Exercise 10.1

Question

Suppose $E$ is a smooth vector bundle over $M$. Show that the projection map $\pi:E 	o M$ is a surjective smooth submersion.

Elu Thingol · Accepted Answer

\begin{proof}
By definition, we must demonstrate that for each $\widetilde{p} \in E$, the differential
\begin{equation*}
    d \pi_{\widetilde{p}}: T_{\widetilde{p}}E 	o T_{\widetilde{p}} M
\end{equation*}
is surjective. Since $E$ is a vector bundle, we can set $p:=\pi(\widetilde{p})$ and find a neighborhood $U$ of $p$ in $M$ and a local trivialization $\Phi:\pi^{-1}(U) 	o U 	imes \mathbb{R}^k$ such that $\pi = \pi_U \circ \Phi$. Therefore, using \href{./exercise_3-7}{Proposition 3.6}(b), we can equivalently show that
\begin{equation*}
    d \pi_{\widetilde{p}} = d(\pi_U \circ \Phi)_{\widetilde{p}} = d(\pi_U)_{\Phi(\widetilde{p})} \circ d\Phi_{\widetilde{p}}
\end{equation*}
is surjective. By \href{./exercise_3-7}{Proposition 3.6}(d), since $\Phi$ is a diffeormorphism, $d\Phi$ is a homeomorphism. Since $\pi_U$ is a submersion, $d(\pi_U)$ is surjective. Thus, $d \pi_{\widetilde{p}}$ must be surjective as a composition of two surjective functions. Since our choice of $\widetilde{p}\in E$ was arbitrary, the assertion follows. Additionally, being a composition of two smooth functions, $\pi$ itself must be smooth.    
\end{proof}

Exercise 10.1

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Exercise 10.1

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