[proofplan] The strategy is to produce a short exact sequence of chain complexes and then invoke the [Snake Lemma](/theorems/1930) to obtain the long exact sequence on homology. In each degree $n$, we form the sequence $0 \to C_n(L) \to C_n(M) \oplus C_n(N) \to C_n(K) \to 0$ with the "diagonal" inclusion on the left and the "difference" on the right. The left map is injective for trivial reasons (the diagonal has zero kernel); the right map is surjective because every simplex of $K$ sits in $M$ or in $N$ by the hypothesis $K = M \cup N$; and exactness in the middle follows from the fact that a pair $(a, b)$ produces zero on the right exactly when $a$ and $b$ record the same simplicial chain on their common part $L$. Verifying that the maps are chain maps (i.e., commute with the differentials) is mechanical: inclusions of subcomplexes commute with the boundary operator. Applying the Snake Lemma produces the Mayer–Vietoris long exact sequence. The tail terminates at $H_0(K) \to 0$ because the Snake Lemma's long sequence does. [/proofplan]

[step:Define the chain-level diagonal and difference maps] Fix $n \geq 0$. The simplicial chain groups $C_n(L), C_n(M), C_n(N), C_n(K)$ are free abelian groups on the $n$-simplices of $L, M, N, K$ respectively. Since $L \subseteq M$, $L \subseteq N$, $M \subseteq K$, $N \subseteq K$ as subcomplexes, every $n$-simplex of $L$ is an $n$-simplex of $M$ and of $N$, and every $n$-simplex of $M$ or of $N$ is an $n$-simplex of $K$. The induced chain maps on the associated chain complexes are \begin{align*} i_n: C_n(L) &\hookrightarrow C_n(M), & j_n: C_n(L) &\hookrightarrow C_n(N), \\ k_n: C_n(M) &\hookrightarrow C_n(K), & \ell_n: C_n(N) &\hookrightarrow C_n(K), \end{align*} each of them a direct-summand inclusion at the level of generators (an $n$-simplex of the smaller complex is sent to the same $n$-simplex viewed in the larger complex). These are chain maps by [Simplicial Maps Induce Chain Maps](/theorems/1926), since each inclusion of subcomplexes is a simplicial map. Define the diagonal map \begin{align*} \Phi_n: C_n(L) &\to C_n(M) \oplus C_n(N) \\ x &\mapsto (i_n(x), j_n(x)) \end{align*} and the difference map \begin{align*} \Psi_n: C_n(M) \oplus C_n(N) &\to C_n(K) \\ (a, b) &\mapsto k_n(a) - \ell_n(b). \end{align*} Both are $\mathbb{Z}$-linear. [/step]

[step:Verify that $\Phi_\bullet$ and $\Psi_\bullet$ are chain maps] To assemble the degree-wise maps into maps of chain complexes, we verify that $\Phi_\bullet$ and $\Psi_\bullet$ commute with the boundary operators, where $C_\bullet(M) \oplus C_\bullet(N)$ carries the boundary $d_n^{M \oplus N}(a, b) := (d_n^M a, d_n^N b)$. For $\Phi$: using that $i_\bullet$ and $j_\bullet$ are chain maps, \begin{align*} d_n^{M \oplus N}(\Phi_n(x)) = (d_n^M i_n(x), d_n^N j_n(x)) = (i_{n-1}(d_n^L x), j_{n-1}(d_n^L x)) = \Phi_{n-1}(d_n^L x). \end{align*} For $\Psi$: using that $k_\bullet$ and $\ell_\bullet$ are chain maps and linearity of $d_n^K$, \begin{align*} d_n^K(\Psi_n(a, b)) &= d_n^K(k_n(a) - \ell_n(b)) = k_{n-1}(d_n^M a) - \ell_{n-1}(d_n^N b) \\ &= \Psi_{n-1}(d_n^M a, d_n^N b) = \Psi_{n-1}(d_n^{M \oplus N}(a, b)). \end{align*} Hence $\Phi_\bullet$ and $\Psi_\bullet$ are chain maps. [/step]

[step:Establish exactness of $0 \to C_n(L) \xrightarrow{\Phi_n} C_n(M) \oplus C_n(N) \xrightarrow{\Psi_n} C_n(K) \to 0$ in each degree]We verify the three exactness conditions. **$\Phi_n$ is injective.** If $\Phi_n(x) = (i_n(x), j_n(x)) = (0, 0)$, then $i_n(x) = 0$. Since $i_n$ is the inclusion of the free abelian group $C_n(L)$ on $n$-simplices of $L$ into the free abelian group $C_n(M)$ on $n$-simplices of $M$, and the generators of $C_n(L)$ are genuine generators of $C_n(M)$ (i.e., distinct $n$-simplices of $M$), the map $i_n$ is an injection of free abelian groups. Hence $x = 0$, giving $\ker \Phi_n = 0$. **$\Psi_n$ is surjective.** Let $\tau \in C_n(K)$ be an $n$-chain, and decompose $\tau = \sum_\sigma m_\sigma \sigma$ over $n$-simplices $\sigma$ of $K$ with $m_\sigma \in \mathbb{Z}$. By hypothesis $K = M \cup N$, so every $n$-simplex of $K$ belongs to $M$ or to $N$ (or both). For each $\sigma$ in the sum, choose $a_\sigma \in \{0, \sigma\} \subseteq C_n(M)$ and $b_\sigma \in \{0, -\sigma\} \subseteq C_n(N)$ with $k_n(a_\sigma) - \ell_n(b_\sigma) = \sigma$: take $a_\sigma = \sigma$, $b_\sigma = 0$ when $\sigma \in M$, and $a_\sigma = 0$, $b_\sigma = -\sigma$ when $\sigma \in N \setminus M$. Then $(a, b) := (\sum_\sigma m_\sigma a_\sigma, \sum_\sigma m_\sigma b_\sigma)$ satisfies $\Psi_n(a, b) = \sum_\sigma m_\sigma \sigma = \tau$. **$\operatorname{im} \Phi_n = \ker \Psi_n$.** *Inclusion $\subseteq$:* For any $x \in C_n(L)$, \begin{align*} \Psi_n(\Phi_n(x)) = k_n(i_n(x)) - \ell_n(j_n(x)) = x - x = 0, \end{align*} using that $k_n \circ i_n$ and $\ell_n \circ j_n$ are both the inclusion $C_n(L) \hookrightarrow C_n(K)$ (as chains, both routes realise the $n$-simplex of $L$ as the same $n$-simplex of $K$). *Inclusion $\supseteq$:* Suppose $(a, b) \in C_n(M) \oplus C_n(N)$ with $\Psi_n(a, b) = k_n(a) - \ell_n(b) = 0$ in $C_n(K)$. Decompose $a = \sum_\sigma m_\sigma \sigma$ over the $n$-simplices of $M$ and $b = \sum_\tau n_\tau \tau$ over the $n$-simplices of $N$. Viewed as chains on $K$, \begin{align*} k_n(a) - \ell_n(b) = \sum_{\sigma \in M} m_\sigma \sigma - \sum_{\tau \in N} n_\tau \tau = 0. \end{align*} In the free abelian group $C_n(K)$, vanishing of this formal sum forces the coefficient at each generator to vanish. Since the simplices of $M$ and the simplices of $N$ overlap exactly in the simplices of $L = M \cap N$: - For $\sigma \in M \setminus L$ (so $\sigma \notin N$), the coefficient in the difference is $m_\sigma$, forcing $m_\sigma = 0$. - For $\tau \in N \setminus L$ (so $\tau \notin M$), the coefficient in the difference is $-n_\tau$, forcing $n_\tau = 0$. - For $\rho \in L$ (appearing in both sums), the coefficient is $m_\rho - n_\rho$, forcing $m_\rho = n_\rho$. Hence $a \in C_n(L) \subseteq C_n(M)$, $b = a \in C_n(L) \subseteq C_n(N)$, and $\Phi_n(a) = (a, a) = (a, b)$, exhibiting $(a, b) \in \operatorname{im} \Phi_n$.[/step]

[guided]The three conditions to verify are standard for exactness: injectivity of the left map, surjectivity of the right map, and agreement of image with kernel in the middle. Injectivity of $\Phi$ is essentially free: the diagonal of two injections is injective. Here both $i_n$ and $j_n$ are inclusions of free abelian groups (the $n$-simplices of $L$ genuinely are $n$-simplices of both $M$ and $N$), so the first coordinate of $\Phi$ already kills any non-trivial element. Surjectivity of $\Psi$ is where the hypothesis $K = M \cup N$ enters. The statement we need is that every $n$-simplex of $K$ sits in $M$ or in $N$, since the chain groups are free abelian on simplices. If $\sigma$ is an $n$-simplex of $M$, we produce it via the "left channel" $(+\sigma, 0) \mapsto \sigma$. If $\sigma$ is only in $N$ (not in $M$), we produce it via the "right channel with a minus sign" $(0, -\sigma) \mapsto \sigma$. The assumption $K = M \cup N$ guarantees one of the two channels is available. Exactness in the middle splits into $\Psi \Phi = 0$ and the converse. The first direction is the tautology that $x$ diagonalised has equal images in $M$ and $N$, which cancel when subtracted in $K$. The second direction — the substantive content — uses the free abelian group structure: a formal sum vanishes iff each coefficient vanishes. The $n$-simplices of $M$ and of $N$ appear as generators of $C_n(K)$, and they are distinct generators except for those in common, namely simplices of $L$. Matching coefficients forces $a$ and $b$ to agree on their common support (so they come from a common $x \in C_n(L)$) and vanish elsewhere. This is the Mayer–Vietoris short exact sequence at the chain level. It encodes the fact that $K$ is "glued" from $M$ and $N$ along $L$ by saying that $C_n(K)$ is the pushout of $C_n(M)$ and $C_n(N)$ over $C_n(L)$ in abelian groups — which is exactly the statement "$C_n(K) = (C_n(M) \oplus C_n(N))/(\Phi_n(C_n(L)))$".[/guided]

[step:Apply the Snake Lemma] By Steps 2 and 3, the diagram \begin{align*} 0 \to C_\bullet(L) \xrightarrow{\Phi_\bullet} C_\bullet(M) \oplus C_\bullet(N) \xrightarrow{\Psi_\bullet} C_\bullet(K) \to 0 \end{align*} is a short exact sequence of chain complexes of free abelian groups, with $\Phi_\bullet$ and $\Psi_\bullet$ chain maps. Applying the [Snake Lemma](/theorems/1930), there is a natural connecting homomorphism $\partial_*: H_n(K) \to H_{n-1}(L)$ such that the sequence \begin{align*} \cdots \to H_n(L) \xrightarrow{\Phi_*} H_n(M) \oplus H_n(N) \xrightarrow{\Psi_*} H_n(K) \xrightarrow{\partial_*} H_{n-1}(L) \xrightarrow{\Phi_*} H_{n-1}(M) \oplus H_{n-1}(N) \to \cdots \end{align*} is exact, where the induced maps on homology are \begin{align*} \Phi_*: H_n(L) &\to H_n(M) \oplus H_n(N), & [x] &\mapsto (i_*[x], j_*[x]) = (i_* \oplus j_*)[x] \\ \Psi_*: H_n(M) \oplus H_n(N) &\to H_n(K), & ([a], [b]) &\mapsto k_*[a] - \ell_*[b] = (k_* - \ell_*)([a], [b]). \end{align*} Here we used that the induced map on homology of a direct sum of chain complexes is the direct sum of the induced maps: $H_n(M \oplus N) = H_n(M) \oplus H_n(N)$, which follows from the boundary acting diagonally on the direct sum. [/step]

[step:Terminate the sequence at $H_0(K) \to 0$] The last nontrivial degree is $n = 0$. The Mayer–Vietoris sequence, as produced by the Snake Lemma, ends as \begin{align*} \cdots \to H_0(L) \xrightarrow{\Phi_*} H_0(M) \oplus H_0(N) \xrightarrow{\Psi_*} H_0(K) \to 0, \end{align*} with the final zero expressing surjectivity of $\Psi_*: H_0(M) \oplus H_0(N) \to H_0(K)$. This surjectivity holds because $\Psi_0: C_0(M) \oplus C_0(N) \to C_0(K)$ is surjective (Step 3): given a cycle class $[c] \in H_0(K)$ with $c \in C_0(K)$, lift $c$ to $(a, b) \in C_0(M) \oplus C_0(N)$ with $\Psi_0(a, b) = c$; then $(a, b)$ is automatically a $C_0$-cycle (every $0$-chain is a cycle, since $d_0 = 0$), and $\Psi_*([a], [b]) = [c]$. Hence the long exact sequence terminates as claimed: \begin{align*} \cdots \to H_0(L) \to H_0(M) \oplus H_0(N) \to H_0(K) \to 0. \end{align*} [/step]

[step:Assemble the Mayer–Vietoris long exact sequence] Renaming $\Phi_* = i_* \oplus j_*$ and $\Psi_* = k_* - \ell_*$ (Step 4), we obtain the Mayer–Vietoris long exact sequence \begin{align*} \cdots \to H_n(L) \xrightarrow{i_* \oplus j_*} H_n(M) \oplus H_n(N) \xrightarrow{k_* - \ell_*} H_n(K) \xrightarrow{\partial_*} H_{n-1}(L) \to \cdots \end{align*} terminating at $\cdots \to H_0(M) \oplus H_0(N) \to H_0(K) \to 0$. This is the claimed exact sequence, completing the proof. [/step]