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Let's consider a smooth sphere bundle over a smooth manifold with structure group is equal to the diffeomorphism group of sphere. Then, can we say that this is a double of some disk bundle? Thank you for your helping.

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No. If a sphere bundle is the double of a disk bundle, then it has a section. You get counterexamples by considering unit sphere bundles of vector bundles with nonvanishing Euler class.

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    $\begingroup$ More precisely, it is not hard to see that a sphere bundle is the double of a disc bundle if and only if it has a section. $\endgroup$
    – Neil Strickland
    Commented May 20, 2016 at 10:44
  • $\begingroup$ @Neil Is it clear that you have a strict double, that is, are both halves automatically isomorphic? Then I guess, another equivalent formulation is that the sphere bundle is a fibrewise suspension of another sphere bundle. And somehow this seems to imply that all sphere bundles with a section are linear, but that seems to be too strong a conclusion. $\endgroup$
    – Sebastian Goette
    Commented May 20, 2016 at 11:39
  • $\begingroup$ @SebastianGoette I guess I was just thinking about the linear case, I agree that there might be more to say in general. $\endgroup$
    – Neil Strickland
    Commented May 20, 2016 at 12:36
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    $\begingroup$ Here's a spherical fibration version of the statement you're after: If a spherical fibration is an unreduced fiberwise suspension of another one, then that fibration comes equipped with two sections which are vertically homotopic. Conversely, if $E \to B$ is a spherical fibration equipped with two sections, represented by a fiberwise map $s:B\times S^0 \to E$, then in a certain metastable range the fibration fiberwise desuspends relative to $s$. This is explained in the paper: Klein, John R.; Williams, E. Bruce Homotopical intersection theory. I. Geom. Topol. 11 (2007), 939–977. $\endgroup$
    – John Klein
    Commented May 20, 2016 at 16:38
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The connected double cover of $S^1$ (boundary of the Möbius strip) is an $S^0$ bundle that is not the double of the unique $0$-disc bundle over $S^1$.

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