Is a sub-bundle of a vector bundle a vector bundle?

Question

Could anyone please help me with this question?

(1) Let (E, p, B) be a vector bundle where E is the total space, B is the base, and p is the structure map, that is, p:E->B. Now suppose E' is a subspace of E, B' is the subspace of B, and p' is the restriction of p to E'. If the image of p' is contained in B', then show (E', p', B') is a vector bundle. (2) Prove or disprove: Let s be a section of (E, p, B). Then restriction of s to B' is a section of (E', p', B') if and only if s(b) is in E' for each b in B'.

THANK YOU SO MUCH IN ADVANCE.

Answer 1

There is a mistake in your formulation of the question, unless you are using the word "subspace" in two different ways. Take $B$ to be a single point, $E = \mathbb{R} \times B$, $B' = B$, and $E' = [0,1] \times B'$.

More likely you meant to assume that $p': E' \to B'$ is such that $p'$ is the restriction of $p$ to $E'$ and additionally $p'^{-1}(b) = p^{-1}(b)$ for every $b \in B'$. If this is the case then (1) is just an exercise in playing around with the subspace topology. Given any $b \in B'$, we need to show that there is a neighborhood $U \subseteq B'$ which trivializes $p': E' \to B'$ in the sense that $p'^{-1}(U)$ is homeomorphic to $U \times \mathbb{R}^n$ in such a way that $p': p'^{-1}(U) \to U$ corresponds to the projection map $U \times \mathbb{R}^n \to U$. But since $p: E \to B$ is a vector bundle there is a trivializing neighborhood $V \subseteq B$ of $b$, and with the assumptions on $p'$ given above one has that $U = V \cap B'$ does the job.

As for (2), you need to check that the restriction $s'$ of $s$ to $B'$ is continuous (obvious) and satisfies $p'(s'(b)) = b$ for every $b \in B'$. But $p'(s'(b)) = p(s(b)) = b$ since $p'$ is the restriction of $p$ and since $s$ is a section, so we're done.