Covering space

Let ${\displaystyle X}$ be a topological space. A covering of ${\displaystyle X}$ is a continuous map

${\displaystyle \pi :{\tilde {X}}\rightarrow X}$

such that for every ${\displaystyle x\in X}$ there exists an open neighborhood ${\displaystyle U_{x}}$ of ${\displaystyle x}$ and a discrete space ${\displaystyle D_{x}}$ such that ${\displaystyle \pi ^{-1}(U_{x})}$ is the disjoint union ${\displaystyle \displaystyle \bigsqcup _{d\in D_{x}}V_{d}}$ and ${\displaystyle \pi |_{V_{d}}:V_{d}\rightarrow U_{x}}$ is a homeomorphism for every ${\displaystyle d\in D_{x}}$. The open sets ${\displaystyle V_{d}}$ are called sheets, which are uniquely determined up to homeomorphism if ${\displaystyle U_{x}}$ is connected.^[2]: 56 For each ${\displaystyle x\in X}$ the discrete set ${\displaystyle \pi ^{-1}(x)}$ is called the fiber of ${\displaystyle x}$. If ${\displaystyle X}$ is connected (and ${\displaystyle {\tilde {X}}}$ is non-empty), it can be shown that ${\displaystyle \pi }$ is surjective, and the cardinality of ${\displaystyle D_{x}}$ is the same for all ${\displaystyle x\in X}$; this value is called the degree of the covering. If ${\displaystyle {\tilde {X}}}$ is path-connected, then the covering ${\displaystyle \pi :{\tilde {X}}\rightarrow X}$ is called a path-connected covering. This definition is equivalent to the statement that ${\displaystyle \pi }$ is a locally trivial fiber bundle.

Some authors also require that ${\displaystyle \pi }$ be surjective in the case that ${\displaystyle X}$ is not connected.^[3]

• For every topological space ${\displaystyle X}$, the identity map ${\displaystyle \operatorname {id} :X\rightarrow X}$ is a covering. Likewise for any discrete space ${\displaystyle D}$ the projection ${\displaystyle \pi :X\times D\rightarrow X}$ taking ${\displaystyle (x,i)\mapsto x}$ is a covering. Coverings of this type are called trivial coverings; if ${\displaystyle D}$ has finitely many (say ${\displaystyle k}$) elements, the covering is called the trivial ${\displaystyle k}$-sheeted covering of ${\displaystyle X}$.

• The map ${\displaystyle r:\mathbb {R} \to S^{1}}$ with ${\displaystyle r(t)=(\cos(2\pi t),\sin(2\pi t))}$ is a covering of the unit circle ${\displaystyle S^{1}}$. The base of the covering is ${\displaystyle S^{1}}$ and the covering space is ${\displaystyle \mathbb {R} }$. For any point ${\displaystyle x=(x_{1},x_{2})\in S^{1}}$ such that ${\displaystyle x_{1}>0}$, the set ${\displaystyle U:=\{(x_{1},x_{2})\in S^{1}\mid x_{1}>0\}}$ is an open neighborhood of ${\displaystyle x}$. The preimage of ${\displaystyle U}$ under ${\displaystyle r}$ is

  ${\displaystyle r^{-1}(U)=\displaystyle \bigsqcup _{n\in \mathbb {Z} }\left(n-{\frac {1}{4}},n+{\frac {1}{4}}\right)}$

and the sheets of the covering are ${\displaystyle V_{n}=(n-1/4,n+1/4)}$ for ${\displaystyle n\in \mathbb {Z} .}$ The fiber of ${\displaystyle x}$ is

${\displaystyle r^{-1}(x)=\{t\in \mathbb {R} \mid (\cos(2\pi t),\sin(2\pi t))=x\}.}$

• Another covering of the unit circle is the map ${\displaystyle q:S^{1}\to S^{1}}$ with ${\displaystyle q(z)=z^{n}}$ for some positive ${\displaystyle n\in \mathbb {N} .}$ For an open neighborhood ${\displaystyle U}$ of an ${\displaystyle x\in S^{1}}$, one has:

${\displaystyle q^{-1}(U)=\displaystyle \bigsqcup _{i=1}^{n}U}$.

• A map which is a local homeomorphism but not a covering of the unit circle is ${\displaystyle p:\mathbb {R_{+}} \to S^{1}}$ with ${\displaystyle p(t)=(\cos(2\pi t),\sin(2\pi t))}$. There is a sheet of an open neighborhood of ${\displaystyle (1,0)}$, which is not mapped homeomorphically onto ${\displaystyle U}$.

• Let ${\displaystyle n\geq 1}$ be odd. The map ${\displaystyle p:\mathrm {O} (n)\to \mathrm {SO} (n)}$ defined by ${\displaystyle p(Q)=(\det Q)Q}$ is a homomorphic double covering.

Since a covering ${\displaystyle \pi :E\rightarrow X}$ maps each of the disjoint open sets of ${\displaystyle \pi ^{-1}(U)}$ homeomorphically onto ${\displaystyle U}$ it is a local homeomorphism, i.e. ${\displaystyle \pi }$ is a continuous map and for every ${\displaystyle e\in E}$ there exists an open neighborhood ${\displaystyle V\subset E}$ of ${\displaystyle e}$, such that ${\displaystyle \pi |_{V}:V\rightarrow \pi (V)}$ is a homeomorphism.

It follows that the covering space ${\displaystyle E}$ and the base space ${\displaystyle X}$ locally share the same properties.

• If ${\displaystyle X}$ is a connected and non-orientable manifold, then there is a covering ${\displaystyle \pi :{\tilde {X}}\rightarrow X}$ of degree ${\displaystyle 2}$, whereby ${\displaystyle {\tilde {X}}}$ is a connected and orientable manifold.^[2]: 234
• If ${\displaystyle X}$ is a connected Lie group, then there is a covering ${\displaystyle \pi :{\tilde {X}}\rightarrow X}$ which is also a Lie group homomorphism and ${\displaystyle {\tilde {X}}:=\{\gamma :\gamma {\text{ is a path in X with }}\gamma (0)={\boldsymbol {1_{X}}}{\text{ modulo homotopy with fixed ends}}\}}$ is a Lie group.^[4]: 174
• If ${\displaystyle X}$ is a graph, then it follows for a covering ${\displaystyle \pi :E\rightarrow X}$ that ${\displaystyle E}$ is also a graph.^[2]: 85
• If ${\displaystyle X}$ is a connected manifold, then there is a covering ${\displaystyle \pi :{\tilde {X}}\rightarrow X}$, whereby ${\displaystyle {\tilde {X}}}$ is a connected and simply connected manifold.^[5]: 32
• If ${\displaystyle X}$ is a connected Riemann surface, then there is a covering ${\displaystyle \pi :{\tilde {X}}\rightarrow X}$ which is also a holomorphic map^[5]: 22 and ${\displaystyle {\tilde {X}}}$ is a connected and simply connected Riemann surface.^[5]: 32

Let ${\displaystyle X,Y}$ and ${\displaystyle E}$ be path-connected, locally path-connected spaces, and ${\displaystyle p,q}$ and ${\displaystyle r}$ be continuous maps, such that the diagram

commutes.

• If ${\displaystyle p}$ and ${\displaystyle q}$ are coverings, so is ${\displaystyle r}$.
• If ${\displaystyle p}$ and ${\displaystyle r}$ are coverings, so is ${\displaystyle q}$.^[6]: 485

Let ${\displaystyle X}$ and ${\displaystyle X'}$ be topological spaces and ${\displaystyle p:E\rightarrow X}$ and ${\displaystyle p':E'\rightarrow X'}$ be coverings, then ${\displaystyle p\times p':E\times E'\rightarrow X\times X'}$ with ${\displaystyle (p\times p')(e,e')=(p(e),p'(e'))}$ is a covering.^[6]: 339 However, coverings of ${\displaystyle X\times X'}$ are not all of this form in general.

Let ${\displaystyle X}$ be a topological space and ${\displaystyle p:E\rightarrow X}$ and ${\displaystyle p':E'\rightarrow X}$ be coverings. Both coverings are called equivalent, if there exists a homeomorphism ${\displaystyle h:E\rightarrow E'}$, such that the diagram

commutes. If such a homeomorphism exists, then one calls the covering spaces ${\displaystyle E}$ and ${\displaystyle E'}$ isomorphic.

All coverings satisfy the lifting property, i.e.:

Let ${\displaystyle I}$ be the unit interval and ${\displaystyle p:E\rightarrow X}$ be a covering. Let ${\displaystyle F:Y\times I\rightarrow X}$ be a continuous map and ${\displaystyle {\tilde {F}}_{0}:Y\times \{0\}\rightarrow E}$ be a lift of ${\displaystyle F|_{Y\times \{0\}}}$, i.e. a continuous map such that ${\displaystyle p\circ {\tilde {F}}_{0}=F|_{Y\times \{0\}}}$. Then there is a uniquely determined, continuous map ${\displaystyle {\tilde {F}}:Y\times I\rightarrow E}$ for which ${\displaystyle {\tilde {F}}(y,0)={\tilde {F}}_{0}}$ and which is a lift of ${\displaystyle F}$, i.e. ${\displaystyle p\circ {\tilde {F}}=F}$.^[2]: 60

If ${\displaystyle X}$ is a path-connected space, then for ${\displaystyle Y=\{0\}}$ it follows that the map ${\displaystyle {\tilde {F}}}$ is a lift of a path in ${\displaystyle X}$ and for ${\displaystyle Y=I}$ it is a lift of a homotopy of paths in ${\displaystyle X}$.

As a consequence, one can show that the fundamental group ${\displaystyle \pi _{1}(S^{1})}$ of the unit circle is an infinite cyclic group, which is generated by the homotopy classes of the loop ${\displaystyle \gamma :I\rightarrow S^{1}}$ with ${\displaystyle \gamma (t)=(\cos(2\pi t),\sin(2\pi t))}$.^[2]: 29

Let ${\displaystyle X}$ be a path-connected space and ${\displaystyle p:E\rightarrow X}$ be a connected covering. Let ${\displaystyle x,y\in X}$ be any two points, which are connected by a path ${\displaystyle \gamma }$, i.e. ${\displaystyle \gamma (0)=x}$ and ${\displaystyle \gamma (1)=y}$. Let ${\displaystyle {\tilde {\gamma }}}$ be the unique lift of ${\displaystyle \gamma }$, then the map

${\displaystyle L_{\gamma }:p^{-1}(x)\rightarrow p^{-1}(y)}$ with ${\displaystyle L_{\gamma }({\tilde {\gamma }}(0))={\tilde {\gamma }}(1)}$

is bijective.^[2]: 69

If ${\displaystyle X}$ is a path-connected space and ${\displaystyle p:E\rightarrow X}$ a connected covering, then the induced group homomorphism

${\displaystyle p_{\#}:\pi _{1}(E)\rightarrow \pi _{1}(X)}$ with ${\displaystyle p_{\#}([\gamma ])=[p\circ \gamma ]}$,

is injective and the subgroup ${\displaystyle p_{\#}(\pi _{1}(E))}$ of ${\displaystyle \pi _{1}(X)}$ consists of the homotopy classes of loops in ${\displaystyle X}$, whose lifts are loops in ${\displaystyle E}$.^[2]: 61

Let ${\displaystyle X}$ and ${\displaystyle Y}$ be Riemann surfaces, i.e. one dimensional complex manifolds, and let ${\displaystyle f:X\rightarrow Y}$ be a continuous map. ${\displaystyle f}$ is holomorphic in a point ${\displaystyle x\in X}$, if for any charts ${\displaystyle \phi _{x}:U_{1}\rightarrow V_{1}}$ of ${\displaystyle x}$ and ${\displaystyle \phi _{f(x)}:U_{2}\rightarrow V_{2}}$ of ${\displaystyle f(x)}$, with ${\displaystyle \phi _{x}(U_{1})\subset U_{2}}$, the map ${\displaystyle \phi _{f(x)}\circ f\circ \phi _{x}^{-1}:\mathbb {C} \rightarrow \mathbb {C} }$ is holomorphic.

If ${\displaystyle f}$ is holomorphic at all ${\displaystyle x\in X}$, we say ${\displaystyle f}$ is holomorphic.

The map ${\displaystyle F=\phi _{f(x)}\circ f\circ \phi _{x}^{-1}}$ is called the local expression of ${\displaystyle f}$ in ${\displaystyle x\in X}$.

If ${\displaystyle f:X\rightarrow Y}$ is a non-constant, holomorphic map between compact Riemann surfaces, then ${\displaystyle f}$ is surjective and an open map,^[5]: 11 i.e. for every open set ${\displaystyle U\subset X}$ the image ${\displaystyle f(U)\subset Y}$ is also open.

Let ${\displaystyle f:X\rightarrow Y}$ be a non-constant, holomorphic map between compact Riemann surfaces. For every ${\displaystyle x\in X}$ there exist charts for ${\displaystyle x}$ and ${\displaystyle f(x)}$ and there exists a uniquely determined ${\displaystyle k_{x}\in \mathbb {N_{>0}} }$, such that the local expression ${\displaystyle F}$ of ${\displaystyle f}$ in ${\displaystyle x}$ is of the form ${\displaystyle z\mapsto z^{k_{x}}}$.^[5]: 10 The number ${\displaystyle k_{x}}$ is called the ramification index of ${\displaystyle f}$ in ${\displaystyle x}$ and the point ${\displaystyle x\in X}$ is called a ramification point if ${\displaystyle k_{x}\geq 2}$. If ${\displaystyle k_{x}=1}$ for an ${\displaystyle x\in X}$, then ${\displaystyle x}$ is unramified. The image point ${\displaystyle y=f(x)\in Y}$ of a ramification point is called a branch point.

Let ${\displaystyle f:X\rightarrow Y}$ be a non-constant, holomorphic map between compact Riemann surfaces. The degree ${\displaystyle \operatorname {deg} (f)}$ of ${\displaystyle f}$ is the cardinality of the fiber of an unramified point ${\displaystyle y=f(x)\in Y}$, i.e. ${\displaystyle \operatorname {deg} (f):=|f^{-1}(y)|}$.

This number is well-defined, since for every ${\displaystyle y\in Y}$ the fiber ${\displaystyle f^{-1}(y)}$ is discrete^[5]: 20 and for any two unramified points ${\displaystyle y_{1},y_{2}\in Y}$, it is: ${\displaystyle |f^{-1}(y_{1})|=|f^{-1}(y_{2})|.}$

It can be calculated by:

${\displaystyle \sum _{x\in f^{-1}(y)}k_{x}=\operatorname {deg} (f)}$ ^[5]: 29

A continuous map ${\displaystyle f:X\rightarrow Y}$ is called a branched covering, if there exists a closed set with dense complement ${\displaystyle E\subset Y}$, such that ${\displaystyle f_{|X\smallsetminus f^{-1}(E)}:X\smallsetminus f^{-1}(E)\rightarrow Y\smallsetminus E}$ is a covering.

• Let ${\displaystyle n\in \mathbb {N} }$ and ${\displaystyle n\geq 2}$, then ${\displaystyle f:\mathbb {C} \rightarrow \mathbb {C} }$ with ${\displaystyle f(z)=z^{n}}$ is a branched covering of degree ${\displaystyle n}$, where by ${\displaystyle z=0}$ is a branch point.
• Every non-constant, holomorphic map between compact Riemann surfaces ${\displaystyle f:X\rightarrow Y}$ of degree ${\displaystyle d}$ is a branched covering of degree ${\displaystyle d}$.

Let ${\displaystyle p:{\tilde {X}}\rightarrow X}$ be a simply connected covering. If ${\displaystyle \beta :E\rightarrow X}$ is another simply connected covering, then there exists a uniquely determined homeomorphism ${\displaystyle \alpha :{\tilde {X}}\rightarrow E}$, such that the diagram

commutes.^[6]: 482

This means that ${\displaystyle p}$ is, up to equivalence, uniquely determined and because of that universal property denoted as the universal covering of the space ${\displaystyle X}$.

A universal covering does not always exist. The following theorem guarantees its existence for a certain class of base spaces.

Let ${\displaystyle X}$ be a connected, locally simply connected topological space. Then, there exists a universal covering ${\displaystyle p:{\tilde {X}}\rightarrow X.}$

The set ${\displaystyle {\tilde {X}}}$ is defined as ${\displaystyle {\tilde {X}}=\{\gamma :\gamma {\text{ is a path in }}X{\text{ with }}\gamma (0)=x_{0}\}/{\text{homotopy with fixed ends}},}$ where ${\displaystyle x_{0}\in X}$ is any chosen base point. The map ${\displaystyle p:{\tilde {X}}\rightarrow X}$ is defined by ${\displaystyle p([\gamma ])=\gamma (1).}$^[2]: 64

The topology on ${\displaystyle {\tilde {X}}}$ is constructed as follows: Let ${\displaystyle \gamma :I\rightarrow X}$ be a path with ${\displaystyle \gamma (0)=x_{0}.}$ Let ${\displaystyle U}$ be a simply connected neighborhood of the endpoint ${\displaystyle x=\gamma (1).}$ Then, for every ${\displaystyle y\in U,}$ there is a path ${\displaystyle \sigma _{y}}$ inside ${\displaystyle U}$ from ${\displaystyle x}$ to ${\displaystyle y}$ that is unique up to homotopy. Now consider the set ${\displaystyle {\tilde {U}}=\{\gamma \sigma _{y}:y\in U\}/{\text{homotopy with fixed ends}}.}$ The restriction ${\displaystyle p|_{\tilde {U}}:{\tilde {U}}\rightarrow U}$ with ${\displaystyle p([\gamma \sigma _{y}])=\gamma \sigma _{y}(1)=y}$ is a bijection and ${\displaystyle {\tilde {U}}}$ can be equipped with the final topology of ${\displaystyle p|_{\tilde {U}}.}$^{[further explanation needed]}

The fundamental group ${\displaystyle \pi _{1}(X,x_{0})=\Gamma }$ acts freely on ${\displaystyle {\tilde {X}}}$ by ${\displaystyle ([\gamma ],[{\tilde {x}}])\mapsto [\gamma {\tilde {x}}],}$ and the orbit space ${\displaystyle \Gamma \backslash {\tilde {X}}}$ is homeomorphic to ${\displaystyle X}$ through the map ${\displaystyle [\Gamma {\tilde {x}}]\mapsto {\tilde {x}}(1).}$

• ${\displaystyle p:\mathbb {R} \to S^{1}}$ with ${\displaystyle p(t)=(\cos(2\pi t),\sin(2\pi t))}$ is the universal covering of the unit circle ${\displaystyle S^{1}}$.
• ${\displaystyle p:S^{n}\to \mathbb {R} P^{n}\cong \{+1,-1\}\backslash S^{n}}$ with ${\displaystyle p(x)=[x]}$ is the universal covering of the projective space ${\displaystyle \mathbb {R} P^{n}}$ for ${\displaystyle n>1}$.

• ${\displaystyle p:\mathrm {SU} (n)\times \mathbb {R} \to U(n)}$ with $${\displaystyle p(A,t)=\exp(2\pi it)A}$$ is the universal covering of the unitary group ${\displaystyle U(n)}$.

• Since ${\displaystyle \mathrm {SU} (2)\cong S^{3}}$, it follows that the quotient map $${\displaystyle p:\mathrm {SU} (2)\rightarrow \mathrm {SU} (2)/\mathbb {Z_{2}} \cong \mathrm {SO} (3)}$$ is the universal covering of ${\displaystyle \mathrm {SO} (3)}$.

• A topological space which has no universal covering is the Hawaiian earring: $${\displaystyle X=\bigcup _{n\in \mathbb {N} }\left\{(x_{1},x_{2})\in \mathbb {R} ^{2}:{\Bigl (}x_{1}-{\frac {1}{n}}{\Bigr )}^{2}+x_{2}^{2}={\frac {1}{n^{2}}}\right\}}$$ One can show that no neighborhood of the origin ${\displaystyle (0,0)}$ is simply connected.^{[6]: 487, Example 1}

Let G be a discrete group acting on the topological space X. This means that each element g of G is associated to a homeomorphism H_g of X onto itself, in such a way that H_{g h} is always equal to H_g ${\displaystyle \circ }$ H_h for any two elements g and h of G. (Or in other words, a group action of the group G on the space X is just a group homomorphism of the group G into the group Homeo(X) of self-homeomorphisms of X.) It is natural to ask under what conditions the projection from X to the orbit space X/G is a covering map. This is not always true since the action may have fixed points. An example for this is the cyclic group of order 2 acting on a product X × X by the twist action where the non-identity element acts by (x, y) ↦ (y, x). Thus the study of the relation between the fundamental groups of X and X/G is not so straightforward.

However the group G does act on the fundamental groupoid of X, and so the study is best handled by considering groups acting on groupoids, and the corresponding orbit groupoids. The theory for this is set down in Chapter 11 of the book Topology and groupoids referred to below. The main result is that for discontinuous actions of a group G on a Hausdorff space X which admits a universal cover, then the fundamental groupoid of the orbit space X/G is isomorphic to the orbit groupoid of the fundamental groupoid of X, i.e. the quotient of that groupoid by the action of the group G. This leads to explicit computations, for example of the fundamental group of the symmetric square of a space.

Let E and M be smooth manifolds with or without boundary. A covering ${\displaystyle \pi :E\to M}$ is called a smooth covering if it is a smooth map and the sheets are mapped diffeomorphically onto the corresponding open subset of M. (This is in contrast to the definition of a covering, which merely requires that the sheets are mapped homeomorphically onto the corresponding open subset.)

Let ${\displaystyle p:E\rightarrow X}$ be a covering. A deck transformation is a homeomorphism ${\displaystyle d:E\rightarrow E}$, such that the diagram of continuous maps

commutes. Together with the composition of maps, the set of deck transformation forms a group ${\displaystyle \operatorname {Deck} (p)}$, which is the same as ${\displaystyle \operatorname {Aut} (p)}$.

Now suppose ${\displaystyle p:C\to X}$ is a covering map and ${\displaystyle C}$ (and therefore also ${\displaystyle X}$) is connected and locally path connected. The action of ${\displaystyle \operatorname {Aut} (p)}$ on each fiber is free. If this action is transitive on some fiber, then it is transitive on all fibers, and we call the cover regular (or normal or Galois). Every such regular cover is a principal ${\displaystyle G}$-bundle, where ${\displaystyle G=\operatorname {Aut} (p)}$ is considered as a discrete topological group.

Every universal cover ${\displaystyle p:D\to X}$ is regular, with deck transformation group being isomorphic to the fundamental group ${\displaystyle \pi _{1}(X)}$.

• Let ${\displaystyle q:S^{1}\to S^{1}}$ be the covering ${\displaystyle q(z)=z^{n}}$ for some ${\displaystyle n\in \mathbb {N} }$, then the map ${\displaystyle d_{k}:S^{1}\rightarrow S^{1}:z\mapsto z\,e^{2\pi ik/n}}$ for ${\displaystyle k\in \mathbb {Z} }$ is a deck transformation and ${\displaystyle \operatorname {Deck} (q)\cong \mathbb {Z} /n\mathbb {Z} }$.
• Let ${\displaystyle r:\mathbb {R} \to S^{1}}$ be the covering ${\displaystyle r(t)=(\cos(2\pi t),\sin(2\pi t))}$, then the map ${\displaystyle d_{k}:\mathbb {R} \rightarrow \mathbb {R} :t\mapsto t+k}$ for ${\displaystyle k\in \mathbb {Z} }$ is a deck transformation and ${\displaystyle \operatorname {Deck} (r)\cong \mathbb {Z} }$.
• As another important example, consider ${\displaystyle \mathbb {C} }$ the complex plane and ${\displaystyle \mathbb {C} ^{\times }}$ the complex plane minus the origin. Then the map ${\displaystyle p:\mathbb {C} ^{\times }\to \mathbb {C} ^{\times }}$ with ${\displaystyle p(z)=z^{n}}$ is a regular cover. The deck transformations are multiplications with ${\displaystyle n}$-th roots of unity and the deck transformation group is therefore isomorphic to the cyclic group ${\displaystyle \mathbb {Z} /n\mathbb {Z} }$. Likewise, the map ${\displaystyle \exp :\mathbb {C} \to \mathbb {C} ^{\times }}$ with ${\displaystyle \exp(z)=e^{z}}$ is the universal cover.

Let ${\displaystyle X}$ be a path-connected space and ${\displaystyle p:E\rightarrow X}$ be a connected covering. Since a deck transformation ${\displaystyle d:E\rightarrow E}$ is bijective, it permutes the elements of a fiber ${\displaystyle p^{-1}(x)}$ with ${\displaystyle x\in X}$ and is uniquely determined by where it sends a single point. In particular, only the identity map fixes a point in the fiber.^[2]: 70 Because of this property every deck transformation defines a group action on ${\displaystyle E}$, i.e. let ${\displaystyle U\subset X}$ be an open neighborhood of a ${\displaystyle x\in X}$ and ${\displaystyle {\tilde {U}}\subset E}$ an open neighborhood of an ${\displaystyle e\in p^{-1}(x)}$, then ${\displaystyle \operatorname {Deck} (p)\times E\rightarrow E:(d,{\tilde {U}})\mapsto d({\tilde {U}})}$ is a group action.

A covering ${\displaystyle p:E\rightarrow X}$ is called normal, if ${\displaystyle \operatorname {Deck} (p)\backslash E\cong X}$. This means, that for every ${\displaystyle x\in X}$ and any two ${\displaystyle e_{0},e_{1}\in p^{-1}(x)}$ there exists a deck transformation ${\displaystyle d:E\rightarrow E}$, such that ${\displaystyle d(e_{0})=e_{1}}$.

Let ${\displaystyle X}$ be a path-connected space and ${\displaystyle p:E\rightarrow X}$ be a connected covering. Let ${\displaystyle H=p_{\#}(\pi _{1}(E))}$ be a subgroup of ${\displaystyle \pi _{1}(X)}$, then ${\displaystyle p}$ is a normal covering iff ${\displaystyle H}$ is a normal subgroup of ${\displaystyle \pi _{1}(X)}$.

If ${\displaystyle p:E\rightarrow X}$ is a normal covering and ${\displaystyle H=p_{\#}(\pi _{1}(E))}$, then ${\displaystyle \operatorname {Deck} (p)\cong \pi _{1}(X)/H}$.

If ${\displaystyle p:E\rightarrow X}$ is a path-connected covering and ${\displaystyle H=p_{\#}(\pi _{1}(E))}$, then ${\displaystyle \operatorname {Deck} (p)\cong N(H)/H}$, whereby ${\displaystyle N(H)}$ is the normaliser of ${\displaystyle H}$.^[2]: 71

Let ${\displaystyle E}$ be a topological space. A group ${\displaystyle \Gamma }$ acts discontinuously on ${\displaystyle E}$, if every ${\displaystyle e\in E}$ has an open neighborhood ${\displaystyle V\subset E}$ with ${\displaystyle V\neq \emptyset }$, such that for every ${\displaystyle d_{1},d_{2}\in \Gamma }$ with ${\displaystyle d_{1}V\cap d_{2}V\neq \emptyset }$ one has ${\displaystyle d_{1}=d_{2}}$.

If a group ${\displaystyle \Gamma }$ acts discontinuously on a topological space ${\displaystyle E}$, then the quotient map ${\displaystyle q:E\rightarrow \Gamma \backslash E}$ with ${\displaystyle q(e)=\Gamma e}$ is a normal covering.^[2]: 72 Hereby ${\displaystyle \Gamma \backslash E=\{\Gamma e:e\in E\}}$ is the quotient space and ${\displaystyle \Gamma e=\{\gamma (e):\gamma \in \Gamma \}}$ is the orbit of the group action.

• The covering ${\displaystyle q:S^{1}\to S^{1}}$ with ${\displaystyle q(z)=z^{n}}$ is a normal coverings for every ${\displaystyle n\in \mathbb {N} }$.
• Every simply connected covering is a normal covering.

Let ${\displaystyle \Gamma }$ be a group, which acts discontinuously on a topological space ${\displaystyle E}$ and let ${\displaystyle q:E\rightarrow \Gamma \backslash E}$ be the normal covering.

• If ${\displaystyle E}$ is path-connected, then ${\displaystyle \operatorname {Deck} (q)\cong \Gamma }$.^[2]: 72

• If ${\displaystyle E}$ is simply connected, then ${\displaystyle \operatorname {Deck} (q)\cong \pi _{1}(\Gamma \backslash E)}$.^[2]: 71

• Let ${\displaystyle n\in \mathbb {N} }$. The antipodal map ${\displaystyle g:S^{n}\rightarrow S^{n}}$ with ${\displaystyle g(x)=-x}$ generates, together with the composition of maps, a group ${\displaystyle D(g)\cong \mathbb {Z/2Z} }$ and induces a group action ${\displaystyle D(g)\times S^{n}\rightarrow S^{n},(g,x)\mapsto g(x)}$, which acts discontinuously on ${\displaystyle S^{n}}$. Because of ${\displaystyle \mathbb {Z_{2}} \backslash S^{n}\cong \mathbb {R} P^{n}}$ it follows, that the quotient map ${\displaystyle q:S^{n}\rightarrow \mathbb {Z_{2}} \backslash S^{n}\cong \mathbb {R} P^{n}}$ is a normal covering and for ${\displaystyle n>1}$ a universal covering, hence ${\displaystyle \operatorname {Deck} (q)\cong \mathbb {Z/2Z} \cong \pi _{1}({\mathbb {R} P^{n}})}$ for ${\displaystyle n>1}$.
• Let ${\displaystyle \mathrm {SO} (3)}$ be the special orthogonal group, then the map ${\displaystyle f:\mathrm {SU} (2)\rightarrow \mathrm {SO} (3)\cong \mathbb {Z_{2}} \backslash \mathrm {SU} (2)}$ is a normal covering and because of ${\displaystyle \mathrm {SU} (2)\cong S^{3}}$, it is the universal covering, hence ${\displaystyle \operatorname {Deck} (f)\cong \mathbb {Z/2Z} \cong \pi _{1}(\mathrm {SO} (3))}$.
• With the group action ${\displaystyle (z_{1},z_{2})*(x,y)=(z_{1}+(-1)^{z_{2}}x,z_{2}+y)}$ of ${\displaystyle \mathbb {Z^{2}} }$ on ${\displaystyle \mathbb {R^{2}} }$, whereby ${\displaystyle (\mathbb {Z^{2}} ,*)}$ is the semidirect product ${\displaystyle \mathbb {Z} \rtimes \mathbb {Z} }$, one gets the universal covering ${\displaystyle f:\mathbb {R^{2}} \rightarrow (\mathbb {Z} \rtimes \mathbb {Z} )\backslash \mathbb {R^{2}} \cong K}$ of the klein bottle ${\displaystyle K}$, hence ${\displaystyle \operatorname {Deck} (f)\cong \mathbb {Z} \rtimes \mathbb {Z} \cong \pi _{1}(K)}$.
• Let ${\displaystyle T=S^{1}\times S^{1}}$ be the torus which is embedded in the ${\displaystyle \mathbb {C^{2}} }$. Then one gets a homeomorphism ${\displaystyle \alpha :T\rightarrow T:(e^{ix},e^{iy})\mapsto (e^{i(x+\pi )},e^{-iy})}$, which induces a discontinuous group action ${\displaystyle G_{\alpha }\times T\rightarrow T}$, whereby ${\displaystyle G_{\alpha }\cong \mathbb {Z/2Z} }$. It follows, that the map ${\displaystyle f:T\rightarrow G_{\alpha }\backslash T\cong K}$ is a normal covering of the klein bottle, hence ${\displaystyle \operatorname {Deck} (f)\cong \mathbb {Z/2Z} }$.
• Let ${\displaystyle S^{3}}$ be embedded in the ${\displaystyle \mathbb {C^{2}} }$. Since the group action ${\displaystyle S^{3}\times \mathbb {Z/pZ} \rightarrow S^{3}:((z_{1},z_{2}),[k])\mapsto (e^{2\pi ik/p}z_{1},e^{2\pi ikq/p}z_{2})}$ is discontinuously, whereby ${\displaystyle p,q\in \mathbb {N} }$ are coprime, the map ${\displaystyle f:S^{3}\rightarrow \mathbb {Z_{p}} \backslash S^{3}=:L_{p,q}}$ is the universal covering of the lens space ${\displaystyle L_{p,q}}$, hence ${\displaystyle \operatorname {Deck} (f)\cong \mathbb {Z/pZ} \cong \pi _{1}(L_{p,q})}$.

Let ${\displaystyle X}$ be a connected and locally simply connected space, then for every subgroup ${\displaystyle H\subseteq \pi _{1}(X)}$ there exists a path-connected covering ${\displaystyle \alpha :X_{H}\rightarrow X}$ with ${\displaystyle \alpha _{\#}(\pi _{1}(X_{H}))=H}$.^[2]: 66

Let ${\displaystyle p_{1}:E\rightarrow X}$ and ${\displaystyle p_{2}:E'\rightarrow X}$ be two path-connected coverings, then they are equivalent iff the subgroups ${\displaystyle H=p_{1\#}(\pi _{1}(E))}$ and ${\displaystyle H'=p_{2\#}(\pi _{1}(E'))}$ are conjugate to each other.^[6]: 482

Let ${\displaystyle X}$ be a connected and locally simply connected space, then, up to equivalence between coverings, there is a bijection:

${\displaystyle {\begin{matrix}\qquad \displaystyle \{{\text{Subgroup of }}\pi _{1}(X)\}&\longleftrightarrow &\displaystyle \{{\text{path-connected covering }}p:E\rightarrow X\}\\H&\longrightarrow &\alpha :X_{H}\rightarrow X\\p_{\#}(\pi _{1}(E))&\longleftarrow &p\\\displaystyle \{{\text{normal subgroup of }}\pi _{1}(X)\}&\longleftrightarrow &\displaystyle \{{\text{normal covering }}p:E\rightarrow X\}\end{matrix}}}$

For a sequence of subgroups ${\displaystyle \displaystyle \{{\text{e}}\}\subset H\subset G\subset \pi _{1}(X)}$ one gets a sequence of coverings ${\displaystyle {\tilde {X}}\longrightarrow X_{H}\cong H\backslash {\tilde {X}}\longrightarrow X_{G}\cong G\backslash {\tilde {X}}\longrightarrow X\cong \pi _{1}(X)\backslash {\tilde {X}}}$. For a subgroup ${\displaystyle H\subset \pi _{1}(X)}$ with index ${\displaystyle \displaystyle [\pi _{1}(X):H]=d}$, the covering ${\displaystyle \alpha :X_{H}\rightarrow X}$ has degree ${\displaystyle d}$.

Let ${\displaystyle X}$ be a topological space. The objects of the category ${\displaystyle {\boldsymbol {Cov(X)}}}$ are the coverings ${\displaystyle p:E\rightarrow X}$ of ${\displaystyle X}$ and the morphisms between two coverings ${\displaystyle p:E\rightarrow X}$ and ${\displaystyle q:F\rightarrow X}$ are continuous maps ${\displaystyle f:E\rightarrow F}$, such that the diagram