Let $\pi:E\longrightarrow M$ be a vector bundle and suppose $U\subset M$ is an open subset. We have an inclusion map $\jmath:U\longrightarrow M$ hence we can consider the pullback bundle $\jmath^*(\pi):\jmath^*(E)\longrightarrow U$.
The sections of $\pi:E\longrightarrow M$ induce sections on $\jmath^*(E)$ by restriction. Is every section of $\jmath^*(E)$ the restriction of a section of $E$?
Thanks
First of all, note that $j^*E = E|_U = \pi^{-1}(U)$.
Let $M = S^1$ and let $E$ be the total space of the Möbius line bundle. The open set $U = S^1\setminus\{p\}$ is contractible. As every vector bundle on a contractible space is trivial, $j^*E$ is trivial. By choosing a Riemannian metric on $E$, we obtain a Riemannian metric on $j^*E$ by restriction. As $j^*E$ is trivial, there is a nowhere zero continuous section $s : U \to j^*E$ such that $|s(x)| = 1$.
If $s$ extended to a continuous section of $E$, then by continuity (of both $s$ and the Riemannian metric) we would have $|s(p)| = 1$; in particular, the extension would be a nowhere zero section of $E$. But this is impossible as the Möbius line bundle is non-trivial.
If you regard the Möbius strip as the elements of the Möbius line bundle with norm less than or equal to one (with respect to the chosen Riemannian metric), then the image of $s$ is given by the red curve in the image below (courtesy of Wolfram MathWorld).
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