Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 17 / 1

Label the six vertices $A,B,C,D,E,F$ clockwise, starting at the upper-left vertex. The arrows identify $AB$ with $ED$, $BC$ with $FE$, and $FA$ with $DC$, preserving the indicated directions. Consequently the quotient space has exactly two vertices,Orient its three topological cells by $a:A\to B$, $b:B\to C$, and $c:F\to A$. The clockwise boundary of the hexagonal topological cell traversesEvery topological cell of dimension one therefore occurs once with each sign. The cellular chain complex, with integral coefficients, isHere the rows correspond to $P,Q$. The image of ${\partial }_1$ is the primitive subgroup generated by $Q-P$, and its kernel is freely generated by $a+b$ and $a+c$. Thus the homology groups areThe two vertex classes are important: identifying all six vertices would give an incorrect cellular chain complex. The result agrees with the homology of a torus.

Write a point of the larger sphere as $(u,v)\in {\mathbb{R}}^{m+1}\times {\mathbb{R}}^{n-m}$. In the complement, $v\ne 0$. There is a homeomorphismwhose inverse sends $(u,w)$ to $(u,\sqrt{1-\Vert u{\Vert }^2}w)$. In particular the complement has a deformation retraction onto its transverse sphere. Explicitly,The denominator stays positive, the homotopy stays in the complement, and points with $u=0$ remain fixed. This establishes the homology of an equatorial sphere complement. By homotopy invariance of homology, its homology is therefore the homology of a sphere of dimension $d=n-m-1$:In the codimension-one case, the complement has two contractible components; this is why degree-zero homology has rank two. Equivalently, in every case its only nonzero reduced homology is ${\tilde{H}}_d(X;\mathbb{Z})=\mathbb{Z}$.

The torus map acts on $H_1(T^2;\mathbb{Z})$ byIts determinant is one, so it is an orientation-preserving homeomorphism and acts as the identity on both $H_0(T^2)$ and $H_2(T^2)$.

To calculate the homology of its mapping torus, cover the time circle by two arcs. Each inverse image has the homotopy type of $T^2$, while their intersection has two components of that homotopy type. The map in the Mayer–Vietoris sequence can be writtenAdding the output coordinates and using the input coordinate $a+b$ separates an identity map from $f_{\ast }-I$. Canceling the identity summands gives the Wang sequence, and henceIn degree one,Its image and kernel are both generated by $(1,-1)$; its cokernel is $\mathbb{Z}$ since this image is primitive. Its Smith normal form is $\mathrm{diag}(1,0)$, so its torsion subgroup is zero. On the other two nonzero homology groups of the torus, $f_{\ast }-I=0$. The displayed short exact sequences split because their right-hand groups are free. ThereforeIn particular, the mapping torus contributes an additional degree-one class from the time circle, as well as the invariant degree-one class in the fiber.