= Solution
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> as
$$
z=m+n,
\qquad m\in C_k(M),\quad n\in C_k(N).
$$
Because $\partial z=0$, we have
$$
\partial m=-\partial n.
$$
The 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 therefore
$$
\boxed{\partial_*[z]=[\partial m]=[-\partial n]\in H_{k-1}(L)}.
$$
Changing the decomposition or the representative changes $\partial m$ only by a boundary in $L$, which is why the construction descends to <homology classes>.
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