direct image map ${\mathcal S}(G)\to {\mathcal S}(G/TU')$ with $\psi'$, which shows that $\tilde \psi$ is also right $TU'$ invariant. Since $G$ is assumed to be almost simple, $U$ and $U'$ together generate $G$, thus we see that $\tilde \psi$ is right $G$ invariant. It follows that $\tilde \psi$ is proportional to the functional $\phi\mapsto \int \phi$, hence the functional $(\psi,\psi')$ factors through the differential in , which yields exactness of .

We can now finish the proof of Conjecture \[imb\](c) in the present case. Breaking into short exact sequences we get $$0\to {\mathcal K}_{\Omega}\to {\mathcal S}(G/T)\to M\to 0,$$ $$0\to M\to {\mathcal S}(G/TU)\oplus {\mathcal S}(G/TU')\to {{\mathbb C}}\to 0.$$ Considering the corresponding long exact sequences on homology we see that it suffices to check that the map ${{\mathbb C}}=H_0(G, {\mathcal S}(G/T))\to H_0(G,M)$ is nonzero while the map $H_1(G, {\mathcal S}(G/T)) \to H_1(G,M)$ has one-dimensional cokernel. The former statement is clear since the composition $H_0(G, {\mathcal S}(G/T))\to H_0(G,M)\to
 H_0(G,{\mathcal S}(G/TU))$ is nonzero. To check the latter recall that $H_i(G,{{\mathbb C}})=0$ for $i>0$ since the resolution of ${{\mathbb C}}$ provided by the simplicial complex for computation of homology of the Bruhat-Tits building ${{\mathfrak B}}$ shows that $H_i(G,{{\mathbb C}})\cong H_i({{\mathfrak B}}/G)$, while ${{\mathfrak B}}/G$ is 