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Any help and hint on how to show that the composition and product of two fibrations are again fibrations. Thanks

David G. Stork
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Gomez
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1 Answers1

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Just verify the homotopy lifting property.

For example, if $p_1 : E \to F$ and $p_2 : F \to B$ are two fibrations, we wish to prove that the composition $p_2 \circ p_1 : E \to B$ is a fibration.

Given a map $f : X \times I \to B$, where $X$ is some other space and $I$ is the closed interval, and given a lift $\tilde f_0 : X \to E$ of $ f|_{X \times 0}$, we want to construct a lift $\tilde f : X \times I \to E$ of $f$ extending $\tilde f_0$.

First, $p_1 \circ \tilde f_0$ is certainly a lift of $f|_{X \times 0}$ to $F$. Since $p_2 : F \to B$ is a fibration, there exists a lift $f' : X \times I \to F$ of $f$ extending $p_1 \circ \tilde f_0$.

Next, $\tilde f_0$ is a lift of $p_1 \circ \tilde f_0 = f'|_{X \times 0}$ to $E$. Since $p_1 : E \to F$ is a fibration, there exists a lift $\tilde f : X \times I \to E$ of $f'$ extending $\tilde f_0$. Clearly, this lift $\tilde f$ is also a lift of $f$ itself. Thus, $p_2 \circ p_1 : E \to B$ obeys the homotopy lifting property, so it is a fibration, and we're done.

Kenny Wong
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  • I have a quick question. How is the product of two fibrations defined? Do you have the proof of how the product of two fibrations is again a fibration? – Jaynot Mar 15 '17 at 22:25
  • Hi Jaynot, I presume the product looks like $(p_1, p_2) : (E_1 \times E_2) \to (B_1 \times B_2)$? If this is the case, then the proof should be not too dissimilar... – Kenny Wong Mar 15 '17 at 22:26
  • Thanks Kenny. You always provide answers to my questions on Lie Algebra :) – Jaynot Mar 15 '17 at 22:29
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    By the way, for products, it might help to know that specifying a continuous map $f : X \to Y \times Z$ is equivalent to specifying a pair of continuous maps $ \pi_1 \circ f: X \to Y$ and $\pi_2 \circ f : X \to Z$, where $\pi_1 : Y \times Z \to Y$ and $\pi_2 : Y \times Z \to Z$ are the canonical projections... – Kenny Wong Mar 15 '17 at 22:31
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    Thank you @KennyWong. That was really helpful. I will sit down and figure out the proof for the product. – Gomez Mar 15 '17 at 22:37
  • @KennyWong I posted a question on fibration and I dont seem to understand clearly the answer that was provided. Do you have time to have a look? http://math.stackexchange.com/questions/2188116/question-on-fibration – Jaynot Mar 15 '17 at 22:58
  • @Jaynot Gosh, I don't even know how some of those maps are defined, let alone what "weak cofibrations" or "Quillen models" are. So sorry I can't help! – Kenny Wong Mar 15 '17 at 23:01
  • @KennyWong. Alright no problem. Thanks for even taking a look. – Jaynot Mar 15 '17 at 23:04
  • @KennyWong I cant seem to see how the specified maps help me figure out the proof of the product. How do I end up with maps that gives $E_1 \times E_2 \rightarrow B_1 \times B_2$. My guess is that I need to use the fact that projections are fibrations and composition of fibrations to show that indeed the product is a fibration. But the maps won't come clear (:. We proved in class that for any arbitrary spaces F and B that $\pi: F \times B \rightarrow B$ is projection. I just need to see how the maps you prescribed will yield the desired products. – Gomez Mar 16 '17 at 08:21
  • @KennyWong I apologize for bothering too much. (: – Gomez Mar 16 '17 at 08:23
  • @Gomez, to put my suggestion into plain English: Define the lift "componentwise". – Kenny Wong Mar 16 '17 at 09:08
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    We are given a map $F : X \times I \to B_1 \times B_2$. Write this as $F(x,s) = (F_{B_1}(x,s), F_{B_2}(x,s))$. We are also given a lift $\tilde F_0(x) $ of $F(x,0)$. Write this lift as $\tilde F_0(x) = ((F_0){E_1}(x), (F_0){E_2}(x))$. By the homotopy lifting criterion for $E_1 \to B_1$ and $E_2 \to B_2$, we can extend $(F_0){E_1}$ to a lift $F{E_1}$ of $F_{B_1}$ and we can extend $(F_0){E_2}$ to a lift $F{E_2}$ of $F_{B_2}$. This gives us a lift $ (F_{E_1}(x,s),F_{E_2}(x,s))$ of $(F_{B_1}(x,s), F_{B_2}(x,s))$ extending $ ((F_0){E_1}(x), (F_0){E_2}(x))$. – Kenny Wong Mar 16 '17 at 09:14
  • I think in the fourth line the codomain of $f$ should be $B,$ am I right? – Intuition Mar 05 '23 at 05:07
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    @Secretly Yes you're right! Sorry about the mistake. – Kenny Wong Mar 05 '23 at 09:33