Let $M^m$ be a smooth manifold. Prove that $\pi:TM\to M$ and $\pi^*:T^*M\to M$ are isomorphic vector bundles using the fact that every manifold admits a riemannian metric.
Here is what I've thought about:
Take $(U_p, X)$ a chart in $p\in U_p\subset M$, then $\left\{\left.\frac{\partial }{\partial x_1} \right|_{u}, \dots, \left.\frac{\partial }{\partial x_m} \right|_{u}\right\}$ is a basis for $T_uM$ and $\{\text{d}x_{1_{u}}, \dots, \text{d}x_{m_{u}}\}$ is a basis for $(T_uM)^*$ for every $u\in U_p$. Define $\phi_p:TU_p \to T^*U_p$ as:
$$\phi_p\left(\sum_{i=1}^{m}a_i\left.\frac{\partial }{\partial x_i} \right|_{u}\right):=\sum_{i=1}^{m}a_i\text{d}x_{i_{u}}.$$
That should give us a local isomorphism. Using a partition of unity, we can define a global isomorphism map $\phi:TM \to T^*M$.
First question: is this correct? Second: how could I possibly use a riemannian metric in this problem?
Any two rank $k$ vector bundles on $X$ are locally isomorphic as they are locally trivial, but these local isomorphisms do not patch together to give a global isomorphism in general.
Instead, let $g$ be a Riemannian metric. Then consider the vector bundle morphism $\Phi : TM \to T^*M$ given by $v \mapsto g(v, \cdot)$. By the non-degeneracy of $g$, you should be able to show that $\Phi$ is injective, then using the fact that $TM$ and $T^*M$ have the same rank, you can conclude $\Phi$ is an isomorphism.