It is well-known that a fiber bundle under some mild hypothesis is a fibration, but I don't know any examples of fiber bundles which aren't (Hurewicz) fibrations (they should be weird examples, I think, because if the base space is paracompact then the bundle is a fibration).

Does anybody know an example?

Thanks!

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    It will still be a Serre fibration, in any case. – David Roberts Jan 17 '13 at 14:09
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    Here's a vague sketch. Take a non-numerable cover of some necessarily non paracompact space, and take a non coboundary, G-valued Čech cocycle. Then this bundle isn't classifiable by a map to BG, as the universal bundle trivialities over a numerable cover (i.e. is numerable), and numerable covers pull back. As soon as your bundle is numerable it is a Hurewicz fibration. Now you have a chance that you'll find an example. – David Roberts Jan 17 '13 at 14:17
up vote 19 down vote accepted

$\newcommand{\RR}{\mathbb{R}} \newcommand{\To}{\longrightarrow} \newcommand{\id}{\mathrm{id}}$The example described in Tom Goodwillie's answer to a related mathoverflow question essentially solves this question. Specifically, Tom defines an orientable, non-trivial, real line bundle $L$ over a contractible space $X$. The principal $GL_1^+(\RR)$-bundle — $GL_1(\RR)$ will work as well — associated to $L$ is then a locally trivial fibre bundle $p:E\to X$. The principal bundle $p:E\to X$ does not admit a section, since the line bundle $L$ is not trivial. It follows that $p$ cannot be a Hurewicz fibration: otherwise it would admit a section, given that $X$ is contractible. For convenience, I will give below the construction of $p:E\to X$ and a few details of the proof — mostly copied from Tom's answer, and following his notation where possible. Please upvote Tom Goodwillie's answer, which is certainly shorter and more readable.

The base space $X$

Let $X$ be the space obtained by gluing two copies of $\RR$ along $\RR^+$: $$ X = (\RR\times\{0,1\})\mathbin{/}{\sim} $$ where ${\sim}$ is generated by $(x,0)\sim(x,1)$ for $x\in\RR^+$. This space is not Hausdorff, and is closely related to the well-known line with two origins. Let $q:\RR\times\{0,1\}\to X$ be the quotient map. Define two open subspaces covering $X$ by $U=q(\RR\times\{0\})$ and $V=q(\RR\times\{1\})$. Finally, let $g:X\to\RR$ be the continuous function determined by $g(q(x,i))=x$, and define $$f=g|_{U\cap V}: U\cap V \To \RR^+$$ Importantly, observe that $f$ does not extend to a continuous map $X\to\RR^+$.

The total space $E$

Consider the topological abelian group $G=GL_1^+(\RR)=(\RR^+,\cdot)$ given by the positive reals with multiplication. Let $E_U=U\times G$ and $E_V= V\times G$ denote the trivial $G$-bundles over $U$ and $V$, respectively. Construct the principal $G$-bundle $E$ over $X$ by gluing $E_U$ and $E_V$ along $U\cap V$ via the $G$-isomorphism $$ \varphi_f : E_U|_{U\cap V}\overset{\simeq}{\To} E_V|_{U\cap V} $$ defined by $$ \varphi_f(x,g)=\bigl(x,f(x)\cdot g\bigr) $$ More concretely, $E$ is obtained from $E_U \amalg E_V$ by identifying $(x,g)\in E_U$ with $\varphi_f(x,g)\in E_V$ for each $x\in U\cap V$. As in Tom Goodwillie's answer, we could just as well use any other continuous function $f:U\cap V\to\RR^+$ which does not extend to a continuous function $X\to\RR^+$.

Non-triviality of the principal bundle $E\to X$

The projection map $p:E\to X$ gives a principal $G$-bundle over $X$, which comes with canonical isomorphisms $E|_U = E_U$ and $E|_V = E_V$. We will now show $p$ does not admit a section. By the construction of $E$, a section of $p:E\to X$ determines:

  • a section of $E_U=U\times G\to U$, and therefore a map $s_U:U\to G=\RR^+$;
  • similarly, a map $s_V:V\to G=\RR^+$;
  • these maps verify $s_V(x)=f(x)\cdot s_U(x)$ for each $x\in U\cap V$.

In particular, $f(x)=s_V(x)/s_U(x)$ for all $x\in U\cap V$. However, this implies that $f$ extends to a continuous function $\overline{f}:X\to\RR^+$ given by $$ \overline{f}(x)=\frac{s_V\bigl(q(g(x),1)\bigr)}{s_U\bigl(q(g(x),0)\bigr)} $$ which contradicts the known non-extension property of $f$.

Conclusion

The projection $p:E\to X$ gives a locally trivial principal $G$-bundle over $X$, since $E|_U\simeq E_U$ and $E|_V\simeq E_V$ are trivial $G$-bundles. Thus, it remains to show that $p$ is not a Hurewicz fibration. Note that $X$ is contractible. So if $p:E\to X$ were a fibration, it would necessarily admit a section. In detail, let $H:X\times I\to X$ be a null-homotopy of $\id_X$. We can obviously lift the constant map $H_1$ to $E$. Assuming $p$ is a Hurewicz fibration, the homotopy lifting property then produces a lift $\widetilde{H}: X\times I\to E$ of $H$ to $E$, and consequently a section $\widetilde{H}_0$ of $p$. But we showed in the previous paragraph that $p$ admits no sections. In conclusion, $p$ is not a Hurewicz fibration.

  • " Specifically, Tom defines an orientable, non-trivial, real line bundle L over a contractible space X" -- I thought fibre bundles over contractible spaces are always trivial? Is there some necessary restriction on the base for this to hold? – ಠ_ಠ Sep 29 '16 at 1:17
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    @ಠ_ಠ if the bundle had a classifying map, then necessarily it would be trivial, because the classifying map would be null-homotopic. But due to the fact the bundle doesn't trivialise over a numerable cover (as the line with two origins is not Hausdorff paracompact) there is no classifying map. – David Roberts Jul 19 '17 at 6:33

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