[proofplan] We work with augmented simplicial chain complexes, since reduced homology is the homology of the augmented complex. The inclusions of subcomplexes give a short exact sequence of chain complexes whose maps are $c \mapsto (c,-c)$ and $(a,b) \mapsto a+b$. We then construct the connecting homomorphism explicitly and verify exactness at all three types of terms in the resulting sequence. This gives the Mayer–Vietoris long exact sequence, including the reduced low-degree terms. [/proofplan]

[step:Build the short exact sequence of augmented chain complexes]For every simplicial complex $X$, let \begin{align*} \widetilde{C}_*(X;k) \end{align*} denote the augmented simplicial chain complex over $k$: for $i \geq 0$, $\widetilde{C}_i(X;k)=C_i(X;k)$ is the $k$-[vector space](/page/Vector%20Space) generated by the oriented $i$-simplices of $X$; $\widetilde{C}_{-1}(X;k)=k$; and $\widetilde{C}_i(X;k)=0$ for $i<-1$. The augmented boundary \begin{align*} \widetilde{\partial}_i^X:\widetilde{C}_i(X;k)\to \widetilde{C}_{i-1}(X;k) \end{align*} is the usual simplicial boundary for $i\geq 1$, the augmentation map for $i=0$, and the zero map for $i\leq -1$. Define a chain map \begin{align*} \iota_*:\widetilde{C}_*(A\cap B;k)\to \widetilde{C}_*(A;k)\oplus \widetilde{C}_*(B;k) \end{align*} by $\iota_*(c)=(c,-c)$ for each chain $c\in \widetilde{C}_*(A\cap B;k)$. Define a chain map \begin{align*} q_*:\widetilde{C}_*(A;k)\oplus \widetilde{C}_*(B;k)\to \widetilde{C}_*(K;k) \end{align*} by $q_*(a,b)=a+b$ for each pair $(a,b)\in \widetilde{C}_*(A;k)\oplus \widetilde{C}_*(B;k)$. Here each chain is viewed in the larger complex by the inclusion of the relevant subcomplex. These maps commute with the augmented boundary maps because simplicial boundaries and the augmentation map are functorial under inclusions of subcomplexes. Therefore $\iota_*$ and $q_*$ are chain maps. For each degree $i\in \mathbb{Z}$, the sequence \begin{align*} 0 \to \widetilde{C}_i(A\cap B;k) \xrightarrow{\iota_i} \widetilde{C}_i(A;k)\oplus \widetilde{C}_i(B;k) \xrightarrow{q_i} \widetilde{C}_i(K;k) \to 0 \end{align*} is exact. Injectivity of $\iota_i$ follows from the first coordinate. The equality $q_i\circ \iota_i=0$ follows from $q_i(c,-c)=c-c=0$. It remains to identify $\ker q_i$ and prove surjectivity of $q_i$. For $i\geq 0$, the oriented $i$-simplices form bases for the simplicial chain groups. Since $K=A\cup B$ as a union of subcomplexes, every oriented simplex of $K$ belongs to $A$ or to $B$, so every basis element of $C_i(K;k)$ lies in the image of $q_i$. Thus $q_i$ is surjective. If $(a,b)\in \ker q_i$, then $a+b=0$ as a chain in $C_i(K;k)$. Comparing coefficients on the basis of oriented simplices of $K$, every simplex lying only in $A$ has coefficient zero in $a$, every simplex lying only in $B$ has coefficient zero in $b$, and on each simplex lying in $A\cap B$ the coefficients of $a$ and $b$ are negatives. Hence $(a,b)=(c,-c)$ for a unique $c\in C_i(A\cap B;k)$. For $i=-1$, the sequence is \begin{align*} 0 \to k \to k\oplus k \to k \to 0, \end{align*} with maps $\lambda\mapsto(\lambda,-\lambda)$ and $(\lambda,\mu)\mapsto \lambda+\mu$, which is exact by direct calculation. For $i<-1$, all three terms are zero. Thus \begin{align*} 0 \to \widetilde{C}_*(A\cap B;k) \xrightarrow{\iota_*} \widetilde{C}_*(A;k)\oplus \widetilde{C}_*(B;k) \xrightarrow{q_*} \widetilde{C}_*(K;k) \to 0 \end{align*} is a short exact sequence of chain complexes.[/step]

[guided]The reduced version of Mayer–Vietoris must use augmented chains, because reduced homology is not computed by the ordinary chain complex alone in degree $0$. Thus, for a simplicial complex $X$, we use the augmented chain complex $\widetilde{C}_*(X;k)$, where $\widetilde{C}_i(X;k)=C_i(X;k)$ for $i\geq 0$, $\widetilde{C}_{-1}(X;k)=k$, and $\widetilde{C}_i(X;k)=0$ for $i<-1$. The boundary in degree $0$ is the augmentation map to $k$. The maps are the algebraic expression of inclusion-exclusion: \begin{align*} \iota_*:\widetilde{C}_*(A\cap B;k)\to \widetilde{C}_*(A;k)\oplus \widetilde{C}_*(B;k) \end{align*} is defined by $c\mapsto(c,-c)$, and \begin{align*} q_*:\widetilde{C}_*(A;k)\oplus \widetilde{C}_*(B;k)\to \widetilde{C}_*(K;k) \end{align*} is defined by $(a,b)\mapsto a+b$. The minus sign in $\iota_*$ is forced: it ensures that $q_*\iota_*=0$. We verify exactness degree by degree. For $i\geq 0$, simplicial chains have a basis given by oriented $i$-simplices. Since $K=A\cup B$ as a union of subcomplexes, every simplex of $K$ is a simplex of $A$ or a simplex of $B$. Therefore every basis chain in $C_i(K;k)$ is in the image of $q_i$, so $q_i$ is surjective. Now suppose $(a,b)\in C_i(A;k)\oplus C_i(B;k)$ satisfies $q_i(a,b)=a+b=0$ in $C_i(K;k)$. Comparing coefficients of each oriented simplex of $K$, any simplex appearing only in $A$ must have coefficient zero in $a$, and any simplex appearing only in $B$ must have coefficient zero in $b$. On a simplex belonging to both $A$ and $B$, the coefficients must be negatives of one another. Hence there is a unique chain $c\in C_i(A\cap B;k)$ such that $(a,b)=(c,-c)$. This proves $\ker q_i=\operatorname{im}\iota_i$. In degree $-1$, the augmented groups are all equal to $k$, and the same maps are \begin{align*} 0 \to k \to k\oplus k \to k \to 0. \end{align*} The first map is $\lambda\mapsto(\lambda,-\lambda)$ and the second map is $(\lambda,\mu)\mapsto\lambda+\mu$. Its kernel is exactly the set of pairs $(\lambda,-\lambda)$, so exactness also holds in degree $-1$. In lower degrees every group is zero. Therefore we have a short exact sequence of augmented chain complexes.[/guided]

[step:Define the connecting homomorphism on reduced homology] For each $i\in \mathbb{Z}$, define \begin{align*} \delta_i:\widetilde{H}_i(K;k)\to \widetilde{H}_{i-1}(A\cap B;k) \end{align*} as follows. Let $[z]\in \widetilde{H}_i(K;k)$ be represented by a cycle \begin{align*} z\in \widetilde{C}_i(K;k), \qquad \widetilde{\partial}_i^K z=0. \end{align*} Since $q_i$ is surjective, choose a pair \begin{align*} (a,b)\in \widetilde{C}_i(A;k)\oplus \widetilde{C}_i(B;k) \end{align*} such that $q_i(a,b)=z$. Because $q_*$ is a chain map, \begin{align*} q_{i-1}\bigl(\widetilde{\partial}_i^A a,\widetilde{\partial}_i^B b\bigr) = \widetilde{\partial}_i^K q_i(a,b) = \widetilde{\partial}_i^K z = 0. \end{align*} Thus \begin{align*} \bigl(\widetilde{\partial}_i^A a,\widetilde{\partial}_i^B b\bigr)\in \ker q_{i-1}. \end{align*} By exactness at the chain level, there is a unique chain \begin{align*} c\in \widetilde{C}_{i-1}(A\cap B;k) \end{align*} such that \begin{align*} \iota_{i-1}(c)=\bigl(\widetilde{\partial}_i^A a,\widetilde{\partial}_i^B b\bigr). \end{align*} Equivalently, \begin{align*} \widetilde{\partial}_i^A a=c,\qquad \widetilde{\partial}_i^B b=-c. \end{align*} The chain $c$ is a cycle. Indeed, \begin{align*} \iota_{i-2}\bigl(\widetilde{\partial}_{i-1}^{A\cap B}c\bigr) = \bigl(\widetilde{\partial}_{i-1}^A\widetilde{\partial}_i^A a,\widetilde{\partial}_{i-1}^B\widetilde{\partial}_i^B b\bigr) = (0,0), \end{align*} and $\iota_{i-2}$ is injective. Hence $\widetilde{\partial}_{i-1}^{A\cap B}c=0$. Set \begin{align*} \delta_i([z])=[c]. \end{align*} This class is independent of the chosen lift $(a,b)$ and of the representative $z$. If another lift of the same $z$ is $(a',b')$, then $(a-a',b-b')\in \ker q_i$, so $(a-a',b-b')=\iota_i(d)=(d,-d)$ for some $d\in \widetilde{C}_i(A\cap B;k)$. Taking boundaries gives the corresponding chains $c$ and $c'$ with \begin{align*} c-c'=\widetilde{\partial}_i^{A\cap B}d, \end{align*} so $[c]=[c']$. If $z$ is replaced by $z+\widetilde{\partial}_{i+1}^K w$, choose a lift $(u,v)$ of $w$ and replace $(a,b)$ by \begin{align*} (a,b)+\bigl(\widetilde{\partial}_{i+1}^A u,\widetilde{\partial}_{i+1}^B v\bigr). \end{align*} The boundary of this new lift is unchanged because boundaries square to zero. Thus $\delta_i$ is a well-defined $k$-[linear map](/page/Linear%20Map). [/step]

[step:Verify exactness at $\widetilde{H}_i(A;k)\oplus\widetilde{H}_i(B;k)$] Let \begin{align*} \alpha_i:\widetilde{H}_i(A\cap B;k)\to \widetilde{H}_i(A;k)\oplus\widetilde{H}_i(B;k) \end{align*} be the map induced by $\iota_i$, and let \begin{align*} \beta_i:\widetilde{H}_i(A;k)\oplus\widetilde{H}_i(B;k)\to \widetilde{H}_i(K;k) \end{align*} be the map induced by $q_i$. Since $q_i\circ \iota_i=0$, [functoriality of homology](/theorems/4530) gives \begin{align*} \beta_i\circ \alpha_i=0. \end{align*} Thus $\operatorname{im}\alpha_i\subset \ker\beta_i$. Conversely, let $([a],[b])\in \ker\beta_i$, where \begin{align*} a\in \widetilde{C}_i(A;k),\qquad b\in \widetilde{C}_i(B;k) \end{align*} are cycles. The condition $\beta_i([a],[b])=0$ means that $a+b$ is a boundary in $\widetilde{C}_i(K;k)$, so there exists \begin{align*} w\in \widetilde{C}_{i+1}(K;k) \end{align*} with \begin{align*} a+b=\widetilde{\partial}_{i+1}^K w. \end{align*} Choose \begin{align*} (u,v)\in \widetilde{C}_{i+1}(A;k)\oplus \widetilde{C}_{i+1}(B;k) \end{align*} such that $q_{i+1}(u,v)=w$. Then \begin{align*} q_i\bigl(a-\widetilde{\partial}_{i+1}^A u,b-\widetilde{\partial}_{i+1}^B v\bigr) = a+b-\widetilde{\partial}_{i+1}^K w = 0. \end{align*} By exactness of the chain sequence, there is \begin{align*} c\in \widetilde{C}_i(A\cap B;k) \end{align*} such that \begin{align*} \iota_i(c)=\bigl(a-\widetilde{\partial}_{i+1}^A u,b-\widetilde{\partial}_{i+1}^B v\bigr). \end{align*} The pair on the right is a cycle, so injectivity of $\iota_{i-1}$ implies $\widetilde{\partial}_i^{A\cap B}c=0$. Therefore $[c]\in \widetilde{H}_i(A\cap B;k)$ and \begin{align*} \alpha_i([c])=([a],[b]). \end{align*} Hence $\ker\beta_i\subset \operatorname{im}\alpha_i$, and exactness holds at $\widetilde{H}_i(A;k)\oplus\widetilde{H}_i(B;k)$. [/step]

[step:Verify exactness at $\widetilde{H}_i(K;k)$] We first have $\delta_i\circ \beta_i=0$. Indeed, if a class in $\widetilde{H}_i(K;k)$ is represented by $q_i(a,b)$ with $a$ and $b$ cycles, then the lift $(a,b)$ has boundary $(0,0)$, so the connecting construction gives $\delta_i([q_i(a,b)])=0$. Conversely, let $[z]\in \ker\delta_i$, where \begin{align*} z\in \widetilde{C}_i(K;k) \end{align*} is a cycle. Choose a lift \begin{align*} (a,b)\in \widetilde{C}_i(A;k)\oplus \widetilde{C}_i(B;k) \end{align*} with $q_i(a,b)=z$. Let \begin{align*} c\in \widetilde{C}_{i-1}(A\cap B;k) \end{align*} be the cycle obtained from the connecting construction, so that \begin{align*} \iota_{i-1}(c)=\bigl(\widetilde{\partial}_i^A a,\widetilde{\partial}_i^B b\bigr). \end{align*} Since $\delta_i([z])=0$, there exists \begin{align*} d\in \widetilde{C}_i(A\cap B;k) \end{align*} with \begin{align*} c=\widetilde{\partial}_i^{A\cap B}d. \end{align*} Replace $(a,b)$ by \begin{align*} (a,b)-\iota_i(d)=(a-d,b+d). \end{align*} Its boundary is \begin{align*} \bigl(\widetilde{\partial}_i^A a,\widetilde{\partial}_i^B b\bigr)-\iota_{i-1}\bigl(\widetilde{\partial}_i^{A\cap B}d\bigr) = \iota_{i-1}(c)-\iota_{i-1}(c) = (0,0). \end{align*} Thus $a-d$ and $b+d$ are cycles, and \begin{align*} q_i(a-d,b+d)=q_i(a,b)=z. \end{align*} Therefore \begin{align*} [z]=\beta_i([a-d],[b+d]), \end{align*} so $[z]\in \operatorname{im}\beta_i$. Hence $\ker\delta_i=\operatorname{im}\beta_i$. [/step]

[step:Verify exactness at $\widetilde{H}_{i-1}(A\cap B;k)$ and assemble the sequence] We first show $\alpha_{i-1}\circ \delta_i=0$. Let $[z]\in \widetilde{H}_i(K;k)$, choose a lift $(a,b)$ of a cycle representative $z$, and let $c\in \widetilde{C}_{i-1}(A\cap B;k)$ satisfy \begin{align*} \iota_{i-1}(c)=\bigl(\widetilde{\partial}_i^A a,\widetilde{\partial}_i^B b\bigr). \end{align*} Then \begin{align*} \alpha_{i-1}([c]) = \bigl([\widetilde{\partial}_i^A a],[\widetilde{\partial}_i^B b]\bigr) = (0,0), \end{align*} because boundaries represent zero homology classes. Hence $\operatorname{im}\delta_i\subset \ker\alpha_{i-1}$. Conversely, let $[c]\in \ker\alpha_{i-1}$, where \begin{align*} c\in \widetilde{C}_{i-1}(A\cap B;k) \end{align*} is a cycle. The condition $\alpha_{i-1}([c])=0$ means that there exist chains \begin{align*} a\in \widetilde{C}_i(A;k),\qquad b\in \widetilde{C}_i(B;k) \end{align*} such that \begin{align*} \widetilde{\partial}_i^A a=c,\qquad \widetilde{\partial}_i^B b=-c. \end{align*} Therefore \begin{align*} q_{i-1}\bigl(\widetilde{\partial}_i^A a,\widetilde{\partial}_i^B b\bigr) = q_{i-1}(c,-c) = 0. \end{align*} Equivalently, \begin{align*} \widetilde{\partial}_i^K q_i(a,b)=0, \end{align*} so $z:=q_i(a,b)$ is a cycle in $\widetilde{C}_i(K;k)$. By the definition of the connecting map, $\delta_i([z])=[c]$. Hence $\ker\alpha_{i-1}\subset \operatorname{im}\delta_i$. We have proved exactness at each type of term: \begin{align*} \widetilde{H}_i(A\cap B;k),\qquad \widetilde{H}_i(A;k)\oplus\widetilde{H}_i(B;k),\qquad \widetilde{H}_i(K;k). \end{align*} Since the construction uses augmented chain complexes in all degrees, the same argument includes the reduced low-degree terms. Thus the natural long exact sequence \begin{align*} \cdots \to \widetilde{H}_i(A \cap B;k) \xrightarrow{\alpha_i} \widetilde{H}_i(A;k)\oplus\widetilde{H}_i(B;k) \xrightarrow{\beta_i} \widetilde{H}_i(K;k) \xrightarrow{\delta_i} \widetilde{H}_{i-1}(A\cap B;k) \to \cdots \end{align*} is exact. This is the Mayer–Vietoris sequence for reduced simplicial homology. [/step]