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A filled triangle is a disk with three vertices, three edges and one two-dimensional topological cell, so $\chi =3-3+1=1$. Its simplicial homology has $(b_0,b_1,b_2)=(1,0,0)$: the filled triangle is contractible, and the three-edge chain cycle is the chain boundary of its face. Keeping only the cycle graph on those three vertices gives $\chi =3-3=0$ and $(b_0,b_1)=(1,1)$, since that chain cycle no longer bounds a face. The surrounding plane's exterior is not part of either simplicial complex; do not add it to $F$.

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 $H_i(A\cap B)$.