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Suppose that $p:E \to B$ and $q:B \to B'$ are fibrations. Is it true that $qp:E \to B'$ is a fibration?

I thought this might just be 'abstract-nonsense'. There is a diagram (possibly correct?) that looks like


which is basically just the two fibrations, but I can't seem to construct a map $H:X \times I \to E$


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1 Answer 1

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I don't quite see the problem: $X\times I$ can be lifted from $B'$ to $B$ (since $B\to B'$ is a fibration) and then from $B$ to $E$ (since $E\to B$ is a fibration).

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did you mean $E \to B$ in your last sentence? (The whole problem is to show $E \to B'$ is a fibration) –  Juan S Jun 15 '11 at 9:25
Actually I think I was just over-complicating this. In reality $\tilde{F}$ and $G$ are the same thing, so as you say, lift $X \times I$ to $B$ and then lift to $E$ –  Juan S Jun 15 '11 at 9:30
@Qwirk Yes, definitely (fixed the typo). It would be more clear with a diagram, but I'm afraid I can't draw one in MathJaX. (And yes, it means that F=G in your diagram.) –  Grigory M Jun 15 '11 at 9:30
thanks got it. Nothing ever seems easy past 10pm :) –  Juan S Jun 15 '11 at 9:33

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