I think I understand what you mean by "Borel" now. I was brought up to interpret it as something different ("Borel equivariant: the (singular) cohomology of the Borel construction associated to an action). Did you wind up working this out?

I'm not familiar with this definition of Borel cohomology, so I'm not certain if this helps any, or if it will come as news, but an explicit generator for the de Rham cohomology space $H^3(G)$, for $G$ compact simple Lie, is given by the Cartan 3-form (see, e.g <mathoverflow.net/questions/62998/…).

This might not be of interest to you since it's usually not normal, but if $H$ is maximal compact, then the coset space is Euclidean, and there is a section inducing a diffeomorphism between $H \times G/H$ and $G$.

@user2275150: Presumably that the restrictions of $g, g'$ to the set $\{q \in \mathbb Q:q \geq f(0)\}$ agree (but the values on rationals less than $f(0)$ are not the same).

@TarasBanakh I see, thanks for clarifying.

Noticed where? Assuming it has an affirmative answer, it'd be reassuringly pro-social to let us see it.

I can write something like an answer if you'd like, but the main issue with your proposed formula is sort of a category error. Your quotation from Weibel is a subquotient of C, but the claimed equivalent set you provide lies in C itself. It would at best be a set of representatives for the nonzero classes in the quotient. If you've quoted correctly, it is: the idea is right, but the levels are mixed up.

There's a notion of cell decomposition distinct from that of CW complexes, so there's an outside possibility that this is what was meant, a decomposition of some model of $BSU(n)$ into some union of Euclidean spaces, all of dimension $>4$ (although this would be more of a curiosity than particularly useful).

Are you certain your source deliberately meant to forgo 0-cells, rather than simply assuming but omitting mention of them?

I'm not sure... I was trying to ask you what the notation meant.

For $q$, it is the $d$-fold suspension of a map $(\mathbb R P^{d-1})^d \to {*}^d = S^d$. By the naturality of the Thom isomorphism, the map induced on $H^d$ by this map will be an isomorphism if the map induced on $H^0$ by the collapse map $\mathbb R P^{d-1} \to {*}$, and we do know this. What is $\mathbb R P^{d-1}_d$?

It might help me a bit to see what $\xi$ is, and I don't know for sure what is meant by the superscript $d$ on the upper right. Is it meant to be $d$-skeleton of the suspension? That of the projective space itself is the whole space.