Classification of Covering Spaces

For every $x\in X$, [F1] gives an open neighbourhood whose image is open. The union of these images is $f[X]$, so $f[X]$ is open in $Y$; it is nonempty because $X$ is nonempty.

By [F2], $f[X]$ is compact, and by [F3] it is closed in the Hausdorff space $Y$.

Fix $y\in Y$. The fibre $K=f^{-1}(y)$ is closed because $f$ is continuous and $\{y\}$ is closed by [F7]. It is discrete: for each $x\in K$, a local-homeomorphism chart $U_x$ is injective, hence $U_x\cap K=\{x\}$, so every singleton is open in the subspace $K$.

The nonempty subset $f[X]$ is both open and closed. By connectedness in [F6], it must equal $Y$, so $f$ is surjective.

By [F4], the closed subspace $K$ of compact $X$ is compact.

The open singleton family $\{\{x\}:x\in K\}$ covers the discrete space $K$. Compactness and [F5] give a finite subcover, so $K$ is finite. It is nonempty by surjectivity from step 2.1.

List the finite fibre as $f^{-1}(y)=\{x_1,\dots ,x_m\}$ with $m\ge 1$ by [L1]. Using [F1] finitely many times and [F2] for the finite selections, choose pairwise disjoint open neighbourhoods $N_i$ of the $x_i$. Intersect each $N_i$ with a local-homeomorphism chart at $x_i$; its image is still an open neighbourhood of $y$. Let $W$ be the finite intersection of these images, and replace each chart by its inverse image of $W$. We obtain pairwise disjoint open sets $U_i$ with $f{\mathrm{\mid }}_{U_i}:U_i\to W$ a homeomorphism.

The set $K=X\setminus {\bigcup }_i{U}_i$ is closed and therefore compact by [F3]. Its image $f[K]$ is compact by [F4] and closed in $Y$ by [F5]. No point of the fibre over $y$ lies in $K$, so $y\mathrm{\notin }f[K]$. Hence $V:=W\setminus f[K]$ is an open neighbourhood of $y$.

Put $V_i=U_i\cap f^{-1}(V)$. Each $V_i$ is open and $f{\mathrm{\mid }}_{V_i}:V_i\to V$ is a homeomorphism. If $x\in f^{-1}(V)$ then $x\mathrm{\notin }K$, so $x$ lies in exactly one $U_i$ and hence in exactly one $V_i$. Thus $f^{-1}(V)$ is the disjoint union of the finitely many $V_i$. Since $y$ was arbitrary and $f$ is surjective by [L1], [F6] makes $f$ a finite-sheeted covering.

For the forward implication, [F7] applied to $p_2\circ f=p_1$ gives the displayed subgroup inclusion. For the reverse implication, [F2] makes $E_1$ locally path-connected and [F5] makes it path-connected. Apply [F1] to the map $p_1:E_1\to B$ and the covering $p_2:E_2\to B$; the inclusion produces a based lift $f:E_1\to E_2$, and $p_{2}f=p_1$ says exactly that it is a map of coverings.

Uniqueness is part of [F1], and also follows from [F3] because any two such maps lift $p_1$ and agree at $e_1$.

First, $f$ is surjective. Indeed, [F2] and [F5] make $E_2$ path-connected. Join $f(e_1)$ to any $z\in E_2$ by a path, project that path through $p_2$, and lift the projection through $p_1$ from $e_1$. The image of this lift under $f$ is a lift with the original initial point, so uniqueness in [F6] makes it the original path and its endpoint maps to $z$. Now fix $b\in B$. Intersect evenly covered neighbourhoods of $b$ for $p_1$ and $p_2$, then use local path-connectedness to choose a path-connected open neighbourhood $O$ inside that intersection. For a $p_1$-sheet $U$ over $O$, choose $x\in U$ and let $V$ be the $p_2$-sheet containing $f(x)$. The maps $f{\mathrm{\mid }}_U$ and $(p_2{\mathrm{\mid }}_V{)}^{-1}\circ p_1{\mathrm{\mid }}_U$ are lifts of $p_1{\mathrm{\mid }}_U$ through $p_2$, agree at $x$, and have connected domain $U$; hence [F3] makes them equal. Thus $f{\mathrm{\mid }}_U:U\to V$ is a homeomorphism. Conversely, every point of $f^{-1}(V)$ lies in one such $U$. Hence $f^{-1}(V)$ is the disjoint union of exactly those $p_1$-sheets sent to $V$, each mapped homeomorphically onto $V$. Surjectivity makes this family nonempty for every $V$, so every point of $E_2$ has an evenly covered neighbourhood and [F4] makes $f$ a covering map.

For the direction from subgroup equality to isomorphism, [L1] gives unique based morphisms $f:E_1\to E_2$ and $g:E_2\to E_1$.

The composite $g\circ f$ and ${\mathrm{id}}_{E_1}$ are lifts of $p_1$ through $p_1$ and agree at $e_1$, so [F1] makes them equal. Likewise $f\circ g={\mathrm{id}}_{E_2}$. Hence $f$ and $g$ are inverse based covering isomorphisms, and uniqueness follows from [L1].

For the converse direction, a based isomorphism and its inverse are covering morphisms, so [L1] gives both subgroup inclusions and therefore equality.

Fix a universal cover $p:(\tilde{B},{\tilde{b}}_0)\to (B,b_0)$ by [F1], and use [F2] to regard $H$ as a subgroup of its deck group. Over an evenly covered neighbourhood of $b\in B$, choose the sheet containing a given $\tilde{b}$. A nonidentity deck transformation sends it to a different sheet: otherwise it would send the unique point over $b$ in that sheet to itself and hence be the identity by [F7]. Thus [F5] holds, so the restricted $H$-action is a covering-space action.

Let $q:\tilde{B}\to E_H:=\tilde{B}\mathrm{/}H$ be the orbit map, which is a covering by [F3]. Since $p$ is constant on $H$-orbits, it induces $p_H:E_H\to B$. Over an evenly covered $W\subseteq B$, the $H$-orbits of the sheets of $p^{-1}(W)$ have disjoint images under $q$, and on each such image $p_H$ is identified with the homeomorphism from any representative sheet to $W$. Hence $p_H$ is a covering by [F6]. The path-connected space $\tilde{B}$ maps continuously and surjectively to $E_H$, so $E_H$ is path-connected, with basepoint $e_H=q({\tilde{b}}_0)$.

For a loop $\alpha$ at $b_0$, its lift to $E_H$ from $e_H$ is $q\circ \tilde{\alpha }$, where $\tilde{\alpha }$ is its universal lift. This lift closes exactly when the universal endpoint lies in the $H$-orbit of ${\tilde{b}}_0$, which by [F2] and [F4] holds exactly when $[\alpha ]\in H$. A loop class is in $(p_H{)}_{\ast }{\pi }_1(E_H,e_H)$ exactly when it has a closed lift to $E_H$, so the induced subgroup is precisely $H$.

If $[\lambda ]\in {\pi }_1(E,e_1)$, then $\tilde{\gamma }\ast \lambda \ast \overline{\stackrel{~}{\gamma }}$ is a loop at $e_0$. Its projection represents $g{p}_{\ast }[\lambda ]g^{-1}$ by [F2], so $g{H}_1{g}^{-1}\subseteq H_0$.

Apply step 1.1 to the reversed path from $e_1$ to $e_0$. This gives $g^{-1}{H}_{0}g\subseteq H_1$, while conjugating the first inclusion by $g^{-1}$ and $g$ gives the reverse containment. Hence $H_1=g^{-1}{H}_{0}g$, with the displayed direction fixed by traversal order.

For an arbitrary $e_1\in p^{-1}(b_0)$, path-connectedness and [F3] supply a path from $e_0$ to $e_1$; its projection begins and ends at $b_0$, hence is a loop. Conversely, [F1] says the endpoint of the lift of that loop from $e_0$ is the prescribed $e_1$.

Every connected covering under consideration has locally path-connected total space by [F1], because $B$ is locally path-connected, and therefore has path-connected total space by [F2]. For the based correspondence, [L1] proves surjectivity: every subgroup occurs.

For the based correspondence, [L2] applies under the base hypotheses and the path-connectedness established in step 1.1, and proves injectivity: two based connected coverings determine the same subgroup exactly when they are based-isomorphic. Thus claim 1 is a bijection.

For claim 2, the path-connectedness from step 1.1 licenses [L3], which shows that changing the chosen point over $b_0$ replaces the subgroup by a conjugate. Hence the conjugacy class depends only on the unbased covering. Every conjugacy class occurs by step 1.1.

Suppose two unbased connected coverings determine the same conjugacy class. Choose fibre points with induced subgroups $H_1,H_2$, and write $H_1=g^{-1}{H}_{2}g$. By [F3], lift a loop representing $g$ from the second fibre point. By [L3], its endpoint gives a new fibre point whose induced subgroup is $H_1$; [L2] then gives a based isomorphism and hence an unbased isomorphism. Conversely, any unbased isomorphism carries a chosen fibre point to a fibre point of the other cover, so [L2] and [L3] make the subgroups conjugate. This proves injectivity and completes claim 2.

Local path-connectedness of the base lifts to $E$, and connectedness then makes $E$ path-connected by [F6].

By [L1], now licensed by step 1.1, the same covering based at $e_g$ has induced subgroup $g^{-1}Hg$.

A deck transformation taking $e_0$ to $e_g$ is exactly a based isomorphism from $(E,e_0)$ to $(E,e_g)$. By [L2], it exists exactly when $H=g^{-1}Hg$, which by [F1] is exactly $g\in N_G(H)$; uniqueness follows from [F2].

For $g,g^{\mathrm{\prime }}\in N_G(H)$, [F2] gives ${\tau }_g={\tau }_{g^{\mathrm{\prime }}}$ exactly when $e_0\cdot g=e_0\cdot g^{\mathrm{\prime }}$. Applying the action by $g^{\mathrm{\prime }-1}$ reduces this to $e_0\cdot (g{g}^{\mathrm{\prime }-1})=e_0$, which holds exactly when the lifted loop closes at $e_0$, equivalently when $g{g}^{\mathrm{\prime }-1}\in p_{\ast }{\pi }_1(E,e_0)=H$. Thus ${\tau }_g={\tau }_{g^{\mathrm{\prime }}}$ exactly when $Hg=H{g}^{\mathrm{\prime }}$. By step 1.1, given any point $e$ in the fibre, choose a path from $e_0$ to $e$; its projection is a loop at $b_0$, and uniqueness in [F7] makes the lifted endpoint $e_0\cdot g$ equal to $e$. Hence the monodromy orbit is the whole fibre, so step 3.1 and [F2] make $\mathrm{\Theta }$ surjective.

By [F5], $N_G(H)$ is a group. Deck transformations commute with lifted endpoints: ${\tau }_g(e_0\cdot h)={\tau }_g(e_0)\cdot h$. Hence $({\tau }_g\circ {\tau }_h)(e_0)={\tau }_g(e_0\cdot h)=(e_0\cdot g)\cdot h=e_0\cdot (gh)={\tau }_{gh}(e_0),$ so [F2] gives ${\tau }_g{\tau }_h={\tau }_{gh}$ and $\mathrm{\Theta }$ is a homomorphism. Its kernel consists of the $g$ with $e_0\cdot g=e_0$. If $g=p_{\ast }[\lambda ]\in H$, the lift of a representative projected loop is the closed loop $\lambda$, so it fixes $e_0$; conversely, if the lift of a representative of $g$ closes at $e_0$, that lifted loop projects to $g$ and puts $g$ in $H$. Thus $\ker \mathrm{\Theta }=H$.

Use [L1] to take the surjective homomorphism $\mathrm{\Theta }:N_G(H)\to \mathrm{Deck}(E\mathrm{/}B)$.

By [F3], $N_G(H)$ is a group. By [F2], $nH{n}^{-1}=H$ for every $n\in N_G(H)$, so $H⊴N_G(H)$ and the quotient $N_G(H)\mathrm{/}H$ is defined. By [L1], $\ker \mathrm{\Theta }=H$.

Applying [F1] to $\mathrm{\Theta }$ gives $N_G(H)\mathrm{/}H=N_G(H)\mathrm{/}\ker \mathrm{\Theta }\cong \mathrm{im}\mathrm{\Theta }=\mathrm{Deck}(E\mathrm{/}B)\mathrm{.}$

By [F2], every point of $p^{-1}(b_0)$ has the form $e_0\cdot g$ for some $g\in G$, and [L1] records the subgroup at that point.

By [L2], a deck transformation reaches $e_0\cdot g$ from $e_0$ exactly when $g$ normalizes $H$. Hence the deck action on $p^{-1}(b_0)$ is transitive exactly when $N_G(H)=G$, which by [F1] is exactly when $H⊴G$. This proves the equivalence of clauses 2 and 3.

For the implication from normality to regularity, clause 2 gives clause 3 by step 2.1. Let $e,e^{\mathrm{\prime }}$ lie over an arbitrary $b\in B$, choose a path from $b$ to $b_0$, and lift it from $e,e^{\mathrm{\prime }}$ to points $u,u^{\mathrm{\prime }}$ over $b_0$. Clause 3 gives a deck transformation $\tau$ with $\tau (u)=u^{\mathrm{\prime }}$. Applying $\tau$ to the reverse lift from $u$ produces a lift from $u^{\mathrm{\prime }}$, so uniqueness in [F3] gives $\tau (e)=e^{\mathrm{\prime }}$. Thus the deck group is transitive on every fibre and the covering is regular.

For the converse implication from regularity, the definition makes the deck action transitive on $p^{-1}(b_0)$, so clause 3 holds. Step 2.1 then gives $N_G(H)=G$, and [F1] gives $H⊴G$. Thus clauses 1, 2, and 3 are equivalent.

By [L2], regularity gives $H⊴G$, so every $g\in G$ preserves $H$ under conjugation and [F1] gives $N_G(H)=G$.

Substitution of $N_G(H)=G$ in [L1] yields $\mathrm{Deck}(E\mathrm{/}B)\cong G\mathrm{/}H$.

The map $p$ is a covering by [F1].

The real line is a nonempty convex subset of itself, so [F2] makes it simply connected.

Steps 1.1 and 1.2 satisfy both clauses of [F3], hence $p$ is a universal covering.

Let $[a]$ be a circle point and let $N$ be an open neighbourhood of it. The inverse image of $N$ is open and contains $a$, so it contains an interval $J$ about $a$ of length below one. By [F8], $p[J]$ is an open neighbourhood of $[a]$ inside $N$ and is homeomorphic to the convex interval $J$. Thus [F9], [F10], and [F11] show that the circle is locally path-connected and semilocally simply connected; [F7] supplies nonemptiness and path-connectedness. The classification theorem [L1] therefore applies. Transporting its subgroups through [F1], [F2] says that every induced subgroup is uniquely $n\mathbb{Z}$ for one $n\in \mathbb{N}$.

By [L1], this gives one based-isomorphism class for each $n$. Since [F4] makes every conjugate of $n\mathbb{Z}$ equal to itself, the same parameter gives the unbased-isomorphism classes. Conversely, [L1] realizes every $n\mathbb{Z}$, so both correspondences are bijections.

Every classified covering has connected total space. Since step 1.1 establishes that the circle is locally path-connected, [F12] and [F13] make each such total space path-connected, licensing [F3]. For $n\ge 1$, [F5] and [F6] give $[\mathbb{Z}:n\mathbb{Z}]=n$, so [F3] gives $n$ sheets. For $n=0$, the subgroup is trivial, [F5] and [F6] give infinite index, and [L2] realizes this class by the real-line universal cover. At $n=1$ the index is one.

By [F4] and [F5], the base is path-connected and locally path-connected. Since the covering total space is connected, [F6] and [F7] make it path-connected. By [F1] and [F3], its induced subgroup corresponds to a subgroup of an abelian group, so [F2] makes it normal.

Applying [L1] to step 1.1 shows that the covering is regular.