each $\bar{U}_i$ bijectively (in fact homeomorphically, being a quotient by a free and properly discontinuous action restricted to each sheet) onto $\tilde{U}_j$, and $q$ then restricts the homeomorphism $\bar{p}|_{\bar{U}_i}: \bar{U}_i \to U$ down the quotient. [/guided] [/step] [step:Verify that $\tilde{X}$ is path connected] Since $\bar{X}$ is path connected (it is the universal cover of a path-connected space) and $\pi: \bar{X} \to \tilde{X}$ is a continuous surjection, the image $\tilde{X} = \pi(\bar{X})$ is path connected: given $\tilde{y}_1, \tilde{y}_2 \in \tilde{X}$, pick preimages $\bar{y}_i \in \pi^{-1}(\tilde{y}_i)$, join them by a path $\gamma$ in $\bar{X}$, and $\pi \circ \gamma$ is a path in $\tilde{X}$ from $\tilde{y}_1$ to $\tilde{y}_2$. [/step] [step:Compute $q_*\pi_1(\tilde{X}, \tilde{x}_0) = H$] We show $q_*\pi_1(\tilde{X}, \tilde{x}_0) = H$ as subgroups of $\pi_1(X, x_0)$. Let $[\gamma] \in \pi_1(X, x_0)$. By the [path lifting property](/theorems/???) applied to the covering $\bar{p}: \bar{X} \to X$, there is a unique lift $\bar{\gamma}: [0, 1] \to \bar{X}$ with $\bar{\gamma}(0) = \bar{x}_0$. Then $[\gamma] \in q_*\pi_1(\tilde{X}, \tilde{x}_0)$ if and only if $[\gamma]$ is represented by a loop in $X$ whose lift to $\tilde{X}$ starting at $\tilde{x}_0$ is a loop, i.e., closes up at $\tilde{x}_0$. The lift of $\gamma$ to $\tilde{X}$ starting at $\tilde{x}_0$ is precisely $\pi \circ \bar{\gamma}$: it is continuous, starts at $\pi(\bar{x}_0) = \tilde{x}_0$, and $q \circ (\pi \circ \bar{\gamma}) = \bar{p} \circ \bar{\gamma} = \gamma$. So $\pi \circ \bar{\gamma}$ closes up at $\tilde{x}_0$ iff $\pi(\bar{\gamma}(1)) = \tilde{x}_0 = \pi(\bar{x}_0)$, which by definition of the quotient holds iff $\bar{\gamma}(1) \in H \cdot \bar{x}_0$ (the $H$-orbit of $\bar{x}_0$, where $H$ acts via $\Phi$). Under the deck transformation identification, the endpoint $\bar{\gamma}(1)$ of the unique lift starting at $\bar{x}_0$ equals $\Phi([\gamma])(\bar{x}_0)$. Since the action is free, $\bar{\gamma}(1) \in H \cdot \bar{x}_0$ iff $\Phi([\gamma]) \in \Phi(H)$ iff $[\gamma] \in H$. Hence $q_*\pi_1(\tilde{X}, \tilde{x}_0) = H$, completing the proof of surjectivity. [guided] To compute $q_*\pi_1(\tilde{X}, \tilde{x}_0)$, we use the characterisation: $[\gamma] \in q_*\pi_1(\tilde{X}, \tilde{x}_0)$ iff the lift of $\gamma$ to $\tilde{X}$ starting at $\tilde{x}_0$ is a loop. We use the universal cover as an intermediary. Lift $\gamma$ all the way to $\bar{X}$ as $\bar{\gamma}$, then project down to $\tilde{X}$. By uniqueness of lifts, this gives the lift to $\tilde{X}$. The lift closes up in $\tilde{X}$ iff the endpoints agree modulo $H$, i.e., $\bar{\gamma}(0) = \bar{x}_0$ and $\bar{\gamma}(1)$ lie in the same $H$-orbit. Now we bring in the key identification: for the universal cover, the map $[\gamma] \mapsto \bar{\gamma}(1)$ sets up a bijection $\pi_1(X, x_0) \xrightarrow{\sim} \bar{p}^{-1}(x_0)$ where $\bar{\gamma}(1) = \Phi([\gamma])(\bar{x}_0)$. This is the content of the universal-cover/deck-group isomorphism. Therefore $\bar{\gamma}(1) \in H \cdot \bar{x}_0$ iff the corresponding deck transformation lies in $\Phi(H)$, which (since $\Phi$ is an isomorphism) holds iff $[\gamma] \in H$. This gives both containments at once: $[\gamma] \in q_*\pi_1(\tilde{X}, \tilde{x}_0) \iff [\gamma] \in H$. [/guided] [/step] [step:Derive the index-sheet count formula] Let $p: (\tilde{X}, \tilde{x}_0) \to (X, x_0)$ be a path-connected based covering with $H = p_*\pi_1(\tilde{X}, \tilde{x}_0)$. The [sheet count of a covering equals the index of the associated subgroup](/theorems/???): there is a bijection between the fibre $p^{-1}(x_0)$ and the set of left cosets $\pi_1(X, x_0)/H$. Explicitly, given $[\gamma] \in \pi_1(X, x_0)$, lift $\gamma$ starting at $\tilde{x}_0$ to $\tilde{\gamma}$, and map $[\gamma] \mapsto \tilde{\gamma}(1) \in p^{-1}(x_0)$. Two classes $[\gamma_1], [\gamma_2]$ give the same endpoint iff $[\gamma_1^{-1} \gamma_2] \in H$, i.e., iff they lie in the same left coset of $H$. Therefore $|p^{-1}(x_0)| = [\pi_1(X, x_0) : H]$, which is the number of sheets. [/step] [step:Identify the universal cover with the trivial subgroup] When $H = \{e\}$, the construction in the previous steps gives $\tilde{X} = \bar{X}/\{e\} = \bar{X}$, the universal cover itself, with $q = \bar{p}$. Conversely, if $p: \tilde{X} \to X$ is a simply connected covering then $p_*\pi_1(\tilde{X}, \tilde{x}_0) = p_*(\{e\}) = \{e\}$, and by injectivity of the assignment, $\tilde{X}$ is isomorphic to the universal cover. This confirms that the trivial subgroup corresponds to the universal cover. [/step] [step:Conclude] Combining the previous steps: the assignment $p \mapsto p_*\pi_1(\tilde{X}, \tilde{x}_0)$ is injective (Steps 1-2), surjective with explicit inverse $H \mapsto \bar{X}/H$ (Steps 3-6), the sheet count equals the index (Step 7), and the universal cover corresponds to the trivial subgroup (Step 8). This establishes the Galois correspondence. [/step]