# General Topology/Covering spaces

Definition (covering space):
Let $X$ be a topological space. Another topological space ${\tilde {X}}$ is called a covering space of $X$ if and only along with ${\tilde {X}}$ comes a continuous function $\pi :{\tilde {X}}\to X$ such that the following condition is satisfied:
Definition (covering map):
Let ${\tilde {X}}$ be a covering space of the topological space $X$ and let $\pi :{\tilde {X}}\to X$ be the continuous map that goes along with it. Then $\pi$ is called the covering map of the covering space ${\tilde {X}}$.
Note that more formally, one could have defined a covering space to be equal to a pair $({\tilde {X}},\pi )$, where $\pi$ is the covering map that belongs to the covering space; indeed, many covering maps may be possible if $\pi$ is not specified. But as with group operations, we will mostly suppress $\pi$ from the notation for brevity.
Definition (evenly covered neighbourhood):
Let $X$ be a topological space and let ${\tilde {X}}$ be a covering space of $X$, where $\pi :{\tilde {X}}\to X$ is the covering map. Let $x\in X$. The set $U$ guaranteed by the definition of a covering space (so that $\pi$, restricted to adequate subsets of the preimage, is a homeomorphism) is called an evenly covered neighbourhood of $x$.
Definition (lift):
Let $Z,X$ be topological spaces, and let $f:Z\to X$ be a continuous function. If ${\tilde {X}}$ is a covering space of $X$, then a lift of $f$ to ${\tilde {X}}$ is a continuous function ${\tilde {f}}:Z\to {\tilde {X}}$ so that, when $\pi$ is the covering map that belongs to ${\tilde {X}}$, $\pi \circ {\tilde {f}}=f$.
When the domain of the function that is to be lifted is connected, then lifts are, in a certain sense, unique:
Proposition (maps of connected domain lift uniquely):
Let $f:Z\to X$ be a continuous function between topological spaces $Z,X$ where $Z$ is connected, and suppose that ${\tilde {X}}$ is a covering space of $X$. Then if two lifts ${\tilde {f}}_{1},{\tilde {f}}_{2}:Z\to {\tilde {X}}$ of $f$ coincide in one point $z_{0}\in Z$, they are equal.
Proof: Let $S\subseteq Z$ be the set on which ${\tilde {f}}_{1}$ and ${\tilde {f}}_{2}$ agree. Since ${\tilde {f}}_{1}$ adn ${\tilde {f}}_{2}$ agree on $z_{0}$, $S$ is nonempty. Further, if $z\in S$, set $x:=f(z)$ and let $U$ be an evenly covered neighbourhood of $x$. Let $V_{\alpha }$ be the disjointed component in which ${\tilde {x}}:={\tilde {f}}_{1}(z)={\tilde {f}}_{2}(z)$ lies, so that $\pi |_{V_{\alpha }}$ is a homeomorphism between $V_{\alpha }$ and $U$. Then define $W:={\tilde {f}}_{1}^{-1}(V_{\alpha })\cap {\tilde {f}}_{2}^{-1}(V_{\alpha })$, both of which are open since ${\tilde {f}}_{1}$, ${\tilde {f}}_{2}$ are continuous. We claim that in fact ${\tilde {f}}_{1}={\tilde {f}}_{2}$ on $W$; indeed, we may note that $\pi \circ {\tilde {f}}_{1}=\pi \circ {\tilde {f}}_{2}=f$ implies that $\pi |_{V_{\alpha }}\circ {\tilde {f}}_{1}|_{W}=\pi |_{V_{\alpha }}\circ {\tilde {f}}_{2}|_{W}$, from which identity of ${\tilde {f}}_{1},{\tilde {f}}_{2}$ follows upon composing on the left with $\pi |_{V_{\alpha }}^{-1}$. We conclude that $S$ is open. However, suppose that $z\notin S$, suppose that $U$ is an evenly covered neighbourhood of $f(z)$ and let $V_{\alpha }$ respectively $V_{\beta }$ be the (disjoint) components of $\pi ^{-1}(V)$, so that $\pi |_{V_{\alpha }}$ and $\pi |_{V_{\beta }}$ are homeomorphisms. Then define $W:={\tilde {f}}_{1}^{-1}(V_{\alpha })\cap {\tilde {f}}_{2}^{-1}(V_{\beta })$. Whenever $z\in W$, we will have ${\tilde {f}}_{1}(w)\in V_{\alpha }$ and ${\tilde {f}}_{2}(w)\in V_{\beta }$, so that on $W$ the functions ${\tilde {f}}_{1}$ and ${\tilde {f}}_{2}$ disagree on every point. Thus, we get that $Z\setminus S$ is open, so that $S$ is open and closed, and since $Z$ is connected, $S=Z$. $\Box$
Proposition (lifting paths):
Let $X$ be a topological space, let $\gamma :[0,1]\to X$ be a path, and let ${\tilde {X}}$ (together with a suitable covering map $\pi$) be a covering space of $X$. Set $x:=\gamma (0)\in X$. Then for each ${\tilde {x}}\in \pi ^{-1}(x)$, there exists a unique curve ${\tilde {\gamma }}:[0,1]\to {\tilde {X}}$ such that $\pi \circ {\tilde {\gamma }}=\gamma$.
Proof: Note that $[0,1]$ is compact. For each $x\in \gamma ([0,1])$, pick an evenly covered neighbourhood $U_{x}$ of $x$. Since $[0,1]$ is compact, upon defining $V_{x}:=\gamma ^{-1}(U_{x})$ and then writing each $V_{x}$ as
    $V_{x}=\bigcup _{\beta \in B_{x}}W_{\beta }$
where the $W_{\beta }$ are intervals (note that the intervals form a basis of the Euclidean topology) and then considering the open cover $(W_{\beta })_{\beta \in \cup _{x}B_{x}}$, we may pick a finite number of intervals $W_{1},\ldots ,W_{n}$ which cover $[0,1]$ and whose images via $\gamma$ are evenly covered. We assume that the intervals $W_{1},\ldots ,W_{n}$ are ordered increasingly according to their starting points. Now we define ${\tilde {\gamma }}$ successively on these intervals. For $t\in W_{1}$, we note that $\gamma (W_{1})$ is evenly covered and contains $x$. Suppose that $U$ is an evenly covered neighbourhood of $\gamma (W_{1})$, and let $(V_{\alpha })_{\alpha \in A}$ be the components of $\pi ^{-1}(U)$ so that $\pi$ restricts to a homeomorphism on them. Let $\alpha _{0}\in A$ so that ${\tilde {x}}\in V_{\alpha _{0}}$. Then define ${\tilde {\gamma }}(t):=\pi |_{V_{\alpha _{0}}}^{-1}\circ \gamma (t)$ for $t\in W_{1}$ and observe that ${\tilde {\gamma }}(0)={\tilde {x}}$, since $\pi ({\tilde {x}})=\pi ({\tilde {\gamma }}(0))=\gamma (0)=x$ and ${\tilde {\gamma }}(0)$ and ${\tilde {x}}$ are both in $V_{\alpha _{0}}$. (Note that a homeomorphism is in particular bijective and that the definition of ${\tilde {\gamma }}$ is independent of the choice of interval, therefore ${\tilde {\gamma }}$ is well defined even on the intersection $W_{i}\cap W_{i+1}$.) Proceed similarly for the ensuing intervals $W_{2},\ldots ,W_{n}$. Then ${\tilde {\gamma }}$ will be continuous on all intervals of the form $[a_{j-1},a_{j}]$, where $a_{j-1}$ is the beginning point of $W_{j}$ and $a_{n}=1$, since the interval $[a_{j-1},a_{j}]$ is contained in $W_{j}$. Hence, since a function which is continuous on two closed sets is continuous on their union, we obtain that ${\tilde {\gamma }}$ is continuous on $[a_{0},a_{2}]$, then on $[a_{0},a_{3}]$, and so on, and finally on $[a_{0},a_{n}]=[0,1]$. ${\tilde {\gamma }}$ is then the desired lift, and it is unique since maps of connected domain lift uniquely. $\Box$
Proposition (lifting homotopies):
Let $Z,X$ be topological spaces and let $H:[0,1]\times Z\to X$ be a continuous function. Suppose that ${\tilde {X}}$ is a covering space of $X$, and that we are given a lift ${\tilde {H}}_{0}:Z\to {\tilde {X}}$ of the function $H_{0}:Z\to X$ defined by $H_{0}(z):=H(0,z)$. Then there exists a unique continuous function ${\tilde {H}}:[0,1]\times Z\to X$ so that for one thing ${\tilde {H}}_{0}(z)={\tilde {H}}(0,z)$ for all $z\in Z$ and further $\pi \circ {\tilde {H}}=H$.
Proof: For each $z\in Z$, $\gamma _{z}(t):=H(t,z)$ lifts uniquely to a path ${\tilde {\gamma }}_{z}$ so that ${\tilde {\gamma }}_{z}(0)={\tilde {H}}(0,z)$. Define ${\tilde {H}}(t,z):=\gamma _{z}(t)$. Obviously, $\pi \circ {\tilde {H}}=H$, and further, we claim that ${\tilde {H}}$ is continuous. Let $(t,z)\in [0,1]\times Z$ be arbitrary. Let $U\subseteq X$ be an evenly covered neighbourhood of $H(t,x)$; then $\pi \upharpoonright _{U}$ is a homeomorphism that has a continuous inverse $(\pi \upharpoonright _{U})^{-1}$. Now by definition of the product topology, take $W\subseteq Z$ open and $\epsilon >0$ so that $(t-\epsilon ,t+\epsilon )\times W\subseteq H^{-1}(U)$. By uniqueness of path lifting, we have ${\tilde {H}}=(\pi \upharpoonright _{U})^{-1}\circ H$ on $(t-\epsilon ,t+\epsilon )\times W$, which is continuous. We conclude that ${\tilde {H}}$ is continuous, since it is continuous in an open set about an arbitrary point. $\Box$