The orbit map of a covering-space action is a covering, with the acting group equal to the deck group when the total space is path-connected

Given: The objects, hypotheses, and choice principles stated above.

A left action of a group $G$ on a space $E$ by homeomorphisms is a covering-space action when every $e\in E$ has an open neighbourhood $U$ such that $gU\cap U=\mathrm{\varnothing }$ for every nonidentity $g\in G$ (def-group-action, def-homeomorphism-and-open-maps). Acting by homeomorphisms means that each map $e\mapsto g\cdot e$ is a homeomorphism of $E$; the underlying set action alone would not make the translates $gU$ open. This condition implies freeness (def-free-group-action) and makes the translates of $U$ pairwise disjoint. (Covering-space actions by disjoint translates of neighbourhoods).

A covering map is a continuous surjection $p:E\to B$ such that every $b\in B$ has an open neighbourhood $U$ for which $p^{-1}(U)$ is a disjoint union of open sets $V_j$, called sheets, and each restriction $p{\mathrm{\mid }}_{V_j}:V_j\to U$ is a homeomorphism (def-continuous-map-top, def-homeomorphism-and-open-maps, def-disjoint-union-topology). Such a $U$ is evenly covered, and $p^{-1}(b)$ is the fibre over $b$. A covering is trivial when it is isomorphic over $B$ to a product projection $B\times F\to B$ with $F$ discrete. (Covering maps, evenly covered neighbourhoods, fibres, sheets, and trivial coverings).

The quotient topology. Let $(X,\mathcal{T})$ be a topological space (def-topological-space), let $Y$ be a set and let $q:X\to Y$ be a surjection (def-injection-surjection-bijection). The quotient topology on $Y$ induced by $q$ is the final topology of the one-element family $(q)$ (def-initial-and-final-topology): ${\mathcal{T}}_q:=\{V\subseteq Y:q^{-1}[V]\in \mathcal{T}\}\mathrm{.}$ That this is a topology is discharged in def-initial-and-final-topology, where every final topology is verified to satisfy (T1), (T2) and (T3). Dually, $C\subseteq Y$ is closed in ${\mathcal{T}}_q$ exactly when $q^{-1}[C]$ is closed in $X$, because $q^{-1}[Y\setminus V]=X\setminus q^{-1}[V]$. (The quotient topology of a surjection, quotient maps, saturated sets, and the quotient of a space by an equivalence relation with its canonical projection).

For a covering $p:E\to B$, a deck transformation is an isomorphism $h:E\to E$ over $B$, so $p\circ h=p$ (def-map-and-isomorphism-of-covering-spaces). Deck transformations form the deck group $\mathrm{Deck}(p)$ under composition, and this group acts on $E$ by evaluation (def-group, def-group-action). (Deck transformations and the deck-transformation group of a covering).

For a covering with connected total space, two deck transformations agreeing at one point are equal. Consequently the deck group acts freely on the total space. (On a connected covering space, a deck transformation is determined by one point and the deck action is free).

Let $X$ and $Y$ be topological spaces and let $q:X\to Y$ be continuous (def-continuous-map-top). Each of the following three conditions makes $q$ a quotient map (def-quotient-topology). 1. $q$ is a surjection and an open map (def-homeomorphism-and-open-maps). 2. $q$ is a surjection and a closed map. 3. $q$ admits a continuous section: a continuous $s:Y\to X$ with $q\circ s={\mathrm{i}\mathrm{d}}_Y$. (Surjectivity of $q$ is then automatic and need not be assumed.) Neither clause 1 nor clause 2 is necessary: a quotient map need be neither open nor closed. A witness that is a quotient map by clause 3 while failing clauses 1 and 2 is worked on the companion page, and is named in the remarks below. (A continuous open surjection, a continuous closed surjection, and a continuous surjection admitting a continuous section are all quotient maps).