 Solution

Use simplicial homology with a fixed coefficient group or ring. The Mayer–Vietoris sequence isThe inclusion maps define $\alpha ([z])=([z],-[z])$ and $\beta ([a],[b])=[a+b]$. To define $\delta$, split a simplicial cycle $c\in C_{i+1}(X)$ as $c=a+b$ with $a$ a chain in $A$ and $b$ in $B$. Since $\partial a=-\partial b$, this boundary lies in $C_i(A\cap B)$, and $\delta [c]=[\partial a]$. At degree zero the unreduced sequence ends with $H_0(A)\oplus H_0(B)\to H_0(X)\to 0$.

If $[z]=\delta [c]$, its representative is a boundary in $A$ and the negative of a boundary in $B$, so $\alpha [z]=0$. Thus $\mathrm{im}\delta \subseteq \ker \alpha$.

Conversely if $\alpha [z]=0$, choose $a\in C_{i+1}(A)$ and $b\in C_{i+1}(B)$ with $\partial a=z=\partial b$. The chain $c=a-b$ is a simplicial cycle in $X$. Its splitting into the $A$ chain $a$ and $B$ chain $-b$ gives $\delta [c]=[z]$. Hence $\mathrm{im}\delta =\ker \alpha$, the requested exactness at $H_i(A\cap B)$.