# composition of covering map and bundle projection

Hello, can somebody help with the following question that I have thought over for quite some time, to no avail?

Suppose f: X--->Y is a universal cover and g: Y--->Z a fiber bundle, where X, Y and Z are manifolds. Is the composition gof: X--->Z necessarily a fiber bundle?

THanks!

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This question is more interesting than I thought. I probably just repeated several of your mistakes in attempting to answer, before realizing the issue, though I can't come up with a counterexample quickly. Could you say a little about what you have done to help others avoid the traps? – Elizabeth S. Q. Goodman Aug 24 '11 at 5:05
Thanks for your comment Elizabeth. Please see Torsten's answer to avoid all traps at once. – Michael Aug 24 '11 at 7:59

It is better to just assume that $f$ is a covering space. By shrinking $Z$ we may assume that that $Z$ is a ball and that $Y=Z\times F$. As $X\to Y$ is a covering space and $Z$ is simply-connected there is a covering space $X'\to F$ such that $X\to Y$ is isomorphic to $Z\times X'\to Z\times F$ which gives what you want.
Addendum: The reason that $X'$ exists is that if $h\colon T\to T'$ is a homotopy equivalence (and possibly $T$ and $T'$ fulfil some local niceness conditions which certainly are fulfilled in the case at hand), then pullback along $h$ induces an equivalence between the category of covering spaces of $T$ and that of $T'$. This is true irregardless on whether $T$ and $T'$ are connected or not.
What is $F\hspace{.04 in}$? – Ricky Demer Aug 24 '11 at 6:33
Thanks to Torsten's answer. The existence of $X'$ is especially enlightening. – Michael Aug 24 '11 at 7:55
As I tried to unravel Torsten's concise proof today, I found myself unable to prove the existence of $X'$. The difficulty lies in that one really needs to argue locally and there is no guarantee that the total space of the cover is connected, which needs to be true for the existence of $X'$. Could Torsten elaborate on why $X'$ must exist? Thanks. – Michael Aug 25 '11 at 6:32
Still don't understand. Torsten seems to be saying if $X \rightarrow Z \times F$ is a covering, then $\exists$ a covering $F' \rightarrow F$ such that $X \cong Z \times F'$. I can prove the existence of $F'$, but am unable to demonstrate a homeomorphism $X \cong Z \times F'$, unless $X$ is connected (so that covering space theory applies). – Michael Aug 26 '11 at 1:48