Covering space

A covering map $p:\tilde{X}\to X$ is locally a disjoint union of homeomorphisms onto the same evenly covered open neighbourhood.

A branched covering is a map that is a covering space away from a branch set and has a specified local covering model at that set. For a smooth two-fold covering branched over a codimension-two submanifold, the local model is $(z,u)\mapsto (z^2,u)$. A two-fold branched cover of a knot is the basic three-dimensional example.

The connected covering space corresponding to $\ker \alpha$ has deck transformation group $\mathbb{Z}$. Choosing a deck generator makes its integral homology a module over the Laurent polynomial ring $\mathbb{Z}[t,t^{-1}]$.

Given an integral epimorphism $\alpha :{\pi }_1(X)\to \mathbb{Z}$, the Alexander module is the first integral homology of the corresponding infinite cyclic cover associated to an epimorphism, with $t$ acting by its chosen deck transformation. Its order is the greatest common divisor of maximal presentation matrix minors when it is finitely presented and torsion. For the meridional epimorphism of a knot exterior, this is the Alexander module of a knot.

For a connected space, this is the connected covering corresponding to the kernel of the homomorphism from its fundamental group to its abelianization. Its deck transformation group is $H_1(Y;\mathbb{Z})$. Lifted cells make its cellular chain complex a complex over the group ring $\mathbb{Z}[H_1(Y)]$.

For $C_2\ast \mathbb{Z}\to S_3$ taking the two generators to a transposition and a $3$-cycle, the kernel's six-sheeted regular covering space consists of three spheres and two circles. Each sphere has two marked points, and each circle meets one point on each sphere. It retracts onto a rank-four graph by replacing each sphere by a meridian arc, but admits no deformation retraction onto any graph because its degree-two homology is ${\mathbb{Z}}^3$.

For $n\ge 1$, mapby sending the projective-plane generator and the first circle generator to the two standard generators. The kernel defines a connected regular four-sheeted cover ${\tilde{X}}_n$. Its lifted cellular boundary in degree two is multiplication by $1+v$ on $\mathbb{Z}[C_2\times C_2]$, so

Let $p:\tilde{X}\to X$ be a covering map and let $Y$ be path-connected. If two lifts ${\tilde{f}}_1,{\tilde{f}}_2:Y\to \tilde{X}$ of the same map $f:Y\to X$ agree at one point, they agree everywhere. Join that point to any other by a path and apply uniqueness in the path lifting theorem.

Lifting loops based at $x_0$ permutes the fibre $p^{-1}(x_0)$ and defines the monodromy action of ${\pi }_1(X,x_0)$ on that fibre. A connected covering has transitive monodromy.

A deck transformation of a covering $p:\tilde{X}\to X$ is a homeomorphism $g:\tilde{X}\to \tilde{X}$ satisfying $p\circ g=p$. Every double covering has a canonical nonidentity deck transformation that exchanges the two points in each fibre.

The deck transformation group consists of all deck transformations under composition. For a connected regular covering corresponding to a normal subgroup $H◃{\pi }_1(X)$, it is isomorphic to ${\pi }_1(X)/H$.

Every finitely generated group acts freely and properly discontinuously by Riemannian isometries on a connected manifold in any dimension $d\ge 3$. Choose a surjection $F_r\to T$ and the regular covering of ${\#}_r(S^1\times S^{d-1})$ associated with its kernel. Pulling back a metric makes the deck transformation group isometric. Uniqueness of lifts makes its action free. The cover can be noncompact and need not be simply connected; no claim about realizing every group as a closed three-manifold fundamental group is involved.

For a connected covering, restriction to one fibre identifies the deck transformation group with the centralizer in the fibre's symmetric group of the monodromy action of a covering space. A deck transformation commutes with every lifted loop, and every commuting fibre permutation extends uniquely over the covering by path lifting.

Every two-sheeted covering of a path-connected base has a nonidentity deck transformation that exchanges the two sheets. In monodromy terms, every subgroup of $S_2$ is centralized by its unique transposition.

For a double covering $p:X\to Y$, the mod-two transfer sends a singular simplex of $Y$ to the sum of its two lifts. Together with the chain map $p_{\#}$, it gives a short exact sequence

A universal cover is a simply connected covering space. The universal cover of a cylinder is the plane, with the angular coordinate unwrapped to a real coordinate.

For a locally path-connected subspace $A\subseteq X$, a selected component of the inverse image in a universal cover covers the corresponding component $A_0$ of $A$. Its covering subgroup is $\ker ({\pi }_1(A,a_0)\to {\pi }_1(X,a_0))$; surjectivity onto all of $A$ additionally requires $A$ to be path connected.

For a path-connected locally path-connected base, any two based universal covers lift to one another. The two composites and the respective identity maps are based lifts of the same covering maps, so uniqueness of lifts makes the composites identities. The lift is therefore a homeomorphism.

Let $X$ have contractible universal cover, let $K$ be a path-connected simplicial complex, and let $i:K^1\hookrightarrow K$. A map $f:|K^1|\to X$ extends to $|K|$ exactly when its induced fundamental-group map factors through $i_{\ast }$. The factorization kills every 2-simplex boundary; all higher-dimensional boundary maps lift to the contractible universal cover and are null-homotopic.

Under the usual local hypotheses, based connected coverings of $X$ correspond to subgroups of ${\pi }_1(X)$; changing the point above the base point conjugates the subgroup. Normal subgroups correspond to regular coverings.

A connected covering is regular when its deck transformation group acts transitively on every fibre. Under the classification of coverings, this is equivalent to the corresponding subgroup of the fundamental group being normal.

For a connected covering corresponding to $H\le {\pi }_1(X)$, every fibre has cardinality $[{\pi }_1(X):H]$.

If the base of a covering is path-connected, a path from $x_0$ to $x_1$ gives a bijectionby sending each initial lift point to the endpoint of its unique lifted path. Lifting the reversed path gives the inverse bijection.

For a presentation complex, the covering associated with a subgroup has one vertex for each coset. A generator gives directed edges according to its coset action, and every relator has one lifted 2-cell beginning at each vertex.

Choose permutations ${\alpha }_1,\dots ,{\alpha }_r\in S_n$. A covering of a wedge of $r$ oriented circles has fibre $\{1,\dots ,n\}$ and an edge labelled $j$ from $i$ to ${\alpha }_j(i)$ above the $j$th circle. It is connected exactly when the generated permutation group is transitive, and its deck group is the centralizer of that group in $S_n$.

For a connected regular covering corresponding to $H\le {\pi }_1(X,x_0)$, the subgroup $H$ is normal andThe quotient acts on a fibre by endpoints of lifted loops, and regularity makes this action transitive with kernel $H$.

A connected covering is normal, or regular, when its deck transformations act transitively on each fibre. Under the subgroup classification, this is equivalent to the associated subgroup of the fundamental group being normal.

A universal covering has simply connected total space, so its associated subgroup of the base fundamental group is trivial. The trivial subgroup is normal; hence every universal covering map is a normal covering map.

For a connected degree-$n$ cover of a closed orientable surface ${\Sigma }_g$, normality is forced when $n=1$, when $n=2$, or when $g=1$. For $g=0$, existence itself forces $n=1$. For every $g\ge 2$ and $n\ge 3$, nonnormal connected degree-$n$ covers exist.

For $g\ge 2$ and $n\ge 3$, sendwhere $\sigma =(12\cdots n)$ and $\tau =(12)$ in $S_n$, and send all remaining surface generators to the identity. The two commutators cancel, and $\sigma ,\tau$ generate $S_n$. The inverse image of a point stabilizer is a nonnormal subgroup of index $n$, hence defines a connected nonnormal degree-$n$ cover.

A path in the base of a covering has a unique lift after its initial point in the fibre is chosen. Subdivide its compact parameter interval so that each segment lies in an evenly covered set, then use the inverse sheet maps successively.

For a covering $p:(\tilde{X},{\tilde{x}}_0)\to (X,x_0)$ and a based map $f:(Y,y_0)\to (X,x_0)$ from a path-connected locally path-connected space, a based lift $\tilde{f}$ exists exactly when

An action of $G$ on $X$ is a covering space action when every point has a neighbourhood $U$ disjoint from $gU$ for every nonidentity $g$. The quotient map is then locally a covering map, although without stronger properness assumptions its quotient need not be Hausdorff.

If $\Gamma$ is a discrete subgroup of a topological group $G$, then left translation by $\Gamma$ is a covering-space action. Choose an identity neighbourhood $U$ with $U{U}^{-1}\cap \Gamma =\{1\}$; the translates $\gamma U$ are pairwise disjoint, and the quotient map $G\to \Gamma \backslash G$ restricts to a homeomorphism on each translate.

On ${\mathbb{R}}^2\setminus \{0\}$, the powers of $T(x,y)=(2x,y/2)$ act freely with locally disjoint translates. Nevertheless $(2^{-n},1)\to (0,1)$ while $T^n(2^{-n},1)\to (1,0)$, so the two distinct limiting orbits cannot be separated in the quotient.

Lifting the hyperbolic-scaling action to the universal cover of ${\mathbb{C}}^{\ast }$ preserves the two inseparable limiting orbits. The covering surface is biholomorphic to $\mathbb{C}$, so a simply connected Riemann surface can still have a non-Hausdorff quotient by a covering space action of homeomorphisms.