代數拓撲/基本群和覆疊空間

命題（覆疊對映誘導的對映是單射的）:

令 $p:Z\to Y$ 為一個覆疊空間，令 $z_{0}\in Z$。那麼誘導對映

$p_{*}:\pi (Z,z_{0})\to \pi (Y,y_{0})$

是單射的。

證明： 假設 $\rho ,\eta :[0,1]\to Z$ 是在 $z_{0}$ 處的環路，使得 $p_{*}([\rho ])=p_{*}([\eta ])$。 那麼存在一個從 $p\circ \rho$ 到 $p\circ \eta$ 的同倫，它們都是 $x_{0}$ 處的環路。 這個同倫 可以唯一地提升 到一個同倫 ${\tilde {H}}$，使得 ${\tilde {H}}_{0}=\rho$。 此外，${\tilde {H}}$ 固定基點，因為對映 $t\mapsto {\tilde {H}}_{t}(0)$ 和 $t\mapsto {\tilde {H}}_{t}(1)$ 是連續的，因此將連通集對映到連通集，並且覆蓋對映下一點的反像具有離散拓撲。 根據 路徑提升的唯一性，${\tilde {H}}_{1}$ 將等於 $\eta$，使得 $[\rho ]=[\eta ]$。 $\Box$

命題（存在提升的基本群判據）:

令 $p:Z\to Y$ 為覆蓋對映，令 $f:X\to Y$ 為連續函式，其中 $X$ 是道路連通且強區域性連通的。令 $y_{0}\in Y$ 且 $x_{0}\in X$ 使得 $f(x_{0})=y_{0}$，令 $z_{0}$ 是 $y_{0}$ 的提升（即， $p(z_{0})=y_{0}$）。那麼以下等價

• 存在 $f$ 的唯一提升 ${\tilde {f}}:X\to Z$ 使得 ${\tilde {f}}(x_{0})=z_{0}$
• $f_{*}(\pi (X,x_{0}))\subseteq p_{*}(\pi (Z,z_{0}))$

證明：假設 ${\tilde {f}}$ 提升 $f$，使得 ${\tilde {f}}(x_{0})=z_{0}$。 那麼 $f_{*}(\pi (X,x_{0}))=p_{*}{\tilde {f}}_{*}(\pi (X,x_{0}))$，因此，由於 ${\tilde {f}}_{*}$ 將 $\pi (X,x_{0})$ 對映到 $\pi (Z,z_{0})$，我們確實有 $f*(\pi _{X},x_{0})\subseteq p_{*}(\pi (Z,z_{0}))$。

Conversely, suppose that $f_{*}(\pi (X,x_{0}))\subseteq p_{*}(\pi (Z,z_{0}))$. Then we define a lift ${\tilde {f}}:X\to Y$ of $f$ as follows: For each $x\in X$, choose a path $\gamma :[0,1]\to X$ so that $\gamma (0)=x_{0}$ and $\gamma (1)=x$ by path-connectedness of $X$. By path lifting, $f\circ \gamma$ lifts uniquely to a path $\rho :[0,1]\to Z$. Then set ${\tilde {f}}(x):=\rho (1)$. We have to show that this definition does not depend on the choice of $\gamma$. Indeed, let $\gamma '$ be another path like $\gamma$. Then $\gamma *{\overline {\gamma '}}$ is a loop at $x_{0}$ that induces an equivalence class $[\gamma *{\overline {\gamma '}}]\in \pi (X,x_{0})$. This in turn induces an equivalence class $[(f\circ \gamma )*{\overline {(f\circ \gamma ')}}]\in \pi (Y,y_{0})$; indeed, ${\overline {(f\circ \gamma ')}}=(f\circ {\overline {\gamma '}})$, since both are composition with the map $t\mapsto 1-t$. By the assumption, there exists a loop $\eta :[0,1]\to Z$ in $z_{0}$ such that $p_{*}([\eta ])=[(f\circ \gamma )*{\overline {(f\circ \gamma ')}}]$. Moreover, $(f\circ \gamma )*{\overline {(f\circ \gamma ')}}$ is a path in $Y$ that may be lifted to a path $\eta '$ in $Z$. Since we may lift homotopies, we may lift a homotopy between $p_{*}([\eta ])$ and $(f\circ \gamma )*{\overline {(f\circ \gamma ')}}$ to a homotopy between $\eta$ and $\eta '$, which, similarly to the proof of injectivity of $p_{*}$, leaves the endpoints fixed. Hence, $\eta '$ is a loop, and in particular $\eta '$ restricted to $[1/2,1]$ yields, when direction is reversed, a lift of $\gamma '$ that connects $z_{0}$ to $\rho (1)$. We conclude well-definedness.

It remains to prove continuity. Hence, let $x_{1}\in X$ be arbitrary; we shall prove continuity at $x_{1}$. Pick an evenly covered neighbourhood $V$ about $f(x_{1})$. Let $W\subseteq Z$ be the open set mapping homeomorphically to $V$ which contains ${\tilde {f}}(x_{1})$. Let $A$ be any open neighbourhood of ${\tilde {f}}(x_{1})$. Set ${\tilde {A}}\subseteq Y$ to be the image of $A\cap W$ (which is open) under $p$, so that ${\tilde {A}}$ is itself open. By continuity of $f$, there exists an open neihbourhood $B$ of $x_{1}$ that is mapped by $f$ into ${\tilde {A}}$. By strong local connectedness, choose $C\subseteq B$ to be connected. Recall that maps from connected domains lift uniquely; this shows that on $C$, we have ${\tilde {f}}=(p|_{V})^{-1}\circ f$. Hence, ${\tilde {f}}$ maps $C$ into $A$.

最後，$X$ 的連通性意味著 ${\tilde {f}}$ 是唯一的。 $\Box$