Why is the fundamental group of the Volodin space $X(R)$ the Steinberg group $St(R)$? The Volodin space $X(R)$ is defined in A.A. Suslin's "On the Equivalence of K-Theories" (https://www.tandfonline.com/doi/abs/10.1080/00927878108822666) as the union of classifying spaces $\bigcup_{\sigma,n} BT^\sigma_n(R) \subset BE(R)$. He states that the fundamental group of this space is $\pi_1 X(R) \cong St(R)$; the Steinberg group. Is there a proof of this? Suslin's argument is that $X(R)$ can be written as a quotient space of $V(St(R), \{T^\sigma_n(R)\})$; the simplicial complex with vertices $St(R)$ (as a set) where elements $(x_0,...,x_p)$ form a $p$-simplex if each $x_i x_j^{-1}$ lies in the same $T^\sigma_n(R)$.
In particular, Suslin states that it can be easily seen that $V(St(R), \{T^\sigma_n(R)\})$ is simply connected and one can quotient out by the natural, free action of $St(R)$ on vertices to yield $X(R)$. This makes sense to me but I can't find a reason why $V(St(R), \{T^\sigma_n(R)\})$ is simply connected.
Weibel in his K-book notes that the map $\iota_*: \pi_1 X(R) \to E(R)$ induced by the inclusion $X(R) \hookrightarrow BE(R)$ is surjective since the $T^\sigma_n(R)$ generate $E(R)$. In light of this an alternative approach I've tried is to show that the kernel of $\iota_*$ is contained in the centre of $\pi_1 X(R)$ (i.e. $\iota_*$ is a central extension of $E(R)$). Since $X(R)$ is acyclic we have $H_n(\pi_1 X(R), \mathbb{Z}) = 0$ for $n =1,2$ (see Weibel's K-book, IV, Lemma 1.3.1 for this result). Altogether this would show that $\pi_1 X(R)$ is a universal central extension of $E(R)$, giving us that $St(R) \cong \pi_1 X(R)$. Only trouble is I can't show that $\iota_*: \pi_1 X(R) \to E(R)$ is a central extension.
Any ideas to the end of showing $\pi_1 X(R) \cong St(R)$ would be much appreciated!
 A: Since thinking about this problem some more, I've figured out a solution so I'm answering my own question. By the remarks in my question, all we have to do is show that $\pi_1 X(R) \to E(R)$ is a central extension.
We first bear three things in mind;

*

*For subgroups $H,H' \leq G$ of some group $G$, we have continuous inclusions $BH, BH' \hookrightarrow BG$ that allow us to make the identification $BH \cap BH' = B(H \cap H')$ (I can include details on this if anyone asks). This also shows that
$BH \cap BH'$ is path connected.

*Since CW complexes are locally contractible, we can find open sets $BT^\sigma_n(R)' \supset BT^\sigma_n(R)$ that deformation retract onto the $BT^\sigma_n(R)$ and hence $X(R) = \bigcup_{n, \sigma} BT^\sigma_n(R)'$ and in particular Van-Kampen's theorem holds.

*The proof that $St(R) \to E(R): x_{ij}(r) \mapsto e_{ij}(r)$ is a central extension only relies on the generators $x_{ij}(r)$ of $St(R)$ satisfying the Steinberg relations (see Srinivas' "Algebraic K-Theory" and/or Weibel's K-book for this proof), so we can generalize this proof to show that for any group $\Theta(R)$ generated by symbols $\theta_{ij}(r)$ that satisfy the Steinberg relations, the homomorphism $\Theta(R) \to E(R): \theta_{ij}(r) \to e_{ij}(r)$ (provided it is well-defined) is a central extension.

Now for the proof we were after:
By Van-Kampen's theorem we have an isomorphism
$$ \rho: (*_{n, \sigma} \pi_1 BT^\sigma_n(R)')/N = (*_{n, \sigma} \pi_1 BT^\sigma_n(R))/N \overset{\sim}{\to} \pi_1 X(R) $$
where $*_{n, \sigma}$ denotes the free product of the $\pi_1 BT^\sigma_n(R) = T^\sigma_n(R)$ and $N$ is the normal subgroup generated by elements of the form $\iota_1(\alpha) \iota_2(\alpha)^{-1}$ where $\iota_1$ denotes an inclusion
$T^{\sigma_1}_{n_1}(R) \cap T^{\sigma_2}_{n_2}(R) \hookrightarrow T^{\sigma_1}_{n_1}(R)$ and $\iota_2$ is defined similarly. In particular,
$\theta_{ij}(r) := \rho(e_{ij}(r))$ is well defined and the symbols $\theta_{ij}(r)$ generate $\pi_1 X(R)$ and satisfy the
Steinberg relations since each $T^\sigma(n,R)$
has the Steinberg relations.
As an aside, one might suppose that the $\theta_{ij}(r)$ satisfy any relation that the $e_{ij}(r)$ satisfy in $E(R)$. However this is not
necessarily the case as, for example, relations in $E(R)$ written in the form $ e_{ij}(r_1) ... e_{ji}(r_2) ... = 1$ will not hold on
any of the $T^\sigma(n,R)$ since $e_{ij}(r_1), e_{ji}(r_2)$ cannot simultaneously exist in any one of these subgroups. By what follows it turns out that no further relations are actually satisfied by the $\theta_{ij}(r)$.
Now the composite map $\pi_1 BT^\sigma(n,R) \to \pi_1 X(R) \to \pi_1 BE(R)$ induced from inclusions is an inclusion of groups,
so the map $\pi_1 X(R) \to E(R)$ sends $\theta_{ij}(r)$ to $e_{ij}(r)$ and by our third remark above this map is in fact a central extension.
As I've already said, the isomorphism $\pi_1 X(R) \cong St(R)$ follows from the remark in the original question.
