Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 13 / 1

Let $M$ be the given closed three-dimensional manifold. It is connected by the meaning of simply connected. An orientation chosen at one point can be transported along paths, and its sign change around loops defines a homomorphism ${\pi }_1(M)\to \{\pm 1\}$. Since the fundamental group is trivial, this monodromy vanishes; thus $M$ is orientable.

The first homology group is the abelianization of the fundamental group, giving $H_1(M;\mathbb{Z})=0$. Also $H_0(M;\mathbb{Z})=\mathbb{Z}$. The universal coefficient theorem for cohomology in degree one givesBoth outer groups vanish: the first because $\mathbb{Z}$ is free and the second because $H_1=0$. Hence $H^1=0$. Integral Poincare duality now gives $H_2(M;\mathbb{Z})\cong H^1(M;\mathbb{Z})=0$ and $H_3(M;\mathbb{Z})\cong H^0(M;\mathbb{Z})=\mathbb{Z}$. Homology in degrees greater than three vanishes by the same duality with negative-degree cohomology. ConsequentlyThese are precisely the homology of a sphere for $S^3$, proving that simply connected closed three-manifolds are homology spheres. No homeomorphism classification of three-manifolds is needed.

The universal cover of $S^1\times S^2$ isBecause $S^3$ is simply connected, any continuous map $f:S^3\to S^1\times S^2$ lifts to $\tilde{f}:S^3\to \mathbb{R}\times S^2$ after choosing a lift of one base point. The lifting criterion for a covering space applies because $f_{\ast }{\pi }_1(S^3)=0$. Equivalently, path lifting defines the lift and simple connectedness makes its value independent of the chosen path.

The cover deformation retracts onto $\{0\}\times S^2$, so $H_3(\mathbb{R}\times S^2;\mathbb{Z})=0$. Since $f=p\circ \tilde{f}$, functoriality of homology givesTherefore $\boxed{f_{\ast }=0\text{ and }\mathrm{deg}f=0}$, with the last equality using the oriented three-dimensional fundamental classes.