Let me recall a standard construction.

Up to homotopy equivalence, any map $f: X \to Y$ is a fibering. Take the special case where $X=E$ the total space of a fiber bundle, and $Y$=B, the basespace of a fiber bundle and $f$ is the projection map.

The associated fibration is given by the projection $f'$ from $P_{X \to M_f}$ to $M_f$. Here $P_{X \to M_f}$ is the set of unbasepointed paths from $X$ to the mapping cylinder $M_f$ attaching $E \times [0,1]$ to $B$ at $E\times {1}$.

Now if there is $\gamma$ in the fiber over a point $b$ in the mapping cylinder, which we can choose to be in $Y=B$, i.e. $P_{E \times 0 \to b \times 0}$, then the obstruction to local triviality is the image of the loop $\gamma$ not being contained in $(f^{-1} U_\alpha )\times 0$.

Thus it is not clear to me that this construction, yields a fiber bundle.

When is it true that a fiber bundle can be turned into a fibration that is a fiber bundle?

  • $\begingroup$ Are are mixing up the standard constructions of a map turning it into a cofibration (using the mapping cylinder) and turning it into a fibration (using a space of paths)? $\endgroup$ – archipelago Feb 29 '16 at 12:04
  • $\begingroup$ I don't think so - I just combined the construction turning any map into an embedding and turning any embedding into a fibration from bott and tu. I had not even read the construction for cofibrations until you made this comment so I don't think I could have mixed them up. Having read about the construction for cofibrations from hatcher after reading your comment, I don't think that I mixed the two up. $\endgroup$ – Hari Rau-Murthy Mar 2 '16 at 3:30
  • $\begingroup$ Shit. Fiber bundles already satisfy the homotopy lifting property. So there is no question to ask on whether the fiber bundle projection can be turned into a map that is a fibration. Now as a separate point, the construction for turning a map into a fibration will not, I don't think, take the fiber bundle projection onto another fiber bundle projection. $\endgroup$ – Hari Rau-Murthy Apr 26 '16 at 23:57

The map $E \to B$ that is turned into a fibration $E' \to B$ can be turned into a fibration in such a way that $E \to E'$ is a deformation retract. Thus the local triviality condition will still be satisfied on $E'$. See Mosher and Tangora page 84.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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