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I'm confused about whether or not the there is a manifold structure here. The set of linear bundle homomorphisms between two vector bundles over the same base is clearly a vector bundle. If you take the set of linear bundles morphisms from the projection bundle $M \times R \to M$ into any vector bundle $q: E \to M$, this is a vector bundle for the stronger reason above and it looks like it should be isomorphic to the set of sections of $q$.

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  • $\begingroup$ The sections of $q$ will be in correspondence with the sections of the bundle you constructed. Remember that an element of a vector bundle is just a single vector in a single fibre. $\endgroup$ – Anthony Carapetis Nov 7 '17 at 3:13
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The smooth sections of the trivial bundle $\mathbb{R}\times\mathbb{R}$ is just the set of all smooth real functions. It is infinite dimensional. Just as linear maps from $\mathbb{R}$ to a vector space is the vector space itself (each map picks a vector), we can say that bundle maps from the trivial vector bundle $\mathbb{R}$ to any bundle $V$ is isomorphic to the bundle itself. And there is a map of sections. But there isn't a map from the set of sections to the bundle, that's mixing categories.

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  • $\begingroup$ But for every smooth section of $s: B \to E$ of $q$, there's a corresponding linear bundle morphism $\hat{s}: B \times R \to E$. Similarly, for every linear bundle morphism $t: B \times R \to E$ there's a corresponding smooth section $t(x,0) : B \to E$. Shouldn't there be an isomorphism of sets between the two? $\endgroup$ – username12121212 Nov 7 '17 at 19:27

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