in the paper 


Foundations of the theory of bounded cohomology,


by N.V. Ivanov, the author considers the complex of bounded singular cochains on a simply connected CW-complex $X$, and constructs a chain homotopy between the identity and the null map. The construction of this homotopy involves the description of a Postnikov system for the space considered. In some sense, $S^2$ represents the easiest nontrivial case of interest for this construction, and I was just trying to figure out what is happening in this case. Since the existence of a contracting homotopy obviously implies the vanishing of bounded cohomology, this is somewaht related to understanding why the bounded cohomology of $S^2$ vanishes.


A first step in constructing the needed Postnikiv system is the computation of the homotopy groups of $X$, so the following question came into my mind:


Do there exists integers $n\neq 0,1$ such that $\pi\_n(S^2)=0$? 


I gave a look around, and I did not find the answer to this question, but I am not an expert of the subject, so I don't even know if this is an open problem.


In 


Berrick, A. J., Cohen, F. R., Wong, Y. L., Wu, J.,
Configurations, braids, and homotopy groups,
 J. Amer. Math. Soc. 19 (2006), no. 2, 265–326


it is stated that $\pi\_n(S^2)$ is known for every $n\leq 64$, and Wikipedia's table 
<http://en.wikipedia.org/wiki/Homotopy_groups_of_spheres#Table_of_homotopy_groups>
shows that $\pi\_n (S^2)$ is non-trivial for $n\leq 21$. 



Homotopy groups of $S^2$
SERGEI O. IVANOV, ROMAN MIKHAILOV, AND JIE WU have recently(2nd June 2015) published a paper in arxive giving a proof that for $n\geq2$, $\pi\_n(S^2)$ is non-zero.
You can look at it in the following link.


Sergei O. Ivanov, Roman Mikhailov, Jie Wu, *On nontriviality of homotopy groups of spheres*, [arXiv:1506.00952](http://arxiv.org/abs/1506.00952)


