On $S^4$, we know that rank 2 complex vector bundles are classified by $\pi\_3(U(2))=\mathbb Z$. Any element $g\in\pi\_3(U(2))=\mathbb Z$ determines a complex vector bundle $E$ over $S^4$. 


**Question**: Can we say that the corresponding second Chern class $c\_2(E)$ equals $g$, i.e. $\left<c\_2(E),[S^4]\right>=\deg(g)$.



Rank 2 complex vector bundles over $S^4$
Choose a homotopy equivalence $U(2)\simeq \Omega BU(2)$ to use as an identification and let $g:S^3\rightarrow U(2)\simeq \Omega BU(2)$ represent a homotopy class in $\pi\_3U(2)$. Then the $U(2)$-bundle $E\rightarrow S^4$ corresponding to $g$ is classified by the map $\tilde{g}:S^4\rightarrow BU(2)$ which is adjoint to $g$. Then $g$ is homotopic to the composite 


$g:S^3\xrightarrow{E}\Omega S^4\xrightarrow{\Omega\tilde{g}}\Omega BU(2)\xrightarrow{\simeq} U(2)$


where $E:S^3\rightarrow\Omega\Sigma S^3$ is the suspension map. 


Now since the fibration $SU(2)\rightarrow U(2)\xrightarrow{det}S^1$ splits and there is a homotopy equivalence $U(2)\simeq S^1\times S^3$, there is a spherical generator $x\_3\in H^3U(2)=\mathbb{Z}$ in the sense that 


$g^\*x\_3=\text{deg}(g)\cdot s\_3\in H^3S^3$


where $s\_3$ generates $H^3S^3=\mathbb{Z}$. Moreover we may assume that $x\_3$ is the cohomology suspension of the second Chern class


$x\_3=\sigma (c\_2)$


Now all of this together gives us a second equation for $g^\*x\_3$


$g^\*x\_3=g^\*\sigma (c\_2)=E^\*(\Omega\tilde{g})^\*\sigma (c\_2)=E^\*\sigma (\tilde{g}^\*c\_2)=\text{deg}(\tilde{g})\cdot E^\*\sigma(s\_4)=\text{deg}(\tilde{g})\cdot s\_3$


Finally combine the two expressions for $g^\*x\_3$ to get


$g^\*x\_3=\text{deg}(g)\cdot s\_3=\text{deg}(\tilde{g})\cdot s\_3$


and conclude that $\text{deg}(g)=\text{deg}(\tilde{g})$. That is, in the wording of your question, "the corresponding second Chern class $c\_2(E)$ equals $g$".


