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Consider a smooth manifold $\mathcal{M}$ of dimension $d$ and some smooth $\mathbb{R}$-vector bundle $E$ of rank $d$. On wikipedia, it is claimed that a bundle-morphisms of the type $T\mathcal{M}\to E$ are one and the same as $E$-valued $1$-forms, i.e. there is an isomorphism

$$\Omega^{1}(\mathcal{M},E)\cong\mathrm{Hom}(T\mathcal{M},E).$$

However, I cannot see how this isomorphism is defined explicitely. I know that there is a $C^{\infty}(\mathcal{M})$-module isomorphism $\Omega^{1}(\mathcal{M},E)\cong \mathrm{Hom}(\mathfrak{X}(\mathcal{M},\Gamma^{\infty}(E))$, but I do not know if this helps.

Furthermore, is it true that the set of bundle-isomorphisms is in one-to-one correspondence with the non-degenerate 1-forms, i.e.

$$\Omega^{1}_{\mathrm{nd}}(\mathcal{M},E)\cong\mathrm{Iso}(T\mathcal{M},E)?$$

I think I have read this somewhere, but I can't remember where exactly.

B.Hueber
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  • There is a natural map $\mathrm{Hom}(T\mathcal{M};E)\rightarrow\mathrm{Hom}(\mathfrak{X}(M),\Gamma^{\infty}(E))$, given by post-composing sections with this bundle morphism. Have you tried checking this is an isomorphism? Injectivity should not be hard; surjectivity requires a bit of playing around. Abstractly speaking, the point is that vector bundles are understood by their local sections (as sheaves) and global sections really suffice in the case of smooth vector bundles, since the respective shaves are soft then. – Thorgott Feb 01 '22 at 22:31
  • I don't see why we're passing to sections at the beginning. On the vector bundle level, this is just $\text{Hom}(TM,E) \cong T^*M\otimes E$, as always. In the smooth category, then there's the next step that global sections are globally-defined $E$-valued $1$-forms. This is clear locally and we globalized with partitions of unity. But perhaps I'm still missing your point. – Ted Shifrin Feb 02 '22 at 00:30
  • I remember writing stuff about this a few years ago, but cannot find it. However, this may be helpful. – Ted Shifrin Feb 02 '22 at 00:39

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