The pullback fibre product is a smooth vector bundle

Give $f^{\ast }E=\{(q,e)\in N\times E:f(q)=\pi (e)\}$ the subspace topology from the product manifold $N\times E$ of [L3]. It is Hausdorff and second countable by [L3]. For a bundle chart ${\mathrm{\Phi }}_{\alpha }:E{\mathrm{\mid }}_{U_{\alpha }}\to U_{\alpha }\times {\mathbb{R}}^r$, put $V_{\alpha }=f^{-1}(U_{\alpha })$, an open subset of $N$. The pullback over $V_{\alpha }$ is open in $f^{\ast }E$ because it is the intersection with $V_{\alpha }\times E$.

For $(q,e)$ over $V_{\alpha }$, define ${\tilde{\mathrm{\Phi }}}_{\alpha }(q,e)=(q,v)$ when ${\mathrm{\Phi }}_{\alpha }(e)=(f(q),v)$. Its inverse is explicitly $(q,v)\mapsto (q,{\mathrm{\Phi }}_{\alpha }^{-1}(f(q),v))$. The forward map is continuous as the restriction of the product of the identity on $N$ with ${\mathrm{pr}}_2\circ {\mathrm{\Phi }}_{\alpha }$; the inverse is continuous because $q\mapsto f(q)$ and ${\mathrm{\Phi }}_{\alpha }^{-1}$ are continuous. Thus ${\tilde{\mathrm{\Phi }}}_{\alpha }$ is a homeomorphism onto the open product manifold $V_{\alpha }\times {\mathbb{R}}^r$. These homeomorphisms cover $f^{\ast }E$.

On overlaps, ${\tilde{\mathrm{\Phi }}}_{\beta }\circ {\tilde{\mathrm{\Phi }}}_{\alpha }^{-1}(q,v)=(q,g_{\beta \alpha }(f(q))v)$. This is smooth because $g_{\beta \alpha }$ and $f$ are smooth, and it is linear in each fibre coordinate. Combine the homeomorphisms of step 2.1 with product charts on $V_{\alpha }\times {\mathbb{R}}^r$; their smooth overlap maps give a smooth atlas on the already Hausdorff second-countable space $f^{\ast }E$. In these charts the projection $(q,e)\mapsto q$ is the first projection, hence smooth and surjective, and each fibre is linearly identified with $E_{f(q)}\cong {\mathbb{R}}^r$. Consequently [L4] gives a smooth rank-$r$ vector bundle over $N$. Empty bases and rank zero obey the same construction.