Linear operations on space of sections of vector bundle

Question

Given vector bundles $E, E'$ over the same base manifold $M$, we can always construct other spaces such as $E^*$, $E \oplus E'$, $E \otimes E'$ etc. by using linear operations on the fibres. I would like to know if this necessarily induces a correspondence between the sections of the new bundle and the original spaces of sections $\Gamma(E)$, $\Gamma(E')$.

For the direct-sum bundle $E \oplus E'$, it seems obvious that there is a 1-1 correspondence between $\Gamma(E \oplus E')$ and $\Gamma(E) \oplus \Gamma(E')$, since any pair of sections $s \in \Gamma(E)$ and $s' \in \Gamma(E')$ define a unique section $(s,s') \in \Gamma(E \oplus E')$. However, for the tensor product I cannot see how something like $$\Gamma(E \otimes E') \cong \Gamma(E) \otimes \Gamma(E')$$ would hold. Is this even true? And moreover, does this hold in general? As in, if we have some vector bundles $E_i$ and some fibrewise linear operation $\square$ that forms a well-defined vector bundle $\square_i E_i$when applied to the $E_i$, is it always the case that $\Gamma(\square_i E_i) \cong \square_i \Gamma(E_i)$?

Answer 1

The assignment $\Gamma\colon E\mapsto \Gamma(E)$ is a functor from the category of smooth vectorbundles over $M$ to the category of modules over the ring $C^\infty(M).$ Keeping this in mind gives a good framework to study the question you have posed. This statement together with the question of how the image of $\Gamma$ looks like is discussed here.

Then the natural question is whether $\Gamma(E\otimes F)\cong\Gamma(E)\otimes_{{C^\infty}(M)}\Gamma(F)$ hold true. The answer is yes and you can find a proof here.

$\Gamma$ also preserves $\oplus$ as you know and thus all constructions that are built out of $\otimes$ and $\oplus$ transform correctly, for example the assignement of $(r,s)-$Tensors $\mathcal{T}_r^sE=E^{\otimes r}\otimes (E')^{\otimes s}$ or that of the algebra of alternating tensors $\Lambda E=\oplus_i\Lambda^iE$.