I am studying Bestvina and Mess's results on the boundary of hyperbolic groups [The Boundary of Negatively Curved Groups. Journal of the American Mathematical Society Vol. 4, No. 3 (Jul., 1991), pp. 469-481], but I am having trouble with the proof of Corollary 1.4, especially the final part (starting from "Assuming $dim_R\partial\Gamma=i\geq k$...").

Could anybody spell the proof out please? Or even point out a more detailed version of the proof?

Thank you so much!

I am studying Bestvina and Mess's results on the boundary of hyperbolic groups [The Boundary of Negatively Curved Groups. Journal of the American Mathematical Society Vol. 4, No. 3 (Jul., 1991), pp. 469-481], but I am having trouble with the proof of Corollary 1.4, especially the final part (starting from "Assuming $dim_R\partial\Gamma=i\geq k$...").

Could anybody spell the proof out please? Or even point out a more detailed version of the proof?

Question edited according to the comment below.

In particular, these are my doubts:

1. The statement of Corollary 1.4 says "(a) $dim_R\partial\Gamma = \max\{n\mid H^n (\Gamma; R\Gamma) \neq0 \}$". I think, in view of Corollary 1.3, it should be "(a) $dim_R\partial\Gamma = \max\{n\mid H^n (\Gamma; R\Gamma) \neq0 \}-1$", which seems also coherent with Corollary 1.4(c).
2. In the 2nd part of the proof of Corollary 1.4(a), i.e. after the proof of the claim, they write "...and construct closed sets $\partial\Gamma = X \supset B_0 \supset B_1 \supset B_2$ as in the proof of Proposition 2.6". Do they mean $P(\Gamma)\cup\partial\Gamma = X \supset B_0 \supset B_1 \supset B_2$? (Setting $\partial\Gamma=X$ seems incoherent with the following formulas)
3. If this is the case, how do they construct the $B_i$'s? Are $B_2=h_1(X)$, $B_1=h(B_2\times[0,1])$, $B_0=h(B_1\times[0,1])$, where $h:X\times[0,1]$ is a homotopy satisfying $h_0=id$, $h_t|_A=inclusion$, $h_t(X \setminus A) \subset (X \setminus \partial\Gamma)$ for $t>0$?
4. It seems to me that the general scheme of the proof can be summarized in the following diagram (all cohomology groups with coefficients in $R$): $\require{AMScd}$ \begin{CD} H^i(X,B_0) @>a_0>> H^i(\partial\Gamma\cup_AB_0,B_0)@>j_0>> H^{i+1}(X,\partial\Gamma\cup_AB_0) @>\cong>> H_c^{i+1}(P(\Gamma),B_0\cap P(\Gamma))\\ @V b_0 V V @Vc_0V V @Vd_0VV @Ve_0VV\\ H^i(X,B_1) @>a_1>> H^i(\partial\Gamma\cup_AB_1,B_1)@>j_1>> H^{i+1}(X,\partial\Gamma\cup_AB_1) @>\cong>> H_c^{i+1}(P(\Gamma),B_1\cap P(\Gamma))\\ @V b_1 V V @Vc_1V V @. @.\\ H^i(X,B_2) @>a_2>> H^i(\partial\Gamma\cup_AB_2,B_2)@>j_2>> H^{i+1}(X,\partial\Gamma\cup_AB_2) @>\cong>> H_c^{i+1}(P(\Gamma),B_2\cap P(\Gamma))\\ \end{CD}

• The (first 3 terms of the horizontal rows are the cohomology long exact sequences of the triples $(X, \partial\Gamma\cup_AB_i,B_i)$.
• The isomorphisms between groups in the last 2 columns seem to me an application of excision for $H^*_c$, taking into account that $X$ is compact, so $H^*_c(X,...)=H^*(X,...)$: is this correct?
• The vertical maps are induced by the inclusions $B_2\subset B_1\subset B_0$, so they map a cohomology class $[\varphi]$ to $[\varphi]$ (but of course the classes may be different, even if the representative can be chosen to be the same). This is why the diagram commutes.
• $c_0$ and $c_1$ are isomorphisms (by excision, each of the groups in the 2nd column is isomorphic to $H^i(\partial\Gamma, A)$).
• If $b_0=0$ then $j_0$ is injective: this is just diagram chasing. But why is $b_0=0$?
• Analogously, if $b_1=0$ then $j_1$ is injective.
• Hence, if $b_0=b_1=0$, each of $H^{i+1}(X,\partial\Gamma\cup_AB_0)$ and $H^{i+1}(X,\partial\Gamma\cup_AB_1)$ contains a copy of $H^i(\partial\Gamma, A)$; by commutativity of the diagram, $d_0$ maps one copy isomorphically onto the other, so $d_0\neq 0$.
• On the other side, by the preceding claim (first part of the proof of Corollary 1.4), $e_0=0$, which contradicts $d_0\neq 0$. Is this summarization correct?

5. Also, it seems to me that the only role played by the group $\Gamma$ is to ensure that the claim in the previous part of the proof holds, namely that for any $i>k$ there exists an integer $K$ such that every $i$-cocycle $z$ is the coboundary of a cochain whose support is contained in the combinatorial $K$-neighbourhood of the support of $z$. This is needed to ensure that the map $e_0$ is zero. Does indeed the rest of the proof only depend on the fact that the $1$-skeleton of $P(\Gamma)$ is a hyperbolic space and $\partial\Gamma$ is its boundary in purely topological terms?

Thank you so much!