# Fiber bundles that can be turned into a fibration that is a fiber bundle.

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?

• 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)? – archipelago Feb 29 '16 at 12:04
• 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. – Hari Rau-Murthy Mar 2 '16 at 3:30
• 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. – Hari Rau-Murthy Apr 26 '16 at 23:57

## 1 Answer

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.