Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 1 / 21H / i

The assertion needs $E$ connected. Without that assumption, the disjoint union of two tori mapping by the identity on each component is a covering of the torus and is none of the three listed connected spaces. The correct general statement classifies each connected component; here is the intended connected proof.

The universal cover is ${\mathbb{R}}^2\to {\mathbb{R}}^2/{\mathbb{Z}}^2=T$, and ${\pi }_1(T)={\mathbb{Z}}^2$. The classification of connected coverings of a connected, locally path-connected, semilocally simply connected space identifies $E$ with ${\mathbb{R}}^2/H$ for a subgroup $H\le {\mathbb{Z}}^2$, acting by translations. A subgroup of ${\mathbb{Z}}^2$ is free abelian of rank zero, one or two. For rank zero the quotient is the plane. For rank one, a linear coordinate change sends its generator to a translation in one coordinate, yielding $S^1\times \mathbb{R}$. For rank two, a linear coordinate change sends a lattice basis to the standard basis, yielding a torus. Thus a connected covering is a plane, cylinder or torus. For arbitrary $E$, its covering components are disjoint unions of these types, rather than necessarily one such space.