Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose that for a contractible space $A$ we are given a fiber bundle $p:E\to A$ and denote for $B\subset A$ by $E(B)$ the restricted bundle. I have good reason to believe that in this situation the restriction map $r:\Gamma^0(E)\to\Gamma^0(E(B))$ is a Serre fibration when the section spaces are endowed with the compact-open topology. After a few unsuccessful tries of proving this I decided it might not be a bad idea to ask the denizens of StackExchange for some help. Any hints/ideas/comments are very appreciated.

EDIT: To give a concrete example that bothers me consider $A=D^k\times D^{m-k}$ and $B=D^k_{\frac{1}{2}}\times D^{m-k}$ where $D^k$ denotes the unit disc in $k$ dimensions and $D^k_{\alpha}:=\{x\in D^k\,\big|\,\alpha\leq\|x\|\leq 1\}$.

share|cite|improve this question
Here are some hints. In your example, $A$ is contractible, so you can reformulate the problem more concretely. Also, the inclusion of B into A is a cofibration. (Once you figure out the answer to your question, you'll understand the term "cofibration.") For general (non-contractible) base spaces, I'd worry about sections over B that don't extend to sections over all of A. – Dan Ramras Jul 3 '11 at 6:01
@Dan Thank you for the hints, you're starting to become my guardian angel. What I got now is that assuming in my concrete example that the inclusion of $B$ into $A$ is indeed a cofibration it seems quite easy to show that $r$ is a Serre fibration using the facts that $A$ is contractible and the identification $\Gamma^0(E)=C^0(A,F)$ for a suitable fiber $F$. So I'd be correct in assuming that the statement of the first paragraph is correct if I additionally assume that the inclusion for general $B$ in $A$ is a cofibration? – Martin Worsek Jul 3 '11 at 9:07
Additionally I want to say that I haven't yet managed to construct a retraction of $A\times [0,1]$ onto $(A\times\{0\}\cup B\times [0,1])$ for the concrete $A$ and $B$ I gave. I'm sure it must be something simple but it just goes to show how bad I am at computation (amongst other things). – Martin Worsek Jul 3 '11 at 9:09
Chapter 0 of Hatcher's Algebraic Topology is a good place to read about cofibrations. In particular, any subcomplex of a CW complex gives a cofibration. Writing down the desired maps explicitly can certainly be a pain. – Dan Ramras Jul 3 '11 at 22:19
And yes, the first paragraph looks correct once you assume you have a cofibration. – Dan Ramras Jul 3 '11 at 22:20
up vote 1 down vote accepted

In your example, A is contractible, so you can reformulate the problem more concretely. Also, the inclusion of B into A is a cofibration, which implies that the dual map restriction map on mapping spaces is a fibration.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.