$H_k(E,\mathbb{Z})\cong H_k(X,\mathbb{Z})$

Question

Let $\pi:E\rightarrow X$ be a real vector fibrate of rank $n$. How to prove that $$H_k(E,\mathbb{Z})\cong H_k(X,\mathbb{Z}).$$ approach: I consider it true that $X$ is a deformation rectract of $E$, it is true?

Any hint would be appreciated.

$X$ is the zero section of $E$. Then define a map on the fibers of $E$ given by $(v, t) \mapsto tv$. — Noel Lundström, Oct 09 '19 at 20:20

score 1 · Accepted Answer · answered Oct 09 '19 at 03:14

1

The key is homotopy equivalence. Try to show that $E$ can be retracted to the zero section (i.e., the image of the section $s:X\to E$ with $s(p)=0\in E_p$). What's the relationship between the zero section and $X$?

answered Oct 09 '19 at 03:14

rmdmc89

9,939

$H_k(E,\mathbb{Z})\cong H_k(X,\mathbb{Z})$

1 Answers1