MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

That is, if $f: X \rightarrow Y$ and $g:Y \rightarrow Z$ are bundle projections, is $g \circ f: X \rightarrow Z$ a bundle projection? Assume $X$, $Y$ and $Z$ are manifolds.

Here is what I know. The answer is affirmative when (1) $f$ is a covering map and $g$ is bundle projection; (2) $f$ is a bundle projection and $g$ is a covering map of finite degree. What can we say about the most general situation?


share|cite|improve this question
Yes. It follows directly from the definition. – Charlie Frohman Sep 5 '11 at 20:08
To Charlie: Just to make sure we have the same definition in mind, a bundle projection requires a cover of the base space by locally trivial neighborhoods and a typical fiber. Using this definition, it is not obvious to me why the composition should be a bundle projection. Could you elaborate? – Michael Sep 5 '11 at 20:59
@Charlie: it doesn't follow from the definition nor from anything else because it is false; see my answer. – Georges Elencwajg Sep 6 '11 at 0:16
I stand corrected. – Charlie Frohman Sep 6 '11 at 0:40
What about when the spaces involved are locally contractible? – David Roberts Sep 6 '11 at 7:36

No, the composition of two bundle projections needn't be a bundle projection.
It is already not true that the composition of two covering maps is a covering map.
You can find a counterexample in Spanier's classic Algebraic Topology, Chapter 2, §2, Example 8, page 69.

Another counterexample, based (!) on the notorious Hawaiian ring space is given by Jerzac's very nice,detailed paper here.

However, on the positive side, the composition $g\circ f :X\to Z$ of the covering maps $f:X\to Y$ and $g:Y\to Z$ is a covering map in each of the following two cases:
a) The covering $g$ is finite (= has finite fibers)
b) The space $Z$ has a universal covering (for connected $Z$, this means that $Z$ is locally pathwise connected and semi-locally simply connected). For example, CW-complexes have a universal covering space, since they are even locally contractible.

share|cite|improve this answer
Georges -- that's a good example. However, I believe for reasonably nice spaces the statement should be true. I'm not sure what exactly the "reasonably nice" condition is, but I guess it would suffice to require that the 1-point compactifications of all spaces should be finite CW-complexes. – algori Sep 6 '11 at 16:57
Thanks to Georges' for your answer. Would the answer be positive if we assume $X$, $Y$ and $Z$ are manifolds, which is really what I meant to ask but forgot to mention? – Michael Sep 7 '11 at 5:15
Dear Michael: yes, as shown by Torsten's and Ben's fine answers. – Georges Elencwajg Sep 8 '11 at 0:51

I think that the following works: Let $X\to Y$ and $Y\to Z$ be locally trivial fibrations with all spaces paracompact and $Z$ locally contractible (I do not assume that a fibration implies that all fibres are homeo- or diffeomorphic). We want to show that $X\to Z$ is locally trivial. We may then reduce to the case when $Z$ is contractible and $Y=Z\times F$. Under the paracompactness assumption locally trivial fibrations are homotopy invariant (see for instance Husemoller: Fibre bundles GTM 20, Springer, Thm 9.8 plus a simple reduction to the principal bundle case). Hence there is a locally trivial fibre bundle $X'\to F$ such that $X\to Y$ is isomorphic to $Z\times X'$ which is the local trivialtity.

share|cite|improve this answer
Dear Torsten, does your parenthetical remark "(I do not assume...)" just mean that fibers depend on the connected component of the basis? And do you mean Thm 9.8 of Chapter 4 ? – Georges Elencwajg Sep 7 '11 at 11:24
Dear Georges, yes to both questions. – Torsten Ekedahl Sep 7 '11 at 14:59

In the category of finite dimensional manifolds, I proved this in a very long paper (p. 23): the composition of smooth fiber bundle maps is a smooth fiber bundle map. Spanier's counterexample is not a finite dimensional manifold.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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