[proofplan] We use the reduced homology formula for joins over a field, which expresses $\widetilde{H}_m(K*L;k)$ as a direct sum of tensor products $\widetilde{H}_i(K;k)\otimes_k \widetilde{H}_j(L;k)$ with $i+j=m-1$ (citing a result not yet in the wiki: [Join Homology Formula](/theorems/6472)). The vanishing hypotheses force every potentially nonzero factor from $K$ to have degree at least $r+1$ and every potentially nonzero factor from $L$ to have degree at least $s+1$. Therefore every potentially nonzero summand in degree $m$ requires $m\ge r+s+3$, so all degrees $m\le r+s+2$ vanish. [/proofplan] [step:Apply the join homology formula in degree $m$] Fix an integer $m \leq r+s+2$. Since $K$ and $L$ are finite simplicial complexes and the coefficient ring $k$ is a field, the reduced homology formula for joins gives a natural isomorphism \begin{align*} \widetilde{H}_m(K*L;k) \cong \bigoplus_{i+j=m-1} \widetilde{H}_i(K;k)\otimes_k \widetilde{H}_j(L;k), \end{align*} where the direct sum is taken over all integer pairs $(i,j)\in \mathbb{Z}^2$ satisfying $i+j=m-1$. [guided] We fix an arbitrary integer $m \leq r+s+2$ because homological $(r+s+2)$-connectedness means that every reduced homology group in such a degree must vanish. The structural input is the reduced homology formula for joins over a field. Its hypotheses apply here because $K$ and $L$ are finite simplicial complexes and the coefficients are taken in the field $k$. The formula gives the isomorphism \begin{align*} \widetilde{H}_m(K*L;k) \cong \bigoplus_{i+j=m-1} \widetilde{H}_i(K;k)\otimes_k \widetilde{H}_j(L;k). \end{align*} The index condition $i+j=m-1$ records the degree shift in the join: homology in degree $m$ of the join is assembled from homology in degrees whose sum is one less than $m$. [/guided] [/step] [step:Show that every summand in degree $m\leq r+s+2$ is zero] Let $(i,j)\in \mathbb{Z}^2$ satisfy $i+j=m-1$. Since $m \leq r+s+2$, we have \begin{align*} i+j = m-1 \leq r+s+1. \end{align*} If both inequalities $i\ge r+1$ and $j\ge s+1$ held, then \begin{align*} i+j \ge (r+1)+(s+1)=r+s+2, \end{align*} contradicting $i+j\le r+s+1$. Hence either $i\le r$ or $j\le s$. If $i\le r$, then $\widetilde{H}_i(K;k)=0$ by the homological $r$-connectedness of $K$. If $j\le s$, then $\widetilde{H}_j(L;k)=0$ by the homological $s$-connectedness of $L$. In either case, \begin{align*} \widetilde{H}_i(K;k)\otimes_k \widetilde{H}_j(L;k)=0. \end{align*} Thus every summand in the direct sum for $\widetilde{H}_m(K*L;k)$ is zero. [/step] [step:Conclude the required vanishing range] Since every summand in \begin{align*} \bigoplus_{i+j=m-1} \widetilde{H}_i(K;k)\otimes_k \widetilde{H}_j(L;k) \end{align*} is zero, the direct sum is the zero $k$-[vector space](/page/Vector%20Space). Therefore \begin{align*} \widetilde{H}_m(K*L;k)=0. \end{align*} Because $m \leq r+s+2$ was arbitrary, this vanishing holds for every integer $m\le r+s+2$. Hence $K*L$ is homologically $(r+s+2)$-connected over $k$. [/step]