Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 4 / 21H (Algebraic Topology)

Let $i:L\hookrightarrow M$, $j:L\hookrightarrow N$, $r:M\hookrightarrow K$, and $s:N\hookrightarrow K$ be the inclusions. The Mayer–Vietoris sequence is the long exact sequence in homologyThe same sequence continues through degree zero, or uniformly in all degrees when written with reduced homology.

Represent a class in $H_k(K)$ by a chain cycle $z\in Z_k(K)$. Since every simplex of $K=M\cup N$ belongs to at least one of the two simplicial subcomplexes, split the chain in the simplicial chain complex asBecause $\partial z=0$, we haveThe left side lies in $C_{k-1}(M)$ and the right side in $C_{k-1}(N)$, so this common chain is supported in $L=M\cap N$. It is a cycle because ${\partial }^2=0$. The connecting homomorphism is thereforeChanging the decomposition or the representative changes $\partial m$ only by a boundary in $L$, which is why the construction descends to homology classes.

The given homeomorphisms and homotopy invariance of homology givebecause $n\ge 2$, while $M\simeq S^{n-1}$ is path connected and $L$ is the disjoint union of two copies of $S^{n-1}$. HenceThe degree-one and degree-zero part of the Mayer–Vietoris sequence is thereforeTaking the rank of an abelian group throughout this short exact sequence givesso $\mathrm{rank}{H}_0(N)=2$. Since the zeroth homology group is freely generated by the path components,