$SU(2)$-bundles over four-manifolds are determined by top cells

Question

Suppose we have a compact oriented simply-connected four manifold $X$ and a $SU(2)$ bundle $P$ with $2$nd Chern class $=k$ over $X$.

We know $X$ have a top cell $e^4$, if we collapse all lower cells to a point to obtain a $S^4$, and denotes the collapsing map $g$.

Suppose $P$ represent by a map $h$ from $X$ to $BSU(2)$. Does $h$ factor through $g$(i.e. exist $\bar{h}$ from $S^4$ to $BSU(2)$ s.t. $h=\bar{h}·g$)?

Answer 1

More generally, if $X$ and $Y$ are closed orientable four-manifolds and $f : X \to Y$ is a degree one map, then every principal $SU(2)$-bundle over $X$ is of the form $f^*P$ for some principal $SU(2)$-bundle $P \to Y$. This follows from the fact that principal $SU(2)$-bundles over a four-dimensional CW complex are determined up to isomorphism by their second Chern class, see this question. Note, this is no longer true in higher dimensions as the example in this answer demonstrates.

If $f : X \to Y$ is arbitrary, then we have

$$\require{AMScd} \begin{CD} \operatorname{Prin}_{SU(2)}(Y) @>{c_2}>> H^4(Y;\mathbb{Z}) @>{\cong}>> \mathbb{Z}\\ @V{f^*}VV @V{f^*}VV @VV{\times\deg f}V\\ \operatorname{Prin}_{SU(2)}(X) @>{c_2}>> H^4(X;\mathbb{Z}) @>{\cong}>> \mathbb{Z} \end{CD}$$

In particular, if $\deg f \neq 0$, a principal $SU(2)$-bundle over $X$ is of the form $f^*P$ for some principal $SU(2)$-bundle over $Y$ if and only if its second Chern class is divisible by $\deg f$. If $\deg f = 0$, then the only principal $SU(2)$-bundle over $X$ of the form $f^*P$ is the trivial one.

Answer 2

I'm not well versed on 4-manifolds, so hopefully I am not making any foolish mistakes:

Since $X$ is simply-connected, up to homotopy equivalence, it can be taken to have a $2$-skeleton given by a wedge of $2$-spheres. By Poincare duality, its $3$rd homology vanishes, which implies that, up to homotopy equivalence, it can be written without $3$-cells. Meaning that $X^3$ is a wedge of $2$-spheres.

Complex bundles over a $2$-sphere are classified by maps $S^1 \rightarrow SU(2)$ which are known to all be null homotopic, i.e. $\pi_1(SU(2))=0$. This means that the restriction of $X \rightarrow BSU(2)$ is null homotopic on $X^3$, which means that up to homotopy $X \rightarrow BSU(2)$ factors through $X \rightarrow X/X^3$, as you desire.