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

A smooth real vector bundle of vector bundle rank $r$ consists of a smooth manifold $E$, a smooth projection $\pi :E\to M$, real vector space structures on its fibers, and local vector bundle trivializationsthat commute with projection to $U$ and are linear isomorphisms on each fiber. On overlaps the change of trivialization is $(p,v)\mapsto (p,g_{VU}(p)v)$, with $g_{VU}:U\cap V\to G{L}_r(\mathbb{R})$ smooth into the general linear group. The usual smooth manifold convention is Hausdorff and second countable.

For an $n$-dimensional $M$, take the disjoint union of its tangent spaces. A manifold chart $x:U\to {\mathbb{R}}^n$ suppliesIf $y$ is another chart, the resulting change of coordinates on the total space isThis is smooth, its inverse is the same construction with $x,y$ exchanged, and the fiber map is invertible and linear. Thus these charts define a smooth structure with the required bundle trivializations.

For completeness, the topology so constructed is Hausdorff: different base points can be separated downstairs, while vectors over one point can be separated inside one product chart. A countable base atlas for $M$ and countable bases for its products with ${\mathbb{R}}^n$ give second countability. Hence the total space is genuinely a smooth manifold, not just a collection of fibers. We obtain

The fibers must retain their base labels. In particular, a literal untagged union of fibers in $E$ is not sufficient when $f$ is not injective. Use the pullback vector bundleThis makes precise the indexed-union notation in the PDF. For a vector bundle trivialization ${\Phi }_U(e)=(\pi (e),v)$, defineIt is a fiberwise linear isomorphism, with inverse $(p,v)\mapsto (p,{\Phi }_U^{-1}(f(p),v))$. The transition functions are $g_{VU}\circ f$, hence smooth. They satisfy the same cocycle identities as those of $E$ and give a smooth total space of manifold dimension $\dim N+r$.

One may also see that this is an embedded submanifold of $N\times E$: in a bundle chart the constraint is the graph of the smooth map $f$ in the base coordinates, leaving the $r$ fiber coordinates free. Thus the chart construction agrees with its natural subspace smooth structure. No immersion or surjectivity hypothesis on $f$ is required. Therefore

A vector bundle morphism over $M$ is a smooth map $F:E\to E^{\prime }$ with ${\pi }^{\prime }\circ F=\pi$ whose restriction $E_p\to E_p^{\prime }$ is linear for every $p$. A vector bundle isomorphism is such a map with a smooth bundle-morphism inverse. Equivalently, a fiberwise bijective smooth bundle morphism is an isomorphism: in local vector bundle trivializations it is multiplication by an invertible smooth matrix, and its inverse matrix is smooth.

For a diffeomorphism $f:N\to M$, defineThe derivative $d{f}_p$ is a linear isomorphism by the chain rule, since $d(f^{-1}{)}_{f(p)}\circ d{f}_p=\mathrm{id}$. In manifold charts, $F$ is represented by the smooth Jacobian matrix of $f$, so it is a smooth bundle morphism. Its inverse iswhich is also smooth and fiberwise linear. ConsequentlyThis identifies the bundles over the same base $N$; the tangent map by itself is a map from $TN$ to $TM$ covering $f$.

Under the standard smooth manifold hypotheses, the first answer is 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 isIt 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 no. On $S^1=\mathbb{R}/2\pi \mathbb{Z}$, compare the trivial vector bundle $S^1\times \mathbb{R}$ with the Möbius line bundleBoth 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