Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 1 / 21J / Solution

A covering space is a map $p:\tilde{X}\to X$ such that every $x\in X$ has an open neighbourhood $U$ for whichand every restriction $p:U_{\alpha }\to U$ is a homeomorphism.

For path lifting, cover the compact image of $\gamma$ by evenly covered sets. The Lebesgue number lemma supplies a subdivisionfor which each $\gamma ([t_{j-1},t_j])$ lies in one such set. Starting in the sheet containing ${\tilde{x}}_0$, lift the first segment with the inverse of that sheet. Its endpoint selects the sheet for the next segment, and induction constructs a continuous lift. Two lifts starting at the same point agree on the first segment because the relevant sheet map is injective, and the same argument at successive endpoints proves uniqueness.

Write $q:Y\to Y/\varphi$ for the quotient map. The identity $p\circ Q=q\circ {\mathrm{pr}}_Y$ on $Y\times \mathbb{Z}$, where $Q$ is the quotient defining $\widetilde{Y/\varphi }$, proves continuity of $p$ by the quotient property. Away from the two gluing regions, a small open set lifts to one copy in every level. Near a glued point choose $U⋐A$ and use the matched pairAfter the prescribed identifications, each such pair maps homeomorphically to $q(U\cup \varphi (U))$, and the pairs are disjoint for different $i$. Thus $p$ is a covering map.

Each copy of $Y$ is path-connected, and adjacent copies meet through the identified copies of the nonempty set $A$, so $\widetilde{Y/\varphi }$ is path-connected. Translationis a deck transformation. Its powers act freely and transitively on every fibre, so the covering is regular. The covering-space subgroup and deck-group quotient then gives

Solved by gpt-5.6-sol high.