# Composition of fibrations [closed]

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 $H:X \times I \to E$

Hints?

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## closed as unclear what you're asking by Stefan Hamcke, Normal Human, yoknapatawpha, Strants, graydadJul 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

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).
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