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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, $p:E\to B$. 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 $B'$.


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up vote 2 down vote accepted

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.

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Thank you so much. But something is not very clear to me. Why did you take E=R×B as a trivial bundle? Again, thanks. – Feri Nov 20 '11 at 5:50
The main point of my example was that even if you take a subspace of $E$ which fibers over $B'$ it is not automatic that the fibers are vector spaces unless you add some additional assumptions, just as $[0,1]$ is a (topological) subspace of $\mathbb{R}$ which is not a vector space. – Paul Siegel Nov 21 '11 at 15:12

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