- Let $S$ be a smooth immersion in $M$,prove $E|_S \to E$ is smooth immersion.
- Let $S$ be a smooth embedding in $M$,prove $E|_S \to E$ is smooth embedding
- Let $U$ be a open subset of $M$,prove $E|_U \subset E$ is a open subset of vector bundle.
I have tried to prove these three result,but I was confused here,so to make the question clear,I will not provide it here.The key point has been shown in the slice chart lemma:
$$\pi^{-1}(V_p) \to V_p\times \Bbb{R}^k \to \hat{V}_p\times \Bbb{R}^k$$
As we can see,it's not surprising that topology relation and smooth structure of $E|_S$ and $S$ are the same.Since it's built on structure of $S$.(vector bundle inherent the properties of base manifold)
The third one is obvious since $\pi$ is continuous and we know $E|_U = \pi^{-1}(U)$ so it's open in $E$.
The first one is also obvious,to see this just consider the inclusion map $i:E|_S \to E$ under coordinate representation.pick a $p \in S \subset M$ ,then the chart for $E|_S$ in the construction given by chart lemma is some $(\pi^{-1}(V),(\varphi\times\ id)\circ\Phi|_V)$ where($(V,\varphi)$ is some chart for $S$)
while chart for $E$ is $(\pi^{-1}(U),(\psi\times id)\circ \Phi)$ where ($(U,\psi)$ is some chart for $M$). under this chart $\hat{i}(p,v) = (p,v)$ in Euclidean space .so it's immersion.Which is also smooth map as we can see under coordinate.
Finally ,for the second one ,we need to show that $E|_S$ has subspace topology.Recall in the slice chart lemma topology on $E|_S$ is given as :
$$\mathcal{B} = \{\cup_\alpha U_\alpha\mid U_\alpha \subset \pi^{-1}(V_\alpha)\text{ is open}\}$$
Such that $V_\alpha$ is some chart for $S$(the point is $V_\alpha $ is open in $S$ with subspace topology since $S$ has subspace topology).We can show $\mathcal{B}$ equals to subspace topology on $E|_S$ the detail is bit long,But it's rutine.