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I am trying to figure out some commutative diagrams and am having difficulty with this one. We have two fibrations of $B$, $E'$ and $E$ such that $E' \subset E$ and $E' \to E$ is a cofibration. $B$, $E'$ and $E$ are all CW complexes. I want to show that $E/E'$ is a fibration of $B$. I have a bunch of commutative diagrams on this but have no way of inputting them.

Thank you in advance.

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2  
Is this even true? Consider, for example, the case $E = E'$. –  Dylan Wilson Aug 22 '11 at 5:22
5  
Do you mean the fiberwise quotient, which is the pushout of the diagram $B \leftarrow E' \rightarrow E$? Otherwise, there is not a map to $B$. –  Tyler Lawson Aug 22 '11 at 5:25
    
BTW, there is a meta article on how to do diagrams. Sadly, it boils down to some pretty messy latex work, but it may be useful for you to know about: tea.mathoverflow.net/discussion/871/… –  David White Aug 22 '11 at 12:44

1 Answer 1

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There seem to be two issues you're addressing. One is "what's the right notion of quotient" and the comments show that it should really be the fiberwise quotient (i.e. pushout of $B\leftarrow E'\rightarrow E$), which I'll denote $\mathcal{E}$. Under some hypotheses, $\mathcal{E}$ has the property that the fiber space over $b\in B$ is $F_b'/F_b$ where $F_b'$ is the fiber over $b$ of the $E'$ fibration and $F_b$ for the $E$ fibration. See e.g. link

The next question is "when is $\mathcal{E}$ a Serre fibration?" A similar question was asked here for simplicial sets. Turns out the pushout is a Serre fibration when one of the two maps is a cofibration (by the way, a pullback of two fibrations is always a fibration, in any model category). I suspect the same will hold in your setup of CW complexes, i.e. $\mathcal{E}$ should be a fibration because $E'\rightarrow E$ is a cofibration. A fact which might help is that cofibrations of CW complexes are really just retracts. The last link probably answers your question if you dig around inside. I'll edit this if I find an exact reference or work out the details of the proof.

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