# 4 Thom spaces

Let $E \to X$ be a vector bundle. Then we define the *Thom space* to be

Let $Z \hookrightarrow X$ be a closed embedding in $\mathrm{Sm}_ S$ with normal bundle $\nu _ Z$. Then we have an equivalence (in $\mathrm{Spc}_ S^{\mathbb {A}^1}$).

$\frac{X}{X \setminus Z} \to \mathrm{Th}(\nu _ Z).$The first geometric input is the construction of a bundle of closed embeddings over $\mathbb {A}^1$ whose fiber over $\{ 0\}$ is $(\nu _ Z, Z)$ and $(X, Z)$ elsewhere. This has a very explicit description:

$D_ ZX = \mathrm{Bl}_{Z \times _ S \{ 0\} } (X \times _ S \mathbb {A}^1) \setminus \mathrm{Bl}_{Z \times _ S \{ 0\} } (X \times _ S \{ 0\} ).$Indeed, the fiber over $\{ 0\}$ is $\mathbb {P}(\nu _ Z \oplus \mathcal{O}_ Z) \setminus \mathbb {P}(\nu _ Z)$, which is canonically isomorphic to $\nu _ Z$. (This construction is known as “deformation to the normal cone”)

The second step shows that in $\mathcal{H}(S)$, we have a homotopy pushout squares

To prove this, one uses Nisnevich descent to reduce to the affine case.

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