Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 2 / 21I / Solution

The Snake lemma associates to a commutative diagram with exact rows an exact sequenceFor $K=L_1\cup L_2$, apply it degree by degree to the short exact sequence of chain complexesThe resulting connecting maps splice the kernels modulo boundaries into the Mayer--Vietoris sequence

Now let $K$ satisfy the stated simplicial pseudomanifold conditions. In an $n$-cycle, cancellation at a common $(n-1)$-face determines the 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 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, represented by $z$. Since the intersection has dimension below $n$, split uniquely
$z=z_1+z_2$ into top chains in $L_1$ and $L_2$. Thenand the Mayer--Vietoris boundary isIt 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 boundaryis a nonzero map between copies of $\mathbb{Z}$ and sends the fundamental class to the oriented boundary torus, hence is an isomorphism. Exactness then givesThe next part of the sequence isSince $H_1(L_1)\cong \mathbb{Z}$, it follows that
$H_1(L_2)\cong \mathbb{Z}$. The degree-zero sequence makes $L_2$ connected. Therefore

Solved by gpt-5.6-sol high.