Second Chern class classifies rank $r$ vector bundle over $S^4$? On page 16 of Atiyah's Geometry of Yang-Mills Fields (http://plouffe.fr/simon/math/Atiyah%20M.%20Geometry%20of%20Yang-Mills%20Fields%20(Pisa,%201979)(100s).pdf) it stated that 
"...the topological theory of fiber bundles over spaces like $S^4$ which are not contractible tells us that in this case they are precisely classified by the [same] integer $k$ (= second Chern class)".
So I'm just wondering if this is saying that a $SU(r)$-bundles $E \rightarrow S^4$ are classified by $c_2(E)$? In particular, if $E,E'$ are two $SU(r)$-bundles over $S^4$ and $c_2(E) = c_2(E')$ then $E\cong E'$ (isomorphic as vector bundle)? 
How do I see that this result is true?
What is a generalization of this result? Would it remain true for any simple Lie group $G$ instead of $SU(r)$? In particular, how do I know when do Chern classes classifies vector bundles over a topological space?
 A: The result generalises to principal $G$-bundles over a 4-dimensional complex $X$, where $G$ is any simply connected Lie group. 
Recall that any principal $G$-bundle $P\rightarrow X$, over any finite complex $X$, is classified by the homotopy class of map $f:X\rightarrow BG$, into the classifying space of $G$, in that there is a bundle isomorphism $P\cong f^*EG$ over $X$, where $EG\rightarrow BG$ is the universal $G$-bundle with contractible total space.
Now the restrictions on $G$ imply that $\pi_0G=\pi_1G=\pi_2G=0$, and $\pi_3G\cong \mathbb{Z}^r$ for some integer $r$. Since there is a homotopy equivalence $\Omega BG\simeq G$ this implies that $\pi_0BG=\pi_1BG=\pi_2BG=\pi_3BG=0$ and $\pi_4BG\cong\mathbb{Z}^r$. 
Now choose a map $c:BG\rightarrow K(\pi_4BG,4)$, which induces an isomorphism on $\pi_4$. This map is 5-connected, so induces an isomorphism $H^4(K(\pi_4BG,4))\cong H^4(BG)=\mathbb{Z}^r$, and, in particular, takes the fundamental classes in $H^4(K(\pi_4BG,4))$ to the 'Chern' classes in $H^4(BG)$.
In case $X$ is a 4-dimensional complex we get an isomorphism $c_*:[X,BG]\xrightarrow{\cong}[X,K(\pi_4BG,4)]\cong H^4(X;\pi_4BG)\cong H^4(X;\mathbb{Z}^r)\cong \oplus^rH^4(X)$, which  takes the class of the map $f$ classifying the bundle $p$ to the value of the degree 4 characteristic class in $BG$. 
Since $c_*$ is an isomorphism, this class completely characterises the isomorphism class of the bundle $P\rightarrow X$.
In particular when $G=SU(r)$ and $X$ is a 4-dimensional complex, the bundle $P$ is completely characterised by its second Chern class $c_2(P)\in H^4(X)$. If $G=Sp(r)$ and $X$ is as above, then the bundle is completely characterised by its first symplectic pontrjagin class $q_1(P)\in H^4(X)$. If $G=Spin(r)$ and $X$ is as before, then $P$ is characterised by its first Pontrjagin class  $p_1(P)\in H^4(X)$. The construction I outline above holds when $G$ is any product of such simple, simply connected Lie groups (which covers all such compact, simply connected lie groups).
Observe also that when $X$ is 4-dimensional then obstruction theory gives shows that any $SU(r)$-bundle has a reduction of structure to an $SU(2)$-bundle. Similarly $Sp(r)$-bundles over $X$ always reduce to $Sp(1)$-bundles, and $Spin(r)$-bundles always reduce to $Spin(4)$-bundles. Since $SU(2)\cong Spin(3)\cong Sp(1)\cong S^3$, and $Spin(4)\cong Spin(3)\times Spin(3)$ the characteristic classes in degree 4 are all essentially the same.
