Under the standard <smooth manifold> hypotheses, the first answer is \b[yes], although the isomorphism requires a choice. A <partition of unity> combines Euclidean <inner products> in <manifold charts> to produce a <Riemannian metric> $g$. Nonnegative local weights summing to one preserve positivity, and local finiteness preserves smoothness. The <musical isomorphism> is
$$
\boxed{\flat_g:TM\longrightarrow T^*M,\qquad v\longmapsto g(v,\cdot).}
$$
It is smooth and fiberwise invertible, with inverse having local <matrix> $(g_{ij})^{-1}$. It is therefore a <vector bundle isomorphism>. There is no preferred such map before a metric or equivalent additional structure is chosen.

The second answer is \b[no]. On $S^1=\mathbb R/2\pi\mathbb Z$, compare the <trivial vector bundle> $S^1\times\mathbb R$ with the <Möbius line bundle>
$$
L=(\mathbb R\times\mathbb R)/((t+2\pi,a)\sim(t,-a)).
$$
Both have rank one. A <continuous> <section of a vector bundle> of $L$ is represented on $[0,2\pi]$ by a <continuous> function $s$ satisfying $s(2\pi)=-s(0)$. If it were nowhere zero, the <intermediate value theorem> would prevent its endpoint signs from being opposite. Thus $L$ has no <nowhere-zero section>. The trivial bundle has the section $1$, and a bundle isomorphism would carry that section to a nowhere-zero section of $L$, a contradiction. Hence
$$
\boxed{L\not\cong S^1\times\mathbb R.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 115 / 1 / d / Solution

{c}
{title2=$L\to S^1$}

The Möbius line bundle over the circle is
$$
L=(\mathbb R\times\mathbb R)/((t+2\pi,a)\sim(t,-a)),
$$
with projection to $t$ modulo $2\pi$. It is a smooth rank-one <vector bundle>. A section is represented by a function $s$ satisfying $s(t+2\pi)=-s(t)$, so the <intermediate value theorem> forces every <continuous> section to vanish somewhere. Thus it has no <nowhere-zero section> and is not isomorphic to the <trivial vector bundle> of rank one.

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 Incoming links: Möbius line bundle