Vector bundle is homotopy equivalence [duplicate]

If $\pi:E\rightarrow B$ is a vector bundle (I allow it to be of nonconstant rank), then I want to prove that it is a homotopy equivalence. As a homotopy inverse I propose the zero section $s:B\rightarrow E$. So $\pi s=id_B$ and $s\pi$ is homotopic to $id_E$ by $f:E\times[0,1]\rightarrow E$, $f(e,t)=te+(1-t)s\pi(e)$. Is this correct?

• To answer your question, yes, your proof is correct (it's the same as in the linked post). – Najib Idrissi Feb 14 '15 at 12:31