9:17 PM
@kiran The isomorphism $H_{\bullet} BP \simeq H_{\bullet} \otimes _{BP} BP_{\bullet} BP \otimes _{BP} H$ consists of two parts. The first is the observation that since $BP_{\bullet} BP$ is free over $BP_{\bullet}$, then since $H$ is a $BP$-algebra we have $H_{\bullet} BP \simeq H_{\bullet} \otimes _{BP} BP_{\bullet} BP$ (there's a collapsing Kunneth spectral sequence here).
The other one (which has little to do with topology) is that $H_{\bullet} \otimes _{BP} BP_{\bullet} BP \otimes _{BP} H_{\bullet} \simeq H_{\bullet} \otimes _{BP} BP_{\bullet} BP$. On the right hand side, we start with $BP_{\bullet} BP$ and mod out by $\eta_{L} (v_{i})$, the images of $v_{i}$ under the left unit. On the left hand side, we mod out by both $\eta_{L}(v_{i})$ and $\eta_{R}(v_{i})$.
As it turns out, $\eta_{L}(v_{n})$ and $\eta_{R}(v_{n})$ only differ by something coming from lower $v_{i}$, so if you mod out by one set (say, left) $v_{i}$-s, you mod out by the right one, too, so these two tensor products are isomorphic.