Explicit model of BSU(2) in terms of singular complex  What is the explicit model of BSU(2) in terms of singular complex, up to 5 dimensions,
so that one can compute $\pi_5(BSU(2))=\mathbb{Z}_2$ explicitly?
This question is related to another question of $\pi$-cohomology and its computation.
(see $\pi$-cohomology class -- a variant of cohomology class )
 A: It seems that a singular complex is not the best option if you are interesting in computing topological properties of $BSU(2) \approx \mathbb HP^\infty$ by a "finite calculation". A singular $n$-simplex is just a continuous map from the standard $n$-simplex to the manifold, there are thus ucountable infinite number of them for any $n$. A simplicial, $\Delta$ or CW complex is a much simpler and more efficient for calculations, CW complex probably being the best one.
I have tried to construct the CW complex concretely (by extending example 0.4, 0.5 and 0.6 of Hatcher to the quaternionic case), and if I am right then you have the following CW decomposition
$\mathbb HP^n = e^0 \cup e^4 \cup \cdots \cup e^{4n}$.
For example for $\mathbb HP^1$, you just have one $1$-cell and one $4$-cell. For the classifying space, $\mathbb HP^\infty$, you have a countable number of cells, only one for each $4n$-cell.
Let me briefly go through how it's done (thinking about the $\mathbb RP^n$ case first, might make this much easier to understand).
One way do define $\mathbb HP^n$ is to identify any two points on the unit quaternionic sphere $p,q\in S(\mathbb H^{n+1})\approx S^{4n+3}$ by $p\sim q$ iff $p=\lambda q$, where $\lambda\in S^3 \approx SU(2)$ is a unit quaternion.
This is however equivalent to taking a disk ("upper hemisphere")
$D^{4n}_+ = \left\{\left(w,\sqrt{1-|w|^2}\right)\in\mathbb H^n\times\mathbb H | |w|\leq 1\right\}\subset S^{4n+3}$ and identifying the points on the boundary by $a\sim b$ iff $a=\lambda b$. The boundary is just $\partial D^{4n}_+=S^{4n-1} = S^{4(n-1)+3}$ with elements $(w,0)$, where $w\in\mathbb H^n$ and $|w|^2=1$. Therefore this identification on the boundary just leads to $\mathbb HP^{n-1}$, "on the boundary".
In other words, to construct $\mathbb HP^n$, take the disjoint union of a $4n$-disk and $\mathbb HP^{n-1}$, then attach the disk boundary to $\mathbb HP^{n-1}$ by the natural projection
$\pi:\partial D^{4n}_+ = S^{4(n-1)+3}\rightarrow \mathbb HP^{n-1},$ $a\mapsto [a]$.
Thus we can construct a CW complex;


*

*Start with a $0$-skeleton $X^0$ containing only one $0$-cell (one point). Add a $4$-disk and attach its boundary ($S^3$) to the $0$-cell (point) to get the $4$-skeleton $X^4 = S^4 = \mathbb HP^1$ with only one $4$-cell.

*Add a $8$-disk to $X^4$ and glue its boundary ($S^7$) to $X^4=\mathbb HP^1$ by the projection map $\pi$, and obtain $X^8=\mathbb HP^2$.

*Continue until the $n$-skeleton $X^n = \mathbb HP^n$.


So we have the inclusions $\mathbb HP^0 \subset\mathbb HP^1 \subset\mathbb HP^2 \subset \dots \subset \mathbb HP^n$,
given iteratively
$\mathbb HP^n = \frac{\mathbb HP^{n-1}\sqcup D^{4n}_+}{a \sim \pi(a)} = e^0\cup e^4\cup\cdots\cup e^{4n}$.
This construction contain relatively few cells and might be effectively used to calculate the $\pi$-cohomology. If you prefer a triangulation, the appendix of Hatcher is probably good to read in order to see how one might decompose this cell-decomposition into a triangulation.
(I feel I made this seem more formal and technical than it is.)
Edit:
If you are interested in a simplicial/CW complex of $BSU(2)=\mathbb HP^\infty$ which is just accurate enough to get $\pi_5(BSU(2)) = \mathbb Z_2$ right, then $\mathbb HP^1$ might be a good enough approximation to $BSU(2)$. According to the above construction we have that $\mathbb HP^1 = S^4$, and it is well-known that $\pi_5(S^4)=\mathbb Z_2$ (see wikipedia). If $\mathbb HP^1 = S^4$ is enough to calculate the $\pi$-cohomology, then it is very simple to get a simplicial complex for this without the need of a CW complex. 
Edit 2:
By the way, $\mathbb HP^1=S^4$ seems to get everything up to (and including) $\pi_6(BSU(2))$ right but it fails after that. Actually $\pi_{i\geq 7}(S^4)$ has all the right factors from $\pi_{i\geq 7}(BSU(2))$, and some additional factors. I think if you use $\mathbb HP^2$ as an approximation instead, then some of these extra factors vanish and a few more homotopy groups turn out correctly. Thus $\mathbb HP^1 = S^4$ seems to be an okay approximation if you are only interested in homotopy up to $\pi_6(BSU(2))$.
A: The infinite quaternionic sphere $S(\mathbb{H}^\infty)=\bigcup_n{S(\mathbb{H}^n)}$ is contractible and carries a free action of the unit quaternions $SU(2)=Sp(1)$. The quotient $\mathbb{H}P^\infty$ is a model for $BSU(2)$, and $\mathbb{H}P^n$ is then an approximation suitable for computing homotopy groups or the cohomology ring in low degrees.
A: If you care about homotopy groups of classifying spaces, then the relevant general result is $\pi_n(BG)=\pi_{n-1}(G)$.
In your specific case, you are looking at $\pi_4(SU(2))=\pi_4(S^3)=\mathbb Z/2$.
For any $n\ge 3$, one has $\pi_{n+1}(S^n)=\mathbb Z/2$.
This is the first stable homotopy group of spheres $\pi_1^{st}(S^0)$, and its generator is one of the three ``Hopf invariant one'' elements.
