$\beta_{t+1,t+4}(I)\neq 0$, and so $\projdim(I)\geq t+1$, a contradiction. Thus the claim follows.

Now Hochster’s formula implies that $\beta_{t+1,t+4}(I)\neq 0$, and so $\projdim(I)\geq t+1$, a contradiction. Thus the claim follows.

Our claim implies that $(\bar{G})_W$ is connected, if $W\cap V(C)\neq \emptyset$.

Now suppose that $W\cap V(C)= \emptyset$. Then $|W\setminus V(C)|=|W|=t+2\geq 3$. Suppose that there exist $j, j'\in W$ such that $\{j, j'\}\notin E(\bar{G})$. Let $W''=V(C)\cup\{ j, j'\}$. Since $C$ is a minimal cycle and since $j, j'$ are neighbors of all vertices of $C$ we have $\mathcal{F}(\Delta_{W''})=\{\{i, i+1, j\}, \{i, i+1, j'\}:\ 1\leq i\leq t+2\}\cup \{\{1, t+3, j\}, \{1, t+3, j'\}\}$. It follows that $\Delta_{W''}$, as a topological space, is homotopy equivalent to $\mathcal{S}_2$. Therefore $\widetilde{H}_2(\Delta_{W''}; K)\neq 0$ and so $\beta_{t+1, t+5}\neq 0$, by Hochster’s formula. This implies that $\projdim(I)\geq t+1$, a contradiction. So in this case $\{j, j'\}\in E(\bar{G})$ for all $j, j'\in W$. It follows that $(\bar{G})_W$ is a complete graph and so is connected. This completes the proof.

\[beta\] Let $G$ be a simple graph on the vertex set $[n]$ with no isolated vertices, and let $I$ be its edge ideal. Suppose that $\index(I)=\projdim(I)=t$. Then

-   $n=t+3$,

-   $\beta_{t,t+3}(I)=1$.

\(a) Let $\Delta=\Delta(\bar{G})$. By Lemma \[index\], $\beta_{t, t+2}(I)=0$, and 