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$\require{AMScd} \newcommand{\RP}{\mathbb{RP}}$ I want a converse of this fact specialized to the case where I am pushing out a map BY a fibration:

I.e., if I am given a diagram

$ \begin{CD} E_1 @>i>> E_2 \\ @V\text{fibration}Vp_1V @Vp_2V\text{fibration}V\\ B_1 @>cofibration>j> B_2 \\ \end{CD}$,

does it follow that $i$ is a cofibration?

I am doing this to prove that the inclusion $F_p E \to F_{p+1} E$ is a cofibration,for the filtration used in the serre spectral sequence.

Edit

When I made this question, I had the misconception that every pushout square was also a pullback square(I will conveniently blame it on a student lecturer in my class who said "this pushout square" when he was referring to a square that was just a pullback square.). Thus I did not realize that I needed to further specify that the above square be a pullback square. I am embarrassed to admit it. But I created a new question with this new condition.

user062295
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  • This claim is plausible since $E_1 \to E_2$ is also an inclusion - a necessary condition for a map to be a cofibration. I am looking again through the proof that the pullback of a fibration is a fibration, to see if I can use it. After all the proof that the pullback of a cofibration is a cofibration is a dualized argument of this, and when I have both of these objects, somehow I might be able to use something.... Thats all I know for now. Come back later if I think of something else. – user062295 Jul 13 '16 at 12:30

1 Answers1

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In full generality, it's not true, even if $E_1 \to E_2$ is an inclusion. Take $E_1 \to E_2$ to be your favorite inclusion that's not a cofibration and let $B_1 = B_2 = *$. Then $E_i \to B_i= *$, $i = 1, 2$, are fibrations (every space is fibrant), the identity $B_1 \to B_2$ is a cofibration, and the diagram commutes, but $E_1 \to E_2$ is not a cofibration by construction.

However, if $E_1$ and $E_2$ are CW complexes, then it's probably true, since every (cellular) inclusion of CW complexes is a cofibration.

EDIT: The example I gave above ignores the "pushout" condition. If we assume (per OP's comment) that the square is a pullback, then the answer seems to be affirmative. See theorem 14.1 in Strom's Modern Classical Homotopy Theory.

JHF
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  • Thanks. What if I add the requirement that $E_1 \to B_1$ is the pullback fibration of $E_2 \to B_2$? (this counterexample won't work since the identity map is always a cofibration). – user062295 Jul 13 '16 at 16:29
  • Thank you so much for helping me with this. You are the reason this world is a great place. – user062295 Jul 14 '16 at 02:25