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For a vector bundle thom space $T$ is defined as $T=E/A$, where $E$ is the total space and $A$ is the set of vectors in $E$ of length $\geq 1$. Alternatively, $T$ is the mapping cone of the associated sphere bundle, i.e., $(S \times I \cup B)/ S \times {1}$. The base is glued to the total space $S$ via the projection map. However, I cannot see that these are equivalent. Any help..

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We know the cell decomposition of T comes from the base. But how do we know that the attaching map between cells is the projection map of the sphere bundle? –  wqr Jul 4 '12 at 16:04

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I think what you have to show is that there is a relative $CW$-complex structure on $(DE,SE)$ where $DE$ denotes the disc bundle and $SE$ denotes the sphere bundle. This is somewhat technical and has already been asked here, you might want to search for that. As soon as you have this, then you can use the following fact. Given any cofibration $i: A\to X$ (and the inclusion of a subcomplex like $SE \to DE$ is a cofibration) has the property that the quotient $X/A$ is homotopy equivalent to the cone of the inclusion $Ci = CA \cup_{i} X$.

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