一般拓撲/覆蓋空間

定義（覆蓋空間）:

令$X$是一個拓撲空間。另一個拓撲空間${\tilde {X}}$稱為$X$的**覆蓋空間**當且僅當${\tilde {X}}$伴隨著一個連續函式$\pi :{\tilde {X}}\to X$，滿足以下條件

• 對於每個$x\in X$，存在$x$的一個鄰域$U$，使得$\pi ^{-1}(U)$是開子集$(V_{\alpha })_{\alpha \in A}$的不交併，使得對於所有的$\alpha \in A$，函式$\pi |_{V_{\alpha }}$是$V_{\alpha }$到$U$的一個同胚。

定義（覆蓋對映）:

令 ${\tilde {X}}$ 為拓撲空間 $X$ 的覆蓋空間，並令 $\pi :{\tilde {X}}\to X$ 為與其相關的連續對映。則 $\pi$ 被稱為覆蓋空間 ${\tilde {X}}$ 的覆蓋對映。

注意，更正式地說，可以將覆蓋空間定義為對 $({\tilde {X}},\pi )$，其中 $\pi$ 是屬於覆蓋空間的覆蓋對映；實際上，如果 $\pi$ 未指定，則可能存在多個覆蓋對映。但與群運算一樣，為了簡潔起見，我們大多會省略 $\pi$ 的符號。

定義（均勻覆蓋鄰域）:

令 $X$ 為拓撲空間，並令 ${\tilde {X}}$ 為 $X$ 的覆蓋空間，其中 $\pi :{\tilde {X}}\to X$ 為覆蓋對映。令 $x\in X$。覆蓋空間定義保證的集合 $U$ （使得 $\pi$ 限制在原像的適當子集上是同胚）被稱為 $x$ 的均勻覆蓋鄰域。

定義（提升）:

設 $Z,X$ 是拓撲空間，設 $f:Z\to X$ 是一個連續函式。如果 ${\tilde {X}}$ 是 $X$ 的覆蓋空間，那麼 $f$ 到 ${\tilde {X}}$ 的提升是一個連續函式 ${\tilde {f}}:Z\to {\tilde {X}}$，使得當 $\pi$ 是屬於 ${\tilde {X}}$ 的覆蓋對映時，$\pi \circ {\tilde {f}}=f$.

當要提升的函式的定義域是連通的，那麼提升在某種意義上是唯一的

命題（連通定義域的對映的唯一提升）:

設 $f:Z\to X$ 是拓撲空間 $Z,X$ 之間的連續函式，其中 $Z$ 是連通的，並假設 ${\tilde {X}}$ 是 $X$ 的覆蓋空間。那麼如果 $f$ 的兩個提升 ${\tilde {f}}_{1},{\tilde {f}}_{2}:Z\to {\tilde {X}}$ 在一個點 $z_{0}\in Z$ 處重合，那麼它們相等。

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, $S=Z$. $\Box$

命題（提升路徑）:

令 $X$ 為拓撲空間，令 $\gamma :[0,1]\to X$ 為一條路徑，並令 ${\tilde {X}}$（以及一個合適的覆蓋對映 $\pi$）為 $X$ 的覆蓋空間。令 $x:=\gamma (0)\in X$。那麼對於每個 ${\tilde {x}}\in \pi ^{-1}(x)$，存在唯一的曲線 ${\tilde {\gamma }}:[0,1]\to {\tilde {X}}$ 使得 $\pi \circ {\tilde {\gamma }}=\gamma$。

證明：注意 $[0,1]$ 是緊緻的。對於每個 $x\in \gamma ([0,1])$，選擇 $x$ 的一個均勻覆蓋鄰域 $U_{x}$。由於 $[0,1]$ 是緊緻的，定義 $V_{x}:=\gamma ^{-1}(U_{x})$ 然後將每個 $V_{x}$ 寫成

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

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. $\Box$

命題（提升同倫）:

令 $Z,X$ 為拓撲空間，並令 $H:[0,1]\times Z\to X$ 為一個連續函式。假設 ${\tilde {X}}$ 是 $X$ 的覆蓋空間，並且我們給定了一個函式 $H_{0}:Z\to X$ 的提升 ${\tilde {H}}_{0}:Z\to {\tilde {X}}$，該函式由 $H_{0}(z):=H(0,z)$ 定義。然後存在一個唯一的連續函式 ${\tilde {H}}:[0,1]\times Z\to X$ 使得一方面對於所有 $z\in Z$，有 ${\tilde {H}}_{0}(z)={\tilde {H}}(0,z)$，並且進一步 $\pi \circ {\tilde {H}}=H$。

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. $\Box$