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Let $\xi$ and $\eta$ be vector bundles over the same base space $X$. Their Whitney sum is a bundle $\xi\oplus\eta$ over $X$. I read somewhere (without proof) that its Thom space is given by $$ T(\xi\oplus\eta) = T(\eta|_{D(\xi)})/T(\eta|_{S(\xi)}), $$ where the restrictions are really pullbacks of $\eta$ under the projection maps $D(\xi),S(\xi)\to X$.

Can anyone provide a proof, or a reference to a proof?

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    $\begingroup$ $D(\xi\oplus\eta)$ can be interpreted as $D(\eta|_{D(\xi)})$, and the sphere bundle is the union of $S(\eta|_{D(\xi)})$ and $D(\eta|_{S(\xi)})$. All more or less by the definitions. $\endgroup$ – Alex Degtyarev Apr 21 '16 at 11:48
  • $\begingroup$ @Alex: You are right, this does appear to be an easier way to see it, thanks. $\endgroup$ – Mark Grant Apr 22 '16 at 13:21
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The Thom space is the homotopy cofiber of the Unit Sphere Projection: $$ \mathbb{S}_\xi X \to X \to T(\xi) $$ while the unit sphere in a sumspace $\mathbb{S}[U\oplus V]$ is a join $$ \begin{array}{c} \mathbb{S}[U] \times \mathbb{S}[V] & \rightarrow & \mathbb{S}[U] \\ \downarrow & & \downarrow \\ \mathbb{S}[V] & \rightarrow & \mathbb{S}[U\oplus V] \end{array} $$ (homotopy-pushout). By one of "Mather's" Cube Theorems (it's a kind of distrutivity) this fiberwise pushout description integrates to a homotopy pushout of bundles: $$ \begin{array}{c} \mathbb{S}_\xi \times_X \mathbb{S}_\eta &\rightarrow & \mathbb{S}_\xi \\ \downarrow & & \downarrow \\ \mathbb{S}_\eta & \rightarrow & \mathbb{S}_{\xi\oplus \eta} \end{array} $$ The top and the left maps are both, as it happens, sphere fibrations, the ones relevant to your question. Consider the composable arrows $$ \mathbb{S}_\xi \times_X \mathbb{S}_\eta \to \mathbb{S}_\xi \to X $$ and form the following small squares as homotopy pushouts: $$ \begin{array}{c} \mathbb{S}_\xi \times_X \mathbb{S}_\eta &\rightarrow & \mathbb{S}_\xi & \rightarrow & \mathbb{S}_{\xi\oplus\eta} & \to & X \rlap{\simeq D\eta} \\ \downarrow & & \downarrow & & \downarrow & & \downarrow \\ * & \to & T(\eta|_{S\xi}) & \to & T(\eta|_{S\xi}) \vee T(\xi|_{S\eta}) & \to & W \\ & & \downarrow & & \downarrow & & \downarrow \\ & & * & \to & T(\xi|_{S\eta}) & \to & T(\xi) \rlap{\simeq T(\xi|_{D\eta})}\\ && && \downarrow & & \downarrow \\ & & & & * & \to & T(\xi\oplus\eta) \end{array} $$ where $W$ doesn't matter for the rest of the argument.

Now, I've sneakily applied the homotopy pushout pasting lemma, several times: to identify the $-\vee-$ in the second row (which gives the Thom space below it) and to identify the other Thom spaces we want. That says the homotopy type we want is the homotopy cofiber of a map of Thom spaces; and the particular fine result you want is: the inclusion $T(\xi|_{S\eta}) \to T(\xi|_{D\eta})$ is a cofibration and represents the homotopy class of the pushout structure map indicated by the diagram.

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    $\begingroup$ Thanks, this is great! I think I've convinced my self that the cofibre of $\mathbb{S}_\xi \times_X \mathbb{S}_\eta \to \mathbb{S}_{\xi\oplus\eta}$ is $T(\eta|_{S\xi})\vee T(\xi|_{S\eta})$. Now I just have to convince myself that the pushout of $\mathbb{S}_{\xi\oplus\eta}\to D\eta$ and $\mathbb{S}_{\xi\oplus\eta}\to T(\xi|_{S\eta})$ is $T(\xi|_{D\eta})$, and I'll have fully understood your argument. $\endgroup$ – Mark Grant Apr 22 '16 at 13:08
  • $\begingroup$ (This is the kind of proof I was trying to come up with, and failing.) $\endgroup$ – Mark Grant Apr 22 '16 at 13:09
  • $\begingroup$ As I say, the pasting lemma is your friend. $\endgroup$ – Jesse C. McKeown Apr 22 '16 at 17:00
  • $\begingroup$ I'm not sure I know what you mean by "the pasting lemma". I assume you mean the following: in a diagram consisting of two squares side by side sharing a common arrow, if any two of the `squares' are homotopy pushouts, then so is the third. Please let me know if its something else! $\endgroup$ – Mark Grant Apr 23 '16 at 8:41
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    $\begingroup$ @MarkGrant, it's not true that if any two squares are pushouts the third also is. That's 3 cases, two of which are true. Say the horizontal arrows point to the right and the squares share a vertical side. If the two little squares are pushouts so is the outer square (rectangle?). And if the outer square and the square on the left are pushouts so is the square on the right. The other case doesn't hold. And for pullbacks it's swapped: outer and right imply left; but with left and right swapped it would be false. (Oh, and to answer your question: it is what Jesse means by pasting lemma.) $\endgroup$ – Omar Antolín-Camarena Apr 23 '16 at 17:45
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I think Statement (3.6) in the following paper of Becker and Schultz

Becker, J.C.; Schultz, R.E. Equivariant function spaces and stable homotopy theory. I. Comment. Math. Helv. 49, 1-34 (1974).

is in the line of what you are after.

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