# General Topology/Covering spaces

Definition (covering space):

Let ${\displaystyle X}$ be a topological space. Another topological space ${\displaystyle {\tilde {X}}}$ is called a covering space of ${\displaystyle X}$ if and only along with ${\displaystyle {\tilde {X}}}$ comes a continuous function ${\displaystyle \pi :{\tilde {X}}\to X}$ such that the following condition is satisfied:

• For each ${\displaystyle x\in X}$, there exists a neighbourhood ${\displaystyle U}$ of ${\displaystyle x}$ so that ${\displaystyle \pi ^{-1}(U)}$ is the disjoint union of open subsets ${\displaystyle (V_{\alpha })_{\alpha \in A}}$ such that for all ${\displaystyle \alpha \in A}$ the function ${\displaystyle \pi |_{V_{\alpha }}}$ is a homeomorphism between ${\displaystyle V_{\alpha }}$ and ${\displaystyle U}$.

Definition (covering map):

Let ${\displaystyle {\tilde {X}}}$ be a covering space of the topological space ${\displaystyle X}$ and let ${\displaystyle \pi :{\tilde {X}}\to X}$ be the continuous map that goes along with it. Then ${\displaystyle \pi }$ is called the covering map of the covering space ${\displaystyle {\tilde {X}}}$.

Note that more formally, one could have defined a covering space to be equal to a pair ${\displaystyle ({\tilde {X}},\pi )}$, where ${\displaystyle \pi }$ is the covering map that belongs to the covering space; indeed, many covering maps may be possible if ${\displaystyle \pi }$ is not specified. But as with group operations, we will mostly suppress ${\displaystyle \pi }$ from the notation for brevity.

Definition (evenly covered neighbourhood):

Let ${\displaystyle X}$ be a topological space and let ${\displaystyle {\tilde {X}}}$ be a covering space of ${\displaystyle X}$, where ${\displaystyle \pi :{\tilde {X}}\to X}$ is the covering map. Let ${\displaystyle x\in X}$. The set ${\displaystyle U}$ guaranteed by the definition of a covering space (so that ${\displaystyle \pi }$, restricted to adequate subsets of the preimage, is a homeomorphism) is called an evenly covered neighbourhood of ${\displaystyle x}$.

Definition (lift):

Let ${\displaystyle Z,X}$ be topological spaces, and let ${\displaystyle f:Z\to X}$ be a continuous function. If ${\displaystyle {\tilde {X}}}$ is a covering space of ${\displaystyle X}$, then a lift of ${\displaystyle f}$ to ${\displaystyle {\tilde {X}}}$ is a continuous function ${\displaystyle {\tilde {f}}:Z\to {\tilde {X}}}$ so that, when ${\displaystyle \pi }$ is the covering map that belongs to ${\displaystyle {\tilde {X}}}$, ${\displaystyle \pi \circ {\tilde {f}}=f}$.

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 ${\displaystyle f:Z\to X}$ be a continuous function between topological spaces ${\displaystyle Z,X}$ where ${\displaystyle Z}$ is connected, and suppose that ${\displaystyle {\tilde {X}}}$ is a covering space of ${\displaystyle X}$. Then if two lifts ${\displaystyle {\tilde {f}}_{1},{\tilde {f}}_{2}:Z\to {\tilde {X}}}$ of ${\displaystyle f}$ coincide in one point ${\displaystyle z_{0}\in Z}$, they are equal.

Proof: Let ${\displaystyle S\subseteq Z}$ be the set on which ${\displaystyle {\tilde {f}}_{1}}$ and ${\displaystyle {\tilde {f}}_{2}}$ agree. Since ${\displaystyle {\tilde {f}}_{1}}$ adn ${\displaystyle {\tilde {f}}_{2}}$ agree on ${\displaystyle z_{0}}$, ${\displaystyle S}$ is nonempty. Further, if ${\displaystyle z\in S}$, set ${\displaystyle x:=f(z)}$ and let ${\displaystyle U}$ be an evenly covered neighbourhood of ${\displaystyle x}$. Let ${\displaystyle V_{\alpha }}$ be the disjointed component in which ${\displaystyle {\tilde {x}}:={\tilde {f}}_{1}(z)={\tilde {f}}_{2}(z)}$ lies, so that ${\displaystyle \pi |_{V_{\alpha }}}$ is a homeomorphism between ${\displaystyle V_{\alpha }}$ and ${\displaystyle U}$. Then define ${\displaystyle W:={\tilde {f}}_{1}^{-1}(V_{\alpha })\cap {\tilde {f}}_{2}^{-1}(V_{\alpha })}$, both of which are open since ${\displaystyle {\tilde {f}}_{1}}$, ${\displaystyle {\tilde {f}}_{2}}$ are continuous. We claim that in fact ${\displaystyle {\tilde {f}}_{1}={\tilde {f}}_{2}}$ on ${\displaystyle W}$; indeed, we may note that ${\displaystyle \pi \circ {\tilde {f}}_{1}=\pi \circ {\tilde {f}}_{2}=f}$ implies that ${\displaystyle \pi |_{V_{\alpha }}\circ {\tilde {f}}_{1}|_{W}=\pi |_{V_{\alpha }}\circ {\tilde {f}}_{2}|_{W}}$, from which identity of ${\displaystyle {\tilde {f}}_{1},{\tilde {f}}_{2}}$ follows upon composing on the left with ${\displaystyle \pi |_{V_{\alpha }}^{-1}}$. We conclude that ${\displaystyle S}$ is open. However, suppose that ${\displaystyle z\notin S}$, suppose that ${\displaystyle U}$ is an evenly covered neighbourhood of ${\displaystyle f(z)}$ and let ${\displaystyle V_{\alpha }}$ respectively ${\displaystyle V_{\beta }}$ be the (disjoint) components of ${\displaystyle \pi ^{-1}(V)}$, so that ${\displaystyle \pi |_{V_{\alpha }}}$ and ${\displaystyle \pi |_{V_{\beta }}}$ are homeomorphisms. Then define ${\displaystyle W:={\tilde {f}}_{1}^{-1}(V_{\alpha })\cap {\tilde {f}}_{2}^{-1}(V_{\beta })}$. Whenever ${\displaystyle z\in W}$, we will have ${\displaystyle {\tilde {f}}_{1}(w)\in V_{\alpha }}$ and ${\displaystyle {\tilde {f}}_{2}(w)\in V_{\beta }}$, so that on ${\displaystyle W}$ the functions ${\displaystyle {\tilde {f}}_{1}}$ and ${\displaystyle {\tilde {f}}_{2}}$ disagree on every point. Thus, we get that ${\displaystyle Z\setminus S}$ is open, so that ${\displaystyle S}$ is open and closed, and since ${\displaystyle Z}$ is connected, ${\displaystyle S=Z}$. ${\displaystyle \Box }$

Proposition (lifting paths):

Let ${\displaystyle X}$ be a topological space, let ${\displaystyle \gamma :[0,1]\to X}$ be a path, and let ${\displaystyle {\tilde {X}}}$ (together with a suitable covering map ${\displaystyle \pi }$) be a covering space of ${\displaystyle X}$. Set ${\displaystyle x:=\gamma (0)\in X}$. Then for each ${\displaystyle {\tilde {x}}\in \pi ^{-1}(x)}$, there exists a unique curve ${\displaystyle {\tilde {\gamma }}:[0,1]\to {\tilde {X}}}$ such that ${\displaystyle \pi \circ {\tilde {\gamma }}=\gamma }$.

Proof: Note that ${\displaystyle [0,1]}$ is compact. For each ${\displaystyle x\in \gamma ([0,1])}$, pick an evenly covered neighbourhood ${\displaystyle U_{x}}$ of ${\displaystyle x}$. Since ${\displaystyle [0,1]}$ is compact, upon defining ${\displaystyle V_{x}:=\gamma ^{-1}(U_{x})}$ and then writing each ${\displaystyle V_{x}}$ as

${\displaystyle V_{x}=\bigcup _{\beta \in B_{x}}W_{\beta }}$

where the ${\displaystyle W_{\beta }}$ are intervals (note that the intervals form a basis of the Euclidean topology) and then considering the open cover ${\displaystyle (W_{\beta })_{\beta \in \cup _{x}B_{x}}}$, we may pick a finite number of intervals ${\displaystyle W_{1},\ldots ,W_{n}}$ which cover ${\displaystyle [0,1]}$ and whose images via ${\displaystyle \gamma }$ are evenly covered. We assume that the intervals ${\displaystyle W_{1},\ldots ,W_{n}}$ are ordered increasingly according to their starting points. Now we define ${\displaystyle {\tilde {\gamma }}}$ successively on these intervals. For ${\displaystyle t\in W_{1}}$, we note that ${\displaystyle \gamma (W_{1})}$ is evenly covered and contains ${\displaystyle x}$. Suppose that ${\displaystyle U}$ is an evenly covered neighbourhood of ${\displaystyle \gamma (W_{1})}$, and let ${\displaystyle (V_{\alpha })_{\alpha \in A}}$ be the components of ${\displaystyle \pi ^{-1}(U)}$ so that ${\displaystyle \pi }$ restricts to a homeomorphism on them. Let ${\displaystyle \alpha _{0}\in A}$ so that ${\displaystyle {\tilde {x}}\in V_{\alpha _{0}}}$. Then define ${\displaystyle {\tilde {\gamma }}(t):=\pi |_{V_{\alpha _{0}}}^{-1}\circ \gamma (t)}$ for ${\displaystyle t\in W_{1}}$ and observe that ${\displaystyle {\tilde {\gamma }}(0)={\tilde {x}}}$, since ${\displaystyle \pi ({\tilde {x}})=\pi ({\tilde {\gamma }}(0))=\gamma (0)=x}$ and ${\displaystyle {\tilde {\gamma }}(0)}$ and ${\displaystyle {\tilde {x}}}$ are both in ${\displaystyle V_{\alpha _{0}}}$. (Note that a homeomorphism is in particular bijective and that the definition of ${\displaystyle {\tilde {\gamma }}}$ is independent of the choice of interval, therefore ${\displaystyle {\tilde {\gamma }}}$ is well defined even on the intersection ${\displaystyle W_{i}\cap W_{i+1}}$.) Proceed similarly for the ensuing intervals ${\displaystyle W_{2},\ldots ,W_{n}}$. Then ${\displaystyle {\tilde {\gamma }}}$ will be continuous on all intervals of the form ${\displaystyle [a_{j-1},a_{j}]}$, where ${\displaystyle a_{j-1}}$ is the beginning point of ${\displaystyle W_{j}}$ and ${\displaystyle a_{n}=1}$, since the interval ${\displaystyle [a_{j-1},a_{j}]}$ is contained in ${\displaystyle W_{j}}$. Hence, since a function which is continuous on two closed sets is continuous on their union, we obtain that ${\displaystyle {\tilde {\gamma }}}$ is continuous on ${\displaystyle [a_{0},a_{2}]}$, then on ${\displaystyle [a_{0},a_{3}]}$, and so on, and finally on ${\displaystyle [a_{0},a_{n}]=[0,1]}$. ${\displaystyle {\tilde {\gamma }}}$ is then the desired lift, and it is unique since maps of connected domain lift uniquely. ${\displaystyle \Box }$

Proposition (lifting homotopies):

Let ${\displaystyle Z,X}$ be topological spaces and let ${\displaystyle H:[0,1]\times Z\to X}$ be a continuous function. Suppose that ${\displaystyle {\tilde {X}}}$ is a covering space of ${\displaystyle X}$, and that we are given a lift ${\displaystyle {\tilde {H}}_{0}:Z\to {\tilde {X}}}$ of the function ${\displaystyle H_{0}:Z\to X}$ defined by ${\displaystyle H_{0}(z):=H(0,z)}$. Then there exists a unique continuous function ${\displaystyle {\tilde {H}}:[0,1]\times Z\to X}$ so that for one thing ${\displaystyle {\tilde {H}}_{0}(z)={\tilde {H}}(0,z)}$ for all ${\displaystyle z\in Z}$ and further ${\displaystyle \pi \circ {\tilde {H}}=H}$.

Proof: For each ${\displaystyle z\in Z}$, ${\displaystyle \gamma _{z}(t):=H(t,z)}$ lifts uniquely to a path ${\displaystyle {\tilde {\gamma }}_{z}}$ so that ${\displaystyle {\tilde {\gamma }}_{z}(0)={\tilde {H}}(0,z)}$. Define ${\displaystyle {\tilde {H}}(t,z):=\gamma _{z}(t)}$. Obviously, ${\displaystyle \pi \circ {\tilde {H}}=H}$, and further, we claim that ${\displaystyle {\tilde {H}}}$ is continuous. Let ${\displaystyle (t,z)\in [0,1]\times Z}$ be arbitrary. Let ${\displaystyle U\subseteq X}$ be an evenly covered neighbourhood of ${\displaystyle H(t,x)}$; then ${\displaystyle \pi \upharpoonright _{U}}$ is a homeomorphism that has a continuous inverse ${\displaystyle (\pi \upharpoonright _{U})^{-1}}$. Now by definition of the product topology, take ${\displaystyle W\subseteq Z}$ open and ${\displaystyle \epsilon >0}$ so that ${\displaystyle (t-\epsilon ,t+\epsilon )\times W\subseteq H^{-1}(U)}$. By uniqueness of path lifting, we have ${\displaystyle {\tilde {H}}=(\pi \upharpoonright _{U})^{-1}\circ H}$ on ${\displaystyle (t-\epsilon ,t+\epsilon )\times W}$, which is continuous. We conclude that ${\displaystyle {\tilde {H}}}$ is continuous, since it is continuous in an open set about an arbitrary point. ${\displaystyle \Box }$