Given a fiber bundle $p:E\to B$ and a point $x\in B$, is the evaluation map $\varepsilon:\Gamma^0(E)\to p^{-1}(x)$ defined by $\varepsilon(\sigma):=\sigma(x)$ a weak homotopy equivalence when $\Gamma^0(E)$ is endowed with the compact-open topology? I'm having trouble coming up with a good counter example but also failed at assuming it was a weak homotopy equivalence in general and trying to prove it (which could simply hint at me being bad at this). Right now it seems unlikely to me that the evaluation maps are w.h.e.'s in general since surjectivity of the induced maps on homotopy doesn't seem to work in general since there are no global sections of $E$ in general, or am I completely off track right now?

EDIT: To make the context clearer I'm going to describe the particular scenario I'm struggling with right now. For a smooth fiber bundle $p:E\to D^m$ over the closed unit disc consider an open sub-bundle $p^k_{\omega}:E^k_{\omega}\to D^m$ of the $k$-jet bundle of germs of smooth local sections of $E$. What I'm trying to understand is why the restriction map $\rho:\Gamma^0(E^k_{\omega})\to\Gamma^0(E^k_{\omega}(0))$ is a weak homotopy equivalence when both spaces are endowed with the compact-open topology and $E^k_{\omega}(0)$ denotes the restriction of the bundle $E^k_{\omega}$ to $0\in D^m$. The reason why I formulated the question slightly differently above is because obviously $\Gamma^0(E^k_{\omega}(0))\cong (p^k_{\omega})^{-1}(0)$.

EDIT: After being convinced that the first paragraph has a negative answer in general I would like to restate the explicit question that I'm unable to answer: given everything from the second paragraph, why is the evaluation map $\varepsilon:\Gamma^0(E^k_{\omega})\to (p^k_{\omega})^{-1}(0)$ defined by $\varepsilon(\sigma):=\sigma(0)$ a weak homotopy equivalence?

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    $\begingroup$ A fibration may not have any section yet the fibres are non-empty (think a non-trivial principal fibration). $\endgroup$ Jun 28 '11 at 16:40
  • $\begingroup$ @Torsten Thank you for convincing me of the falseness of the statement in my first paragraph in general. $\endgroup$ Jun 28 '11 at 17:44

In your situation, the base space (a disk) is contractible, so your smooth fiber bundle is actually diffeomorphic to a product bundle, i.e. $E = B\times p^{-1} (x)$. (This follows, for example, by passing to the associated principal bundles - with structure group the diffeomorphism group of the fiber, say.) Now the space of sections is just the mapping space $F(B, p^{-1}(F))$, which is homotopy equivalent to $F(*, p^{-1} (x)) = p^{-1} (x)$ since $B$ is homotopy equivalent to a point. (Here $F$ is the unbased mapping space.)

In general, as you point out, there need not be any global sections at all.

  • $\begingroup$ Unfortunately I'm not sure I understand the argument completely. I know that $E^k_{\omega}(0)$ is a subspace of the space $J^k_0(D^m,F)$ of $k$-jets of smooth maps $D^m\to F$ where $F$ is the fiber of $E$. You are arguing that thus $\Gamma^0(E^k_{\omega})$ is homotopy equivalent to $(p^k_{\omega})^{-1}(0)$ which we know is homeomorphic to $\Gamma^0(E^k_{\omega}(0))$. My gripe is that this does not show that the explicit map $\rho$ is a w.h.e. but only that the domain and range are (weak) homotopy equivalent. It's very well possible that I'm missing the point here. $\endgroup$ Jun 28 '11 at 16:56
  • $\begingroup$ @Dan I forgot to include your name for you to receive the notification. $\endgroup$ Jun 28 '11 at 16:58
  • $\begingroup$ I forgot to type that $J^k_0(D^m,F)$ is supposed to be the space of $k$-jets "at $0$" of smooth maps $D^m\to F$. $\endgroup$ Jun 28 '11 at 17:13
  • $\begingroup$ Frankly I was just answering the version of the question in your first paragraph. But I think you just need to look at the various maps in question and see that there is a commutative diagram. Specifically, you have the restriction map from global sections to sections over 0 and you have the homeomorphism from sections over 0 to the fiber over 0. The composite is the evaluation map (evaluating a section at 0). Since two of these maps are homotopy equivalences, so is the third. (Of, just note that evaluation and restriction are the same map, up to the homeomorphism I mentioned.) $\endgroup$
    – Dan Ramras
    Jun 28 '11 at 17:16
  • $\begingroup$ It's possible that I'm missing some subtlety regarding jet spaces, but this seems to me to just be a question about fiber bundles. $\endgroup$
    – Dan Ramras
    Jun 28 '11 at 17:17

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