Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 4 / 21H (Algebraic Topology)

For an open cover $X=A\cup B$, the Mayer–Vietoris theorem gives the long exact sequenceThe same sequence holds for a union of simplicial subcomplexes. All homology groups here have integer coefficients.

An explicit triangulation of $X_q$ is as follows. Take vertices $b_0,b_1,b_2$ forming a target circle, a circle of $3q$ vertices $a_0,\dots ,a_{3q-1}$, and a cone vertex $c$. With subscripts on $a$ interpreted modulo $3q$ and on $b$ modulo three, include trianglesfor every $j$, together with all their faces. The first two types triangulate the mapping cylinder of the degree-$q$ covering $a_j\mapsto b_j$; the last type fills the inner circle by a disc. A filled disc with an attached annular collar is still a disc, and identifying the outer collar boundary by this covering gives exactly the source quotient. Distinct inner vertices ensure this is a genuine simplicial complex, not just a one-cell presentation.

Let $A$ be the mapping cylinder and $B$ the coned disc. Then $A$ retracts to the target circle, $B$ is contractible and $A\cap B$ is the inner circle. Its inclusion into $A$ induces multiplication by $q$ on $H_1$. The exact sequence therefore reduces toConnectedness and the absence of higher-dimensional simplices giveThis is a two-dimensional Moore space with torsion in degree one.

The suspension of a connected complex is the union of two cones whose intersection retracts to the original complex. Reduced Mayer-Vietoris givesIn particular its $H_1$ vanishes. For the wedge of two connected polyhedra, choose neighborhoods retracting onto the two summands with contractible intersection at the wedge point. ThenDecompose each prescribed $G_i$ into cyclic summands. Realize each infinite cyclic summand in degree $i$ by $S^i$, and each finite nontrivial summand $\mathbb{Z}/q$ by $S^{i-1}{X}_q$, meaning $i-1$ successive suspensions. Take their finite wedge, or a point if there are no summands. The suspension formula shifts the torsion to degree $i$ and creates no other positive-degree homology, while the wedge formula adds summands. Thus the resulting connected polyhedron satisfies $\boxed{H_0(X)=\mathbb{Z},H_i(X)=G_i(1\le i\le n),H_i(X)=0(i>n)}$. The finite triangulations above, their simplicial suspensions and wedges provide an actual polyhedron.