The given homeomorphisms and <homotopy invariance of homology> give
$$
H_1(K)=H_1(S^n)=0
$$
because $n\geq2$, while $M\simeq S^{n-1}$ is path connected and $L$ is the <disjoint union> of two copies of $S^{n-1}$. Hence
$$
H_0(M)\cong\mathbb Z,
\qquad
H_0(L)\cong\mathbb Z^2,
\qquad
H_0(K)\cong\mathbb Z.
$$
The degree-one and degree-zero part of the <Mayer–Vietoris sequence> is therefore
$$
0\longrightarrow\mathbb Z^2
\longrightarrow\mathbb Z\oplus H_0(N)
\longrightarrow\mathbb Z\longrightarrow0.
$$
Taking the <rank of an abelian group> throughout this short exact sequence gives
$$
2=1+\operatorname{rank}H_0(N)-1,
$$
so $\operatorname{rank}H_0(N)=2$. Since the <zeroth homology group> is freely generated by the <path components>,
$$
\boxed{|N|\text{ has exactly two path components}.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 4 / 21H / c / Solution

{title2=$H_0(X)$}

The zeroth homology group is the free <abelian group> on the <path components> of $X$. In particular, if $X$ has finitely many path components, then
$$
H_0(X;\mathbb Z)\cong\mathbb Z^{\#\pi_0(X)}.
$$

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 Incoming links: Zeroth homology group