Let $M$ be a smooth, compact manifold and $\xi: \mathcal B \to M$ a smooth complex [Banach bundle](https://en.wikipedia.org/wiki/Banach_bundle) over $M$. Here, smooth is understood to be in the [Fréchet-sense](https://de.wikipedia.org/wiki/Fr%C3%A9chet-Ableitung). Further, let $p: V \to M$ be an ordinary smooth finite-rank $\mathbb C$-vector bundle over $M$. Then one can form the vector bundle tensor product $\xi \otimes p: \mathcal B \otimes V \to M$, so that for $m \in M$, the fiber $(\mathcal B \otimes V)\_m$ is just the algebraic tensor products $\mathcal B\_m \otimes\_{\mathbb C} V\_m$. It is again a Banach bundle.


Denote by $\Gamma(\mathcal B \otimes V)$ the space of smooth sections of that bundle. It has the natural structure of a $C^\infty(M,\mathbb C)$-module. The map $\Gamma(\mathcal B) \times \mathcal \Gamma(V) \to \Gamma(\mathcal B \otimes V)$, given by $\sigma \times \omega \mapsto \sigma \otimes \omega$ is clearly $C^\infty(M,\mathbb C)$-bilinear, thus extends to a map of $C^\infty(M,\mathbb C)$-modules $$F: \Gamma(\mathcal B) \otimes\_{C^\infty(M,\mathbb C)} \Gamma(V) \to \Gamma(\mathcal B \otimes V).$$


**Question: Under what additional conditions on $\mathcal B$ is $F$ an isomorphism of $C^\infty(M,\mathbb C)$-modules ?**


I know that the answer is positive whenever $\mathcal B$ is also of finite-rank. However, the [proof that I know](https://math.stackexchange.com/questions/492166/global-sections-of-a-tensor-product-of-vector-bundles-on-a-smooth-manifold) heavily relies on the fact that any finite rank bundle over $M$ is isomorphic to the direct summand of a trivial bundle over $M$ (presented in the same proof). I am uncertain if this fact still holds for arbitrary Banach bundles. In fact, I believe that a reasonable assumption on $\mathcal B$ would be that it is a a Hilbert bundle, so that we at least find a metric on $\mathcal B$. 



On sections into Banach bundles over a compact manifold
if you go back to the proof you linked to you realize that you only need to embed one of the vector bundles into a trivial bundle.


In detail: For $V$ a trivial bundle your map is an isomorphism. In general, let $V\oplus V^{\perp}$ be a trivial bundle. Then you have $\Gamma((V\oplus V^{\perp})\otimes \mathcal B)\cong \Gamma(V\otimes \mathcal B)\oplus \Gamma(V^{\perp}\otimes \mathcal B)$ and $\Gamma(\mathcal B)\otimes\_{C^\infty(M,\mathbb C)} \Gamma(V\oplus V^{\perp})\cong \Gamma(\mathcal B)\otimes\_{C^\infty(M,\mathbb C)} \Gamma(V) \oplus \Gamma(\mathcal B)\otimes\_{C^\infty(M,\mathbb C)}\Gamma(V^{\perp})$. Your morphism $F$ for $V\oplus V^{\perp}$ splits as a direct sum of two morphisms which thus both have to be isos. 


