Covering Spaces

Existence. Cover $B$ by evenly covered open sets. Since $\gamma$ is continuous and $[0,1]$ is compact, the collection $\{\gamma^{-1}(U) : U \text{ evenly covered}\}$ is an open cover of $[0,1]$; by the Lebesgue number lemma there is a partition

such that each $\gamma([t_{i-1}, t_i])$ lies inside a single evenly covered open set $U_i$. We construct $\tilde\gamma$ inductively on $[0,t_i]$, starting from $\tilde\gamma(0) = e_0$. Suppose $\tilde\gamma$ has been defined and is continuous on $[0, t_{i-1}]$, with $p(\tilde\gamma(t_{i-1})) = \gamma(t_{i-1})$. Since $\gamma([t_{i-1},t_i]) \subseteq U_i$ and $U_i$ is evenly covered, write $p^{-1}(U_i) = \bigsqcup_\alpha V_\alpha$ with each $p|_{V_\alpha} : V_\alpha \to U_i$ a homeomorphism. The point $\tilde\gamma(t_{i-1})$ lies in exactly one sheet $V_{\alpha_0}$ (the sheets are disjoint). Define, for $t \in [t_{i-1}, t_i]$,

This is continuous (composition of $\gamma|_{[t_{i-1},t_i]}$, continuous into $U_i$, with the homeomorphism $(p|_{V_{\alpha_0}})^{-1}$), agrees with the previous piece at $t_{i-1}$ (both equal $\tilde\gamma(t_{i-1})$, since $(p|_{V_{\alpha_0}})^{-1}(\gamma(t_{i-1})) = \tilde\gamma(t_{i-1})$ by injectivity of $p|_{V_{\alpha_0}}$ and the fact that $\tilde\gamma(t_{i-1}) \in V_{\alpha_0}$ already maps to $\gamma(t_{i-1})$), and satisfies $p \circ \tilde\gamma = \gamma$ on $[t_{i-1}, t_i]$ by construction. By the pasting lemma for finitely many closed sets covering $[0,t_i]$, this extends $\tilde\gamma$ continuously to $[0, t_i]$. After $n$ steps we obtain $\tilde\gamma$ continuous on all of $[0,1]$ with $p\tilde\gamma = \gamma$, $\tilde\gamma(0) = e_0$.

Uniqueness. Suppose $\tilde\gamma_1, \tilde\gamma_2 : [0,1] \to E$ both lift $\gamma$ with $\tilde\gamma_1(0) = \tilde\gamma_2(0) = e_0$. Let

$S$ is nonempty ($0 \in S$). We show $S$ is both open and closed in $[0,1]$; since $[0,1]$ is connected this forces $S = [0,1]$. Fix $t \in [0,1]$ and let $U$ be an evenly covered neighborhood of $\gamma(t)$, with sheets $\{V_\alpha\}$. Both $\tilde\gamma_1, \tilde\gamma_2$ are continuous, so there is an open interval $I \ni t$ in $[0,1]$ with $\tilde\gamma_1(I) \subseteq V_{\alpha_1}$ and $\tilde\gamma_2(I) \subseteq V_{\alpha_2}$ for single sheets $V_{\alpha_1}, V_{\alpha_2}$ (shrink $I$ so both images land in single sheets, using continuity at $t$). If $t \in S$: then $\tilde\gamma_1(t) = \tilde\gamma_2(t) \in V_{\alpha_1} \cap V_{\alpha_2}$, forcing $V_{\alpha_1} = V_{\alpha_2}$ (sheets are disjoint). On $I$, both $\tilde\gamma_1, \tilde\gamma_2$ are lifts of $\gamma|_I$ landing in the same sheet $V_{\alpha_1}$, on which $p$ is injective; since $p\tilde\gamma_1 = p \tilde\gamma_2 = \gamma$ on $I$ and $p|_{V_{\alpha_1}}$ is injective, $\tilde\gamma_1 = \tilde\gamma_2$ on $I$. Hence $I \subseteq S$, so $S$ is open. If $t \notin S$: then $\tilde\gamma_1(t) \ne \tilde\gamma_2(t)$. If $V_{\alpha_1} \ne V_{\alpha_2}$, disjointness gives $\tilde\gamma_1(s) \ne \tilde\gamma_2(s)$ for all $s \in I$ (since $\tilde\gamma_1(s) \in V_{\alpha_1}$, $\tilde\gamma_2(s) \in V_{\alpha_2}$, disjoint sets), so $I \cap S = \varnothing$. If $V_{\alpha_1} = V_{\alpha_2} = V$, then on $I$, injectivity of $p|_V$ together with $p\tilde\gamma_1 = p\tilde\gamma_2$ would force $\tilde\gamma_1 = \tilde\gamma_2$ throughout $I$ — but that contradicts $\tilde\gamma_1(t) \ne \tilde\gamma_2(t)$; hence this case cannot occur when $t \notin S$. So in fact $I \cap S = \varnothing$, showing the complement of $S$ is open, i.e. $S$ is closed. Both open and closed, nonempty, in the connected space $[0,1]$: $S = [0,1]$, so $\tilde\gamma_1 = \tilde\gamma_2$. $\blacksquare$

The construction mirrors path lifting but on the square, using a Lebesgue-number grid rather than a single partition.

Existence. By compactness of $[0,1]\times[0,1]$ and continuity of $F$, using the Lebesgue number lemma applied to the open cover of the square by preimages $F^{-1}(U)$ of evenly covered sets $U$, choose partitions $0=x_0<\cdots<x_m=1$ and $0=y_0<\cdots<y_n=1$ fine enough that each rectangle $R_{ij} = [x_{i-1},x_i]\times[y_{j-1},y_j]$ satisfies $F(R_{ij}) \subseteq U_{ij}$ for some evenly covered $U_{ij}$. Order the rectangles lexicographically, first by $j$ (rows, bottom to top) then by $i$ (left to right within a row). We define $\tilde F$ rectangle by rectangle. On the bottom row ($j=1$), on the segment $[0,1]\times\{0\}$ we already have $\tilde f_0$. Proceed along $i = 1, \dots, m$: having defined $\tilde F$ continuously on the "staircase" region consisting of all previously handled rectangles together with the bottom edge, note that the boundary of the next rectangle $R_{i1}$ already carries a continuous partial lift on (at least) one edge or corner — either the bottom edge (part of $\tilde f_0$) or the shared left edge with $R_{i-1,1}$. The key local step: given a rectangle $R$ with $F(R) \subseteq U$ evenly covered, and given a continuous partial lift already defined and agreeing with $p$ on a connected, nonempty closed subset $K \subseteq R$ containing at least one point (e.g. a single vertex or an edge), there is a unique sheet $V_\alpha$ over $U$ containing the image of that already-lifted point; since $R$ is connected and the existing partial lift takes values in $E$ mapping into $U$, and different sheets are disjoint open sets, continuity of the existing partial lift forces its entire image on $K$ to lie in the single sheet $V_\alpha$ (a connected set mapping into a disjoint union of opens must land in one piece). Define $\tilde F$ on all of $R$ by $\tilde F := (p|_{V_\alpha})^{-1} \circ F$. This agrees with the prior partial lift on $K$ (both are $(p|_{V_\alpha})^{-1}\circ F$ there, using injectivity of $p|_{V_\alpha}$), and is continuous. By the pasting lemma, extending rectangle by rectangle (finitely many, each glued along a shared closed edge with matching values) yields a global continuous $\tilde F : [0,1]^2 \to E$ with $p\tilde F = F$ and $\tilde F(\cdot,0) = \tilde f_0$.

Uniqueness. For each fixed $x \in [0,1]$, the path $s \mapsto \tilde F(x,s)$ is, by Unique Path Lifting, the unique lift of $s \mapsto F(x,s)$ starting at $\tilde f_0(x)$. So $\tilde F$ is uniquely determined pointwise, hence uniquely determined as a map.

Path-homotopy consequence. Suppose $F(0,s) = b_0$ and $F(1,s) = b_1$ constant in $s$. Then $s \mapsto \tilde F(0,s)$ is a lift of the constant path at $b_0$ starting at $\tilde F(0,0) = \tilde f_0(0)$; the constant path at $\tilde f_0(0)$ is a lift of the constant path at $b_0$, and by uniqueness (unique path lifting) it is the lift. Hence $\tilde F(0,s) = \tilde f_0(0)$ for all $s$. The same argument at $x=1$ shows $\tilde F(1,s) = \tilde f_0(1)$ for all $s$. So $\tilde F$ is a homotopy rel endpoints between $\tilde F(\cdot,0) = \tilde f_0$ and $\tilde F(\cdot, 1)$. $\blacksquare$

Well-defined: if $\gamma \simeq \gamma'$ rel endpoints via homotopy $F$, apply Homotopy Lifting to $F$ with initial lift $\tilde\gamma$; the resulting lift $\tilde F$ is a path-homotopy from $\tilde \gamma$ to some lift of $\gamma'$ starting at $e_0$, and by uniqueness of path lifting, that lift of $\gamma'$ starting at $e_0$ is $\widetilde{\gamma'}$. Path-homotopy fixes endpoints, so $\tilde\gamma(1) = \widetilde{\gamma'}(1)$. Group action: $e_0 \cdot [\text{const}_{b_0}] = e_0$ since the constant path at $e_0$ lifts the constant path at $b_0$ and lands at $e_0$. For concatenation, if $\tilde\gamma$ lifts $\gamma$ from $e_0$ to $e_1 := \tilde\gamma(1)$, and $\tilde\delta$ lifts $\delta$ starting at $e_1$, then the concatenation $\tilde\gamma \ast \tilde\delta$ is continuous (pasting lemma, matching at $e_1$) and lifts $\gamma \ast \delta$ starting at $e_0$; by uniqueness this is the lift used to define $e_0 \cdot [\gamma \ast \delta]$, whose endpoint is $\tilde\delta(1) = e_1 \cdot [\delta] = (e_0\cdot[\gamma])\cdot[\delta]$. Hence $e_0 \cdot ([\gamma][\delta]) = (e_0\cdot [\gamma]) \cdot [\delta]$, the defining law of a right action. $\blacksquare$

(1) Injectivity. Suppose $[\alpha] \in \pi_1(E,e_0)$ with $p_*[\alpha] = [p\alpha] = 1$ in $\pi_1(B,b_0)$, i.e. $p\alpha$ is null-homotopic rel basepoint to the constant loop at $b_0$. Let $F : [0,1]^2 \to B$ be this path-homotopy, with $F(\cdot,0) = p\alpha$, $F(\cdot,1) = \text{const}_{b_0}$, endpoints fixed. By Homotopy Lifting with initial lift $\alpha$ (a lift of $F(\cdot,0)$), we get $\tilde F$ lifting $F$ with $\tilde F(\cdot,0) = \alpha$, and (by the path-homotopy consequence of that theorem) $\tilde F(\cdot,1)$ is a loop at the same basepoint $\alpha(0)=e_0$ lifting the constant loop $\text{const}_{b_0}$; by uniqueness of path lifting, the only lift of a constant path starting at $e_0$ is the constant path at $e_0$. So $\tilde F(\cdot,1) = \text{const}_{e_0}$, and $\tilde F$ is a path-homotopy in $E$ rel basepoint from $\alpha$ to $\text{const}_{e_0}$. Hence $[\alpha] = 1$ in $\pi_1(E,e_0)$. This proves $\ker p_* = 1$, i.e. $p_*$ is injective.

(2) Transitivity and stabilizer. Transitivity: given any $e_1 \in p^{-1}(b_0)$, since $E$ is path-connected choose a path $\tilde\eta$ in $E$ from $e_0$ to $e_1$. Then $\eta := p\tilde\eta$ is a loop at $b_0$ (since $p(e_0)=p(e_1)=b_0$), and by construction $\tilde\eta$ is the lift of $\eta$ starting at $e_0$, so $e_0 \cdot [\eta] = \tilde\eta(1) = e_1$. Hence the action is transitive. Stabilizer: $[\gamma] \in \pi_1(B,b_0)$ stabilizes $e_0$ iff the lift $\tilde\gamma$ of $\gamma$ starting at $e_0$ satisfies $\tilde\gamma(1) = e_0$, i.e. iff $\tilde\gamma$ is a loop at $e_0$ in $E$. If so, $[\gamma] = [p\tilde\gamma] = p_*[\tilde\gamma] \in p_*\pi_1(E,e_0)$. Conversely if $[\gamma] = p_*[\beta]$ for a loop $\beta$ at $e_0$ in $E$, then $\beta$ is a lift of $p\beta \simeq \gamma$... more carefully: $p\beta$ is homotopic rel endpoints to $\gamma$ (they represent the same class), so by well-definedness of the monodromy action (part of the corollary above), $e_0\cdot[\gamma] = e_0\cdot [p\beta] = \beta(1) = e_0$ (using that $\beta$, a loop at $e_0$, is trivially the lift of $p\beta$ starting at $e_0$). So $[\gamma]$ stabilizes $e_0$. This proves $\mathrm{Stab}(e_0) = p_*\pi_1(E,e_0)$.

(3) Orbit–stabilizer bijection. For a group $G$ acting on a set $X$ transitively, the orbit map $g \mapsto x_0 \cdot g$ induces a bijection $\mathrm{Stab}(x_0)\backslash G \xrightarrow{\ \sim\ } X$ sending the coset $\mathrm{Stab}(x_0)\, g$ to $x_0\cdot g$ (well-defined and injective because $x_0\cdot g = x_0\cdot g'$ iff $g g'^{-1} \in \mathrm{Stab}(x_0)$ iff $\mathrm{Stab}(x_0) g = \mathrm{Stab}(x_0)g'$; surjective by transitivity). Apply this with $G = \pi_1(B,b_0)$, $X = p^{-1}(b_0)$, $x_0 = e_0$: we obtain a bijection

If $B$ is path-connected and $U$ is an evenly covered neighborhood of $b_0$ with sheets in bijection with $p^{-1}(b_0)$ locally, standard connectivity arguments (each fiber has the same cardinality, via lifting a path between any two base points and using unique path lifting to move fibers bijectively) show every fiber has this same cardinality, so "number of sheets" is well-defined and equals $\big|p_*\pi_1(E,e_0)\backslash \pi_1(B,b_0)\big| = [\pi_1(B,b_0):p_*\pi_1(E,e_0)]$. $\blacksquare$