Algebraic Topology/The fundamental group and covering spaces

Proposition (induced map by covering map is injective):

Let $p:Z\to Y$ be a covering space, and let $z_{0}\in Z$. Then the induced map

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

is injective.

Proof: Suppose that $\rho ,\eta :[0,1]\to Z$ are loops at $z_{0}$ such that $p_{*}([\rho ])=p_{*}([\eta ])$. Then there exists a homotopy $H$ from $p\circ \rho$ to $p\circ \eta$, which are both loops at $x_{0}$. This homotopy lifts uniquely to a homotopy ${\tilde {H}}$ such that ${\tilde {H}}_{0}=\rho$. Moreover ${\tilde {H}}$ fixes the basepoint, because the maps $t\mapsto {\tilde {H}}_{t}(0)$ and $t\mapsto {\tilde {H}}_{t}(1)$ are continuous and hence map connected sets to connected sets, and the preimage of a point under a covering map bears the discrete topology. By uniqueness of path lifting, ${\tilde {H}}_{1}$ will be equal to $\eta$, so that $[\rho ]=[\eta ]$. $\Box$

Proposition (fundamental group criterion for existence of lifting):

Let $p:Z\to Y$ be a covering and let $f:X\to Y$ be a continuous function, where $X$ is path-connected and strongly locally connected. Let $y_{0}\in Y$ and $x_{0}\in X$ so that $f(x_{0})=y_{0}$, and let $z_{0}$ be a lift of $y_{0}$ (ie. $p(z_{0})=y_{0}$). Then the following are equivalent:

• There exists a unique lift ${\tilde {f}}:X\to Z$ of $f$ so that ${\tilde {f}}(x_{0})=z_{0}$
• $f_{*}(\pi (X,x_{0}))\subseteq p_{*}(\pi (Z,z_{0}))$

Proof: Suppose that ${\tilde {f}}$ lifts $f$, so that ${\tilde {f}}(x_{0})=z_{0}$. Then $f_{*}(\pi (X,x_{0}))=p_{*}{\tilde {f}}_{*}(\pi (X,x_{0}))$ so that, since ${\tilde {f}}_{*}$ maps $\pi (X,x_{0})$ to $\pi (Z,z_{0})$, we indeed have $f*(\pi _{X},x_{0})\subseteq p_{*}(\pi (Z,z_{0}))$.

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 $A$.

Finally, connectedness of $X$ implies that ${\tilde {f}}$ is unique. $\Box$