Let $M$ be a smooth, compact manifold and $\xi: \mathcal B \to M$ a smooth complex Banach bundle over $M$. Here, smooth is understood to be in the Fréchet-sense. 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 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 $\mathcal B$.
