= Solution
The <Snake lemma> associates to a <commutative diagram> with <exact sequences>[exact] rows an <exact sequence>
$$
\ker f_1\to\ker f_2\to\ker f_3
\xrightarrow{\delta}\operatorname{coker}f_1
\to\operatorname{coker}f_2\to\operatorname{coker}f_3.
$$
For $K=L_1\cup L_2$, apply it degree by degree to the short exact sequence of chain complexes
$$
0\to C_*(L_1\cap L_2)
\xrightarrow{c\mapsto(c,-c)}
C_*(L_1)\oplus C_*(L_2)
\xrightarrow{(a,b)\mapsto a+b}
C_*(K)\to0.
$$
The resulting <connecting homomorphisms>[connecting maps] splice the <kernels> modulo <chain boundaries>[boundaries] into the <Mayer–Vietoris theorem>[Mayer--Vietoris sequence]
$$
\cdots\to H_n(L_1\cap L_2)\to
H_n(L_1)\oplus H_n(L_2)\to H_n(K)
\xrightarrow{\partial}H_{n-1}(L_1\cap L_2)\to\cdots.
$$
Now let $K$ satisfy the stated <simplicial pseudomanifold> conditions. In an $n$-<chain cycle>[cycle], cancellation at a common $(n-1)$-<face of a simplex>[face] determines the <chain coefficient>[coefficient] of either incident $n$-<simplex> from the other, up to the <orientation> sign. Connectivity through such faces therefore determines every top-simplex coefficient from one <integer>. If the signs are globally compatible, their oriented sum is a cycle and <generator of a group>[generates] $H_n(K)\cong\mathbb Z$; if they are inconsistent, that integer must be zero and $H_n(K)=0$.
Let $x$ be the resulting <fundamental class of an orientable simplicial pseudomanifold>[fundamental class], represented by $z$. Since the intersection has dimension below $n$, split uniquely
$z=z_1+z_2$ into top chains in $L_1$ and $L_2$. Then
$$
\partial z_1=-\partial z_2\in C_{n-1}(L_1\cap L_2),
$$
and the Mayer--Vietoris boundary is
$$
\boxed{\partial x=[\partial z_1]=-[\partial z_2]}.
$$
It is nonzero exactly when both $L_1$ and $L_2$ contain top-dimensional simplices; if one side contains none, the fundamental cycle already lies in the other side, while if both do, connectedness forces a nonempty interface and its oriented boundary represents a nonzero class.
Finally take $K\cong S^3$, $L_1\cong S^1\times D^2$, and
$L_1\cap L_2\cong T^2$. The preceding boundary
$$
H_3(S^3)\longrightarrow H_2(T^2)
$$
is a nonzero map between copies of $\mathbb Z$ and sends the fundamental class to the oriented boundary torus, hence is an isomorphism. Exactness then gives
$$
H_2(L_2)=0.
$$
The next part of the sequence is
$$
0\to H_1(T^2)\cong\mathbb Z^2
\to H_1(L_1)\oplus H_1(L_2)
\to0.
$$
Since $H_1(L_1)\cong\mathbb Z$, it follows that
$H_1(L_2)\cong\mathbb Z$. The degree-zero sequence makes $L_2$ connected. Therefore
$$
\boxed{H_0(L_2)\cong\mathbb Z,\quad
H_1(L_2)\cong\mathbb Z,\quad
H_i(L_2)=0\ (i\geq2)}.
$$
Back to article page
