= Solution
Suppose $X=A\cup B$ and that the <interiors> of $A$ and $B$ cover $X$. The <Mayer-Vietoris theorem> gives the <long exact sequence in homology>
$$
\cdots\longrightarrow H_n(A\cap B)\xrightarrow{(i_*,-j_*)}H_n(A)\oplus H_n(B)\xrightarrow{k_*+l_*}H_n(X)\xrightarrow{\partial}H_{n-1}(A\cap B)\longrightarrow\cdots,
$$
where all four displayed maps except the <connecting homomorphism> $\partial$ are induced by the relevant <inclusion maps>.
To prove it, let $C_*^{A,B}(X)$ be the <chain complex> generated by those <singular simplices> whose images lie wholly in $A$ or wholly in $B$. The sequence
$$
0\longrightarrow C_*(A\cap B)\xrightarrow{c\mapsto(c,-c)}C_*(A)\oplus C_*(B)\xrightarrow{(a,b)\mapsto a+b}C_*^{A,B}(X)\longrightarrow0
$$
is a <short exact sequence> of <chain complexes>. Repeated <barycentric subdivision> makes every singular simplex small enough to lie in $A$ or $B$, and the subdivision operator is <chain homotopic> to the identity. The inclusion $C_*^{A,B}(X)\hookrightarrow C_*(X)$ therefore induces an <isomorphism> on every <homology group>. The <long exact homology sequence of a short exact sequence of chain complexes> now gives the displayed sequence. Explicitly, if an $n$-cycle $z=a+b$ with $a\in C_n(A)$ and $b\in C_n(B)$, then $\partial a=-\partial b$ lies in $C_{n-1}(A\cap B)$ and represents the connecting class $\partial[z]$.
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