I ran into the following sanity check. Is the following statement true?



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> Every smooth fiber bundle (with compact fiber) over $S^2$ can be extended to a smooth fiber bundle over $\mathbb{C}P^\infty$ (or just $\mathbb{C}P^n$).
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For the case of $\mathbb{C}P^2$, if the bundle is a principal bundle, then it can be extended, since $\pi\_3(BG)=\pi\_2(G)=0$. So I thought possible counterexamples might come from a fiber $F$, with $\pi\_2(\text{Diff}(F))\ne 0$, for example
$\text{Diff}(S^1\times S^2)\simeq O(2)\times O(3)\times \Omega SO(3)$, but seems like the fiber bundle can still be extended, since $S^3\rightarrow S^2\rightarrow B(\text{Diff}(S^1\times S^2))$ is still nullhomotopic.


**Update**: The counterexample is provided in Oscar's post, when the fiber is $(D^7,\partial D^7)$. Now my question is, if the fiber is a closed manifold, is the following statment true?



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> Every smooth fiber bundle(with closed fiber) over $S^2$ can be extended to a smooth fiber bundle over $\mathbb{C}P^\infty$ (or just $\mathbb{C}P^n$).
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In fact, this should be my original question, but somehow I wrote "compact" but with "closed" in mind. I am really sorry about that.



All fiber bundles over $S^2$ extendable to $\mathbb{C}P^\infty$?
No it isn't, but I had to dig quite deep to get a counterexample. Let us look at smooth $(D^7, \partial D^7)$-bundles over $S^2$, i.e $D^7 \to E \overset{\pi} \to S^2$ with an identification $\partial E \cong S^2 \times \partial D^7$. These are classified by a map
$$f : S^2 \to BDiff\_\partial(D^7)$$
to the classifying space of the group of diffeomorphisms of $D^7$ relative to the boundary. I will show that there exists an $f$ which does not extend to $\mathbb{CP}^2$.


Smoothing theory gives an identification
$$BDiff\_\partial(D^7) \simeq \Omega^7\left(\frac{TOP(7)}{O(7)}\right)$$
and by page 246 of Kirby--Siebenmann the map
$$\Omega^7\left(\frac{TOP(7)}{O(7)}\right) \to \Omega^7\left(\frac{TOP}{O}\right)$$
is 2-connected, so there is a surjection
$$\pi\_2(BDiff\_\partial(D^7)) \to \pi\_9(\tfrac{TOP}{O}) \quad \quad(\*).$$
To find an example of such a bundle which does not extend over $\mathbb{CP}^2$, it is therefore enough to show that the map
$$- \circ \eta : \pi\_9(\tfrac{TOP}{O}) \to \pi\_{10}(\tfrac{TOP}{O}) \quad \quad(\*\*)$$
given by precomposition with the (suitably suspended) Hopf map is not trivial.


By surgery theory $\pi\_n(\tfrac{TOP}{O}) \cong \Theta\_n$, the group of homotopy $n$-spheres, and a homotopy $n$-sphere has a unique Spin structure, determining a $\pi\_n(\tfrac{TOP}{O}) \cong \Theta\_n \to \Omega^{Spin}\_n$, which we can further compose with the Atiyah--Bott--Shapiro map
$$\alpha : \Omega^{Spin}\_n \to KO\_n.$$
Applying this construction to both sides, the map $(\*\*)$ above becomes the map
$$- \circ \eta : KO\_9 = \mathbb{Z}/2 \to KO\_{10} = \mathbb{Z}/2,$$
which is an isomorphism. Finally, it is well-known that
$$\pi\_n(\tfrac{TOP}{O}) \cong \Theta\_n \to \Omega^{Spin}\_n \to KO\_n$$
is surjective for $n \equiv 1,2 \mod 8$. 


So choose a $f' \in \pi\_9(\tfrac{TOP}{O}) \cong \Theta\_9$ which is non-trivial under the Atiyah--Bott--Shapiro map, and use that $(\*)$ is surjective to lift it to a $f \in \pi\_2(BDiff\_\partial(D^7))$. This has the desired property.


**Addendum**


The following gambit can be used to get a closed fibre example. The product of the natural maps $SO(n+1) \to Diff^+(S^n)$ and $Diff\_\partial(D^n) \to Diff^+(S^n)$ gives a map
$$SO(n+1) \times Diff\_\partial(D^n) \to Diff^+(S^n)$$
which is a weak homotopy equivalence (though not a homomorphism). Thus the map
$$BDiff\_\partial(D^7) \to BDiff^+(S^7)$$
is (split) injective on all homotopy groups, so the example above also gives an oriented $S^7$-bundle over $S^2$ which cannot extend to $\mathbb{CP}^2$.


