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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

For a smooth integral variety, every Weil divisor is a Cartier divisor, so the Picard group equals the divisor class group. If $U$ is obtained by removing irreducible Weil divisors $D_j$, the localization sequence for the divisor class group gives the displayed exact sequence. A rational trivialization of a line bundle on $U$ has divisor supported on the removed divisors, describing the kernel concretely.

Let $A$ be a Noetherian ring that is an integral domain. Ifis exact, $N$ is a finitely generated free module, and $P$ is a torsion-free module, then $M$ is a reflexive module. In particular, the dual of every finitely generated $A$-module is reflexive.

If $0\to \mathcal{F}\to {\mathcal{A}}^0\to {\mathcal{A}}^1\to \cdots$ is an exact sequence of sheaves of abelian groups on $X$ and $H^q(X,{\mathcal{A}}^j)=0$ for every $q>0$ and $j\ge 0$, then sheaf cohomology is computed by the cochain complex of global sections:In particular a flasque resolution computes sheaf cohomology. Acyclicity is required on the space on which cohomology is being computed.

A sheaf of abelian groups is a sheaf of sets whose sections on each open set form an abelian group and whose restriction maps are group homomorphisms. Its stalk of a sheaf is the corresponding group of germs. Kernels are computed on sections, whereas cokernels are obtained by sheafification of the sectionwise cokernel. An exact sequence can therefore be checked on stalks.

The Tor functor is the homology of a tensor product of modules with a free resolution: ${\mathrm{Tor}}_i^R(M,N)=H_i(F_{\bullet }{\otimes }_{R}N)$ for a free resolution $F_{\bullet }$ of $M$. It measures the failure of tensoring to preserve exact sequences. For a commutative ring, one may resolve either argument; the double complex of the two resolutions proves that both computations agree.