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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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closed as unclear what you're asking by Stefan Hamcke, Normal Human, yoknapatawpha, Strants, graydad Jul 23 at 2:11

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

The "diagram" is currently a silhouette of a frog. It needs to be updated if the question is to make any sense. –  Joel Reyes Noche Jul 23 at 1:19

1 Answer 1

up vote 1 down vote accepted

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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