 Solution

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