Note that your statement is only true in rational cohomology. For example, $H^*(SO(5)/T)$ H^\ast(SO(5)/T)$is not generated in degree$2$(though it is rationally). The easiest proof I know starts from equivariant cohomology:$ H_T(K/TH^\ast_T(K/T) = H_{T\times H^\ast_{T\times T}(K) = H_{T\times H^\ast_{T\times K\times T}(K\times K) = H_K(K/T H^\ast_K(K/T \times K/T) $So far this uses$H^*_F(X) H^\ast_F(X) = H^*(X/F)$H^\ast(X/F)$ for free actions. Now use the equivariant K\"unneth Künneth formula:
$... = H^*K(K/TH^\ast_K(K/T) \otimes{H_K} otimes_{H^\ast_K} H_K(K/T) = H_T H^\ast_T \otimes_{H_K} H_T$otimes_{H^\ast_K} H^\ast_T$Rationally, the base ring$H_K$H^\ast_K$ is $(H_T)^W$, (H^\ast_T)^W$, the invariants. Since you didn't want equivariant but ordinary cohomology, kill the left factor, leaving${\mathbb Q} \otimes_{(H^*_T)^W} H^*_T$otimes_{(H^\ast_T)^W} H^\ast_T$, which is your desired ring of coinvariants.
(I'm having a bunch of trouble with $H$ vs. $H^*$ H^\ast$in typesetting here, sorry!) 1 Note that your statement is only true in rational cohomology. For example,$H^*(SO(5)/T)$is not generated in degree$2$(though it is rationally). The easiest proof I know starts from equivariant cohomology:$ H_T(K/T) = H_{T\times T}(K) = H_{T\times K\times T}(K\times K) = H_K(K/T \times K/T) $So far this uses$H^*_F(X) = H^*(X/F)$for free actions. Now use the equivariant K\"unneth formula:$... = H^*K(K/T) \otimes{H_K} H_K(K/T) = H_T \otimes_{H_K} H_T$Rationally, the base ring$H_K$is$(H_T)^W$, the invariants. Since you didn't want equivariant but ordinary cohomology, kill the left factor, leaving${\mathbb Q} \otimes_{(H^*_T)^W} H^*_T$, which is your desired ring of coinvariants. (I'm having a bunch of trouble with$H$vs.$H^*\$ in typesetting here, sorry!)