Restriction of a smooth vector bundle is a smooth bundle?

In John Lee's Smooth manifolds, pg255, he wrote

(Restriction of a vector bundle.) Suppose $$\pi:E \rightarrow M$$ is a rank $$k$$ vector bundle and $$S \subseteq M$$ is any subset. We define the restriction of $$E$$ to $$S$$ to be the set $$E|_S = \bigcup_{p \in S}E_p$$. with projection $$E|S \rightarrow S$$ obtained by restriciting $$\pi$$. A local trivilization $$\Phi: \pi^{-1}(U) \rightarrow U \times \Bbb R^k$$ restricts to a bijection $$\Phi|_S: (\pi|S)^{-1}(U \cap S) \rightarrow (U \cap S) \times \Bbb R^k.$$ If $$E$$ is a smooth vector bundle, and $$S \subseteq M$$ is an immersed or embedded submanifold it follows from chart lemma that $$E|_S$$ is a smooth vector bundle.

1. Where have we used the fact that $$S$$ is an immersed submanifold?

i.e. that $$i$$ is of constant rank and at each point $$d_pi:T_pS \rightarrow T_pM$$ is an injective map.

1. Is $$E|_S$$ an immersed submanifold of $$E$$?
• The short answer to your question 1 is that you need to assume $S$ is an immersed submanifold for the statement "$E|_S$ is a smooth vector bundle" even to make sense. The only kinds of smooth submanifold are embedded ones and immersed ones, and the latter category encompasses the former. Sep 28 '18 at 22:08

At 1: Not really, if one actually interprets $$S \subset M$$ as a mapping $$f \colon S \to M$$ and instead of $$E|_S$$ the the construction of the pullback $$f^*E$$ of a (vector) bundle $$\pi \colon E \to M$$ along a map $$f \colon S \to M$$. I find this construction much easier to grasp, so I sketch it here. Just think of $$f$$ as the immersion; but it can be any other smooth mapping.
The total space of the pullback bundle $$f^*E$$ is defined by $$f^*E : = \{ (x,u) \in S \times E \;|\; f(x) = \pi(u) \} \subset S \times E.$$ Its footpoint map is just the projection onto the first factor: $$p \colon f^*E \to S, \qquad p(x,u) := x.$$ Its addition and multiplication are defined by $$(x,u) + (x,v) := (x, u+v) \quad \text{and} \quad \lambda (x,u) := (x, \lambda u).$$ We have to show that it is locally trivial. We do so by constructing a bundle atlas as follows: For each $$x\in S$$ choose an open neighborhood $$V \subset M$$ of $$f(x)$$ with local trivialization $$\varphi \colon E|_V \to V \times \mathbb{R}^m$$. Then $$U = f^{-1}(V)$$ is an open neighborhood of $$x$$. A local trivialization can be defined as follows $$\psi \colon (f^*E)|_U \to U \times \mathbb{R}^m, \quad \psi(x,u) = (x, \operatorname{pr}_{\mathbb{R}^m}\varphi(f(x),u)),$$ where $$\operatorname{pr}_{\mathbb{R}^m} \colon V \times \mathbb{R}^m \to \mathbb{R}^m$$ is the projection onto the second factor. Indeed, this is a diffeomorphism that is linear in each fiber; it's inverse can be written down directly: $$\psi^{-1}(x,v) = (x, \varphi^{-1}(f(x),v)).$$ These local trivializations are compatible with each other so that they form a bundle atlas follows by construction. It is a bit tediuous to show (but not hard, really), so I skip it here.
At 2.: We can show that the mapping $$F \colon f^*E \to E, \quad F(x,u) = u$$ is an immersion. With the local trivialization $$\varphi$$ from above, we have $$\varphi \circ F(x,u) = (f(x), \operatorname{pr}_{\mathbb{R}^m} \varphi(f(x),u)$$ (locally of course). Then the differential looks like this in block matrix form: $$d(\varphi \circ F)(x,u) = \begin{pmatrix} \mathrm{d}_x f & 0 \\ ?? & \operatorname{pr}_{\mathbb{R}^m} \varphi(f(x), \cdot)\end{pmatrix}.$$ Since $$f \colon S \to M$$ is an immersion, no matter what $$??$$ is, this matrix has rank equal to $$\operatorname{rank}(\mathrm{d}_x f) + \operatorname{rank}(\varphi(f(x), \cdot)) = \operatorname{dim}(M)+ (\operatorname{dim}(E|_{f(x)})).$$ Since $$\varphi$$ is a diffeomorphism, $$F$$ has maximal rank and is thus an immersion.