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Is there a short five lemma for fibrations in algebraic topology (in whatever category where it would be suitable -- the topological category, the homotopy category, whatever).

By short five lemma I mean as follows. Let $E \to B$ is a fibration, with fiber $F$, and $E' \to B$ be another fibration with fiber $F'$. Suppose there are maps from $F \to F'$, $E \to E'$, and $B \to B'$ that all satisfy the obvious commutative diagram. Suppose all of the spaces are connected. If the maps from $F$ and $B$ are isomorphisms in the appropriate category, is the map from $E$ an isomorphism?

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2 Answers

up vote 2 down vote accepted

Associated to a fibration tere is always a long exact sequence in homotopy groups. So if all spaces have the homotopy type of CW complexes the usual 5 Lemma tells you that if two of the three maps are (weak) homotopy equivalences, then so is the third.

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Succinct, and obviously correct. I can't ask for much more from an answer. –  arsmath Dec 31 '12 at 20:38
Glad I could help :) –  mland Jan 1 '13 at 2:12
Strictly speaking, you need a bit more than the usual algebraic 5-lemma because $\pi_1$ can be non-abelian and $\pi_0$ is just a set. But it still works, with some care. –  Dan Ramras Jan 4 '13 at 7:42
@DanRamras : You are right. Sorry for not pointing out the subtleties concerning the low homotopy groups. –  mland Jan 4 '13 at 15:39
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What do you mean by the map between the fibres $F \to F'$ ? The projections $E \to B$ and $E' \to B'$ have lots of fibres. Do you mean that the map $\phi \colon E \to E'$ sends every fibre of $E \to B$ to a fibre of $E' \to B'$ and on each one is an isomorphism or do you just mean it sends one fibre in that way ?

If the former and you are interested in manifolds and isomorphisms are diffeomorphisms the result is true. A proof would be to notice that the map $E \to E'$ is a bijection because it is a bijection on fibres and the map on the base is a bijection. Hence there is a set-theoretic inverse. If $e \in E$ then the condition of being a diffeomorphism on the fibre through $e$ and on the base will be enough to show that the derivative of this map is a bijection. Hence the inverse function theorem shows there is a smooth local inverse which must coincide with the global set-theoretic inverse.

Not sure about the topological category with homeomorphisms as isomorphisms. Does that follow from mland's answer ? Is a weak homotopy inverse which is a bijection a homeomorphism ?

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Sorry, I probably should have put in more detail. I was thinking of spaces with a base point. The fiber there is just the inverse image of the base point. –  arsmath Jan 1 '13 at 13:07
OK I understand. Thanks. In that case I think you can do $E = E' = \mathbb{R}^2$, $B = B' = \mathbb{R}$ and $F = F' = \mathbb{R}$ with the projections to $B$ being the projection on the first factor and the basepoint being $1$. Define $E \to E'$ to be $(x, y) \mapsto (x, xy)$. This is identity map on the fibre and the base but not invertible. –  Michael Murray Jan 3 '13 at 1:45
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