I am trying to proof, that if a vector bundle $\xi$ posses an Euclidean metric, then it is isomorphic to its dual $Hom( \xi, \epsilon^1 )$, where $\epsilon^1 = B( \xi ) \times \mathbb{R}$ denotes the trivial line bundle.
This statement suprises me, as locally the statement should always hold. Does someone have a counterexample of a vector bundle, where this is not true?
My "proof" so far is the following:
Consider the map $\Phi: E(\xi) \to Hom(\xi, \epsilon^1)$, that maps $y=(x,b) \in \mathbb{R}^n \times F$ after the choice of a neighborhood to $g_b(x, -)$, where $g_b(-,-)$ is the metric at the basepoint $b$. It is an isomorphism as $g_b$ is positive definite, so its describing matrix (after the choice of a basis) is invertible, so any linear form $\varphi \in (E(\xi)_b)^*$ is of the form $g_b(z,-)$ of the form $g_b(z,-)$.
I feel uncomfortable with it, as for me it is not clear, if everything glues to a global map, as I only consider everything locally, by using the isomorphism $(x,b) \in U \cong \mathbb{R}^n \times F$. Is it true, that these map glue to a global map, and how can one show this?
Your proof is correct; you do not even need a Euclidean metric, any (fiberwise) nondegenerate continuous bilinear form $\omega$ will suffice. To construct an isomorphism $\xi\to \xi^*$ use the fiberwise isomorphism $$ \xi_b\to \xi^*_b, v\mapsto \omega(v, \cdot). $$ It is clear that this map is well-defined. To check its continuity, use local trivialization, where the bundle is the product $$ U\times F, $$ where $F$ is the fiber. With respect to this product structure, $\omega$ has the form $$ \omega_b(\cdot, \cdot) $$ where $b\mapsto \omega_b$ is a continuous map from the base $U$ to the space of 2-tensors on the fiber $F$. Now, continuity of the map $$ (v,b)\mapsto \omega_b(v, \cdot) $$ is clear.