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Let $\pi: E\to M$ be a smooth vector bundle, and $V$ an open subset of $E$ with the property that $V \cap E_p$ is non empty and convex for all $p\in M$.

By "a section of $V$" we mean a (local or global) section of $E$ whose image lies in $V$.

Show that there exists a smooth global section of $V$.

My Argument

Let $p\in M$ and $(\sigma_i^p:U_p\to E)_{i=1}^k$ a smooth local frame with $p \in U_p\subseteq M$ open subset.

Since $E_p \cap V\ne\emptyset$ then $v^i_p\sigma_i^p|_p\in V \cap E_p$ for a certain $v_p \in \mathbb{R}^k$.

Define a smooth local section $\sigma^p:U_p\to E, q\mapsto v^i_p\sigma^p_i|_q$, and take $(\sigma^p)^{-1}(V)$ which is open in $M$.

Then $\{(\sigma^p)^{-1}(V):p\in M\}$ is an open cover of $M$. Take a smooth partition of unity subordinate to it, say $(\psi_p:p\in M)$.

Consider $\sigma=\sum_p\psi_p\sigma^p$. Then $\sigma$ is a smooth global section of $E$.

It remains to show that the image of $\sigma$ lies in $V$.

Let $q\in M$, then I have $\sigma(q)=\psi_p(q)\sigma^p|_q$ for finite indices $p$, and I know that $\sigma^p|_q$ lies in $V$ for each $p$. It follows that $\sigma(q)=\psi_p(q)\sigma^p|_q$ lies in $V$ by the convexity of $V\cap E_q$. $\qquad\square$

Is my proof correct? It is possible to simply it?

P.S: this is the first part of Problem 13.2 in Lee's book: "Introduction to smooth manifolds, 2 Edition"

Here Exercise verification in John Lee's Introduction to Smooth Manifolds, Part 2 is the link to part 2 of the question.

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