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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

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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
ImageShack seems to have deleted your image and replaced it with an ad banner. If you can, please reupload the image (or something equivalent) using the image upload button in the editor toolbar (which will upload it to Stack Exchange's imgur account). –  Ilmari Karonen Aug 17 at 18:33
@JoelReyesNoche It's no longer a frog; now it's an ad... –  Normal Human Aug 17 at 23:22

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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