Why are $E/F$ and $E^\ast$ smooth manifolds?

Question

In chapter $5$ of Morita's Geometry of Differential Forms, he defines a real vector bundle over a smooth manifold $M$ as a triple $(E,\pi,M)$, where $E$ is another smooth manifold, $\pi \colon E \to M$ is smooth, each fiber $E_p = \pi^{-1}[p]$ is a real vector space, and there's the trivialization condition (which we won't need here I guess).

Then he procceds to do a lot of cool constructions such as the quotient bundle and the dual bundle. I understand everything except why

$$E/F = \bigsqcup_{p \in M} E_p/F_p \quad\mbox{and}\quad E^\ast =\bigsqcup_{p \in M}E_p^\ast$$ are smooth manifolds. Here, of course, $(E,\pi,M)$ is a vector bundle over $M$ and $(F,\pi,M)$ is a subbundle.

I think I must be missing some basic fact about smooth manifolds, but anyway, I need some push here.

Answer 1

Lemma 5.5 in Lee's Introduction to smooth manifolds (the Vector Bundle Construction Lemma) seems to be what you need.

Suppose $M$ is a smooth manifold and $E_p$, $p \in M$ are $k$-dimensional real vector spaces. Let $E$ be the disjoint union of $E_p$ and let $$\pi : E \longrightarrow M, \; \; \pi(x) = p \; \mathrm{for} \; x \in E_p.$$ Suppose we are given

(i) an open cover $\{U_{\alpha}\}_{\alpha}$ of $M$;

(ii) for each $\alpha$, a bijection $\Phi_{\alpha} : \pi^{-1}(U_{\alpha}) \rightarrow U_{\alpha} \times \mathbb{R}^k$ whose restriction to each $E_p$ is a linear isomorphism from $E_p$ to $\mathbb{R}^k$;

(iii) for each $\alpha,\beta$ with $U_{\alpha} \cap U_{\beta},$ a smooth map $$\tau_{\alpha,\beta} : U_{\alpha} \cap U_{\beta} \longrightarrow GL(k,\mathbb{R})$$ such that $$\Phi_{\alpha} \circ \Phi_{\beta}^{-1}(p,v) = (p, \tau_{\alpha,\beta}(p)v).$$ Then $E$ has a unique smooth manifold structure making it a smooth vector bundle of rank $k$ over $M$, with projection $\pi$ and smooth trivializations $\Phi_{\alpha}.$

The point is that the smooth trivializations induce the manifold structure, and the only thing you need to check is that everything glues correctly on the overlaps (condition (iii)). These conditions are straightforward to check for both $E/F$ and $E^*$, assuming they are true for $E$ and $F$.