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$\require{AMScd} \newcommand{\RP}{\mathbb{RP}}$ I am trying to show that for a given fibration $E \xrightarrow{p} B$, and a cell structure $\{B_n\}$ on $B$ that , that $p^{-1}B_n \to p^{-1}B_{n+1}$ is a cofibration. I am trying to deduce it from a more general fact(that I hope is true).

Question:

Let $E_1$ be the pullback of a fibration $E_2 \xrightarrow{p_2} B_2$ along a cofibration $j$:

$ \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?

Note: This question comes from my previous quetsion Partial Converse to "Pushout of a cofibration is a cofibration". I did not realize that I needed to further specify that the above square be a pullback square so I created this question.

Community
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user062295
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  • I edited my answer to your previous question. The answer is yes (but does not follow from formal arguments); see theorem 14.1 in Strom's Modern Classical Homotopy Theory. – JHF Jul 13 '16 at 21:00
  • i think, if $p^{-1}B_n \to p^{-1}B_{n+1}$ is a closed inclusion, it is a cofibration automatically – Andrey Ryabichev Jul 14 '16 at 13:36

1 Answers1

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Without loss of generality assume that $B_1 \to B_2$ is an inclusion(This is satisfied for the cofibration $B_p \to B_{p+1}$ that you are interested in). The only thing I need to use is $B_1 \to B_2$ is a cofibration $\iff$ $(B_2,B_1)$ an NDR Pair. Suppose that its given by the homotopy $h: B_2 \times I \to B_2$ and $u: B_2 \to I$.

For the obvious lift $E_2 \times 0 \to E_2$ $\require{AMScd} \newcommand{\RP}{\mathbb{RP}}$ the diagram $ \begin{CD} E_2 \times 0 @>>> E_2 \\ @VVV @Vp_2VV\\ E_2 \times I @>(h:B_2 \times I \to B_2)\circ p_2>> B_2 \\ \end{CD}$, has a solution $H: E_2 \times I \to E_2$. Its a 5 second mental exercise to check that $(E_2,E_1)$ is an NDR pair given explicitly by the homotopy $H$ and $u \circ p: E_2 \to I$. Therefore $E_1 \hookrightarrow E_2$ is a cofibration.

Hari Rau-Murthy
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  • What if $p_2$ is a Serre fibration? Do you need to assume $E_2$ to be a CW complex, or can you avoid it? – nrkm Dec 02 '17 at 08:21
  • I think you also need to prove that $H$ does not move $E_1$, but it doesn't seem to be true. – nrkm Dec 02 '17 at 08:29