I was trying to learn some motivic homotopy theory and I am stuck at this problem.
Let $p: E \rightarrow X$ be a Vector bundle over a smooth scheme X. We define Thom space $\textbf{Th(E) }$ := $E/E-i(X)$. Where $ i: X \rightarrow E$ is the zero section.
I was trying to compute the Thom space of a trivial bundle.
By definition, it must be $\mathbb{A}^n \times X/ (\mathbb{A}^n - \{ 0 \}) \times X. $. But how to prove that this is the same as $\mathbb{A}^n / \mathbb{A}^n - \{ 0 \} \wedge X_+$ ?
Probably it is trivial and I might be overlooking something. Is there some kinda 'excision' property at play here?
Thank you in advance.
This is a sketch of the claim I made in the comments. It doesn't really have anything to do with (algebraic-geometry) or (homotopy-theory) - just some good old-fashioned (general-topology).
The claim we wish to show is:
First, note that from the definition it is easy to see that $A/B \wedge X_+ \cong (A/B \times X)/(B/B \times X)$. Now consider the composite map $$A \times X \to A/B \times X \to (A/B \times X)/(B/B \times X) \cong A/B \wedge X_+.$$ This map sends $B \times X$ to the basepoint, so we get a map $(A \times X)/(B \times X) \to A/B \wedge X_+$, and I claim that this is a homeomorphism. It's continuous and open by construction, and it's not hard to see that it's bijective. Perhaps drawing a couple of pictures would be more convincing.