TITLE: Why is the set of all smooth sections of a smooth vector bundle closed under addition?
QUESTION [2 upvotes]: Let $\pi:E\to M$ be a smooth vector bundle, where $E$ and $M$ can have nonempty boundaries. Let $\sigma_1$ and $\sigma_2$ be smooth global sections of $E$. I want to show that $\sigma=\sigma_1+\sigma_2$ defined by $\sigma(p)=\sigma_1(p)+\sigma_2(p)$ is also smooth.
It suffices to show that $\sigma$ is smooth on an open neighborhood of each point $p$ in $M$. I can do that when $p$ is an interior point of $M$. Because then there is a smooth local trivialization $(U,\phi)$, where $p\in U\subseteq\operatorname{Int}M$ and $\phi:\pi^{-1}(U)\to U\times\mathbb R^k$ is a diffeomorphism. Then $\sigma|_U$ is smooth because it is a composition of the smooth maps
\begin{equation}
U\xrightarrow{(\sigma_1,\sigma_2)}\pi^{-1}(U)\times\pi^{-1}(U)\xrightarrow{\phi\times\phi}(U\times\mathbb R^k)\times(U\times\mathbb R^k)\xrightarrow{+}U\times\mathbb R^k\xrightarrow{\phi^{-1}}\pi^{-1}(U)\text{,}
\end{equation}
where $(\sigma_1,\sigma_2)(p)=(\sigma_1(p),\sigma_2(p))$ and $+((q_1,v_1),(q_2,v_2))=(q_1,v_1+v_2)$. But if $p$ is a boundary point of $M$, then I cannot argue as above, since $\pi^{-1}(U)$ could have a nonempty boundary, so $\pi^{-1}(U)\times\pi^{-1}(U)$ does not make sense.

REPLY [1 votes]: I found the answer myself. Let $\pi:E\to M$ be a smooth vector bundle. If $(U,\phi)$ is a smooth local trivialization and $\tau:U\to\pi^{-1}(U)$ is a rough local section of $\pi$, then
\begin{equation}
\phi\circ\tau(p)=(p,\tau^1(p),\ldots,\tau^k(p))
\end{equation}
for some functions $\tau^i:U\to\mathbb R$. $\tau$ is smooth if and only if each $\tau^i$ is smooth.
In my case, $\sigma^i=\sigma_1^i+\sigma_2^i$. Since $\sigma_1^i$ and $\sigma_2^i$ are smooth, $\sigma^i$ are
also smooth. Thus $\sigma$ is smooth.
I would be grateful for any feedback.
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