Here we deal with finite-rank real vector bundles over the same base $M$. I focus on order 2 tensor in order to keep the notation simple.
In most DG texts it is proved that
Result 1. Tensors are $C^\infty(M)$-multilinear maps.
More precisely the space of (let's say) $(2,0)$ tensors $\mathcal{T}^{2,0}(M) = \Gamma(T^*M \otimes T^*M)$ is isomorphic, as $C^\infty(M)$-module, with the module of $C^\infty(M)$-bilinear forms $\Gamma(TM) \times \Gamma(TM) \to C^{\infty}(M)$.
The proof consist of showing that a $C^\infty(M)$-multilinear map can be localized and indeed acts pointwise. The other direction is easy.
Then we have
Result 2. Given two vector bundles $E_1,E_2$ over $M$, we have: $$ \mathrm{Hom}(\Gamma(E_1),\Gamma(E_2)) \cong \Gamma(\mathrm{Hom(E_1,E_2)})$$ as $C^\infty(M)$-modules. Where $\mathrm{Hom}(E_1,E_2)$ is the homomorphism bundle.
The proof of this result is essentially the same of result 1, carried out for general vector bundles (even thought in the linear and not multi-linear context)
So I was wondering if I could deduce the first one from the latter.
But, in order to proceed we need the following non-trivial result
Result 3 $$\Gamma(E_1 \otimes_\mathbb{R} E_2) \cong \Gamma(E_1) \otimes_{C^{\infty}(M)} \Gamma(E_2)$$ as $C^\infty(M)$-modules
Now we can exchange $\Gamma$ with $\otimes$:
$$\mathrm{Hom}(\Gamma(TM) \otimes \Gamma(TM), C^{\infty}(M)) \cong \mathrm{Hom}(\Gamma(TM \otimes TM), C^{\infty}(M))$$ by result 3.
Then we can apply result 2 obtaining: $$\mathrm{Hom}(\Gamma(TM \otimes TM), C^{\infty}(M)) \cong \Gamma(\mathrm{Hom}(TM \otimes TM, M \times \mathbb{R})) \cong \Gamma(T^*M \otimes T^*M)$$ using properties of $\mathrm{Hom}, \otimes$ and $\cdot^*$.
The problem
In proving result 1 from result 2, I used result 3 to exchange $\Gamma$ and $\otimes$. In the meantime I found this approach (proving 1 using 2 and 3) also in Conlon's book (Chapter 7).
Considered that proof of 1 and 2 are really much the same and result 3 is not trivial at all:
- Can result 1 be deduced from result 2 without using result 3? In
other words, starting from result 2, do we need the general
result 3 to prove result 1? - How, in the usual proof of result 1, we can avoid to pass from result 3?
The fact is that in the proof we use result 3 only to pass from a multilinear setting to a linear one in order to apply result 2
Thanks