Past exam of the mathematics course of the University of Cambridge / 2007 / ii / Paper 3 / 20H (Algebraic Topology)

An action assigns each $g\in G$ a homeomorphism of $X$, with $e\cdot x=x$ and $g\cdot (h\cdot x)=(gh)\cdot x$. For the covering assertion use a properly discontinuous free action: every $x$ has an open neighbourhood $U$ such that $gU\cap U=\varnothing$ for every $g\ne e$. Then the translates $gU$ are pairwise disjoint. The quotient map $p$ is open because $p^{-1}(p(U))={\bigcup }_{g}gU$ is open. Each restriction $p:gU\to p(U)$ is a continuous open bijection and hence a homeomorphism. Thus $p(U)$ is evenly covered and $p:X\to X/G$ is a covering map.

For the surjective homomorphism one additionally needs $X$ path connected; choose $x_0$ and its image $y_0$. Lift a based loop at $y_0$ from $x_0$. Its endpoint has a unique form $g{x}_0$. Homotopy lifting makes this assignment well-defined on the fundamental group. With concatenation meaning first the first loop and then the second, the second lift from $g{x}_0$ is $g$ times the lift from $x_0$, so the endpoint is $gh{x}_0$. This proves the homomorphism property. For each $g$, a path in $X$ from $x_0$ to $g{x}_0$ projects to a loop realizing $g$. HencePointwise freeness alone is insufficient for a covering: an irrational rotation action of $\mathbb{Z}$ on the circle is free but its orbits are dense. Path connectedness is also essential for the final surjectivity claim as written: the discrete space $X=G$ with left translation is a free properly discontinuous action whose quotient is a point, and its trivial fundamental group cannot surject onto nontrivial $G$. The proof supplies the standard connected covering-space interpretation rather than hiding these hypotheses.