**Definition** Let $E_{1} \xrightarrow{\pi_{1}} M_{1}$, $E_{2} \xrightarrow{\pi_{2}} M_{2}$ be two vector bundles over $M_{1}$ and $M_{2}$ with fibers $V_{1}$, $V_{2}$ respectively. The exterior tensor product $E_{1} \boxtimes E_{2}$ is defined as the vector bundle over $M_{1} \times M_{2}$, whose fiber over $(x, y) \in M_{1} \times M_{2}$ is $E_{1 x} \otimes E_{2 y}$, where $E_{1 x}$ is the fibre of $E_{1}$ over $x$ and $E_{2 y}$ is the fibre of $E_{2}$ over $y$.

In the article Brouder, Dang, Laurent-Gengoux, and Rejzner - Properties of field functionals and characterization of local functionals they have the so called fundamental result on the projective tensor product of sections of a vector bundle

Proposition III.8. Let $\Gamma(M, B)$ be the space of smooth sections of some smooth finite rank vector bundle $B \rightarrow M$ on a manifold $M$. Then $\Gamma(M, B)^{\hat{\otimes}_{\pi} k}=\Gamma\left(M^{k}, B^{\boxtimes k}\right)$.

Where $\boxtimes$ means exterior tensor product and $\otimes_\pi$ is the projective tensor product.

I am trying to prove this theorem. My first step is to prove the algebraic isomorphism. This is how I am trying to prove the algebraic isomorphism by following the same steps as in A nonlinear theory of generalized tensor fields on Riemannian manifolds by Eduard Nigsch at page 89:

Let $V$ be the fiber of the **trivial vector bundle** $B \xrightarrow{\pi} M$.
We know there is a bilinear map from $\alpha :V\times V \rightarrow V\otimes V$ with $$\alpha(v_1,v_2)=v_1 \otimes v_2$$

where $v_1,v_2 \in V$.

Now let $\psi :B \rightarrow U\times V$ be a local trivialization map with $U \subset M$.

Now we define the local sections $\sigma_i$ by $\sigma_i(x) =\psi^{-1}(x,v_i)$.

We also define the local sections $\gamma_{ij}$ by $\gamma_{ij} =\phi^{-1}(x,y,v_i\otimes v_j)$ where $\phi :B{\boxtimes}B \rightarrow U^2 \times V\otimes V$ is a local trivialization map of the bundle $B{\boxtimes}B \rightarrow M \times M$.

Define the map $g :\Gamma\left(M, B\right) \times \Gamma\left(M, B\right) \rightarrow \Gamma\left(M\times M, B{\boxtimes}B\right)$ by $$g=\sum_{i, j} m \circ\left(i d \times m\left(\cdot, \gamma_{i j}\right)\right) \circ\left(\sigma_{i}^{*} \times \sigma_{j}^{*}\right)$$

where $m: C^{\infty}(M) \times \Gamma(B\otimes B)\rightarrow \Gamma(B\otimes B) $ is the module multiplication on $\Gamma(B\otimes B)$ and $\sigma_{i}^{*}$ is the dual of $\sigma_{i}$.

By the properties of the tensor product there is a bilinear map $f: \Gamma(M, B)\otimes \Gamma(M, B)\rightarrow \Gamma\left(M\times M, B{\boxtimes}B\right)$ such that if $g$ is the map $h:\Gamma\left(M, B\right) \times \Gamma\left(M, B\right) \rightarrow \Gamma(M, B)\otimes \Gamma(M, B)$ we have that $$g=f \circ h.$$

For the inverse define $h^{-1}: \Gamma\left(M\times M, B{\boxtimes}B\right)\rightarrow \Gamma(M, B)\otimes \Gamma(M, B)$ $h^{-1}(s)=\sum_{i, j} \gamma_{i j}^{*}(s) \alpha_{i} \otimes \beta_{j}$ for $s \in \Gamma\left(M\times M, B{\boxtimes}B\right)$

Now we have that $$ \begin{aligned} &h^{-1}\left(f\left(t \otimes u\right)\right)=h^{-1}\left(h \left(t^{i} \alpha_{i}, u^{j} \beta_{j}\right)\right)=h^{-1} \left(t^{i} u^{j} \gamma_{i j}\right)=t^{i} u^{j} \alpha_{i} \otimes \beta_{j}\\ &=t^{i} \alpha_{i} \otimes u^{j} \beta_{j}=t \otimes u \text { and }\\ &h(h^{-1}(s))=h\left(s^{i j} \alpha_{i} \otimes \beta_{j}\right)=s^{i j} g\left(\alpha_{i}, \beta_{j}\right)=s^{i j} \gamma_{i j}=s . \end{aligned} $$

Is my proof correct?

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