[ I asked the same question on stackexchange but attracted little attention. Besides, I made some progress after I posted it. So I decided to move it here. ]

Consider path-connected CW-complexes $A$, $B$, $C$, $X$. I would like to show the equivalence of the following two conditions.
(i) They fit into fibrations $A\to X\to Y_1$ and $B\to Y_1\to C$ for some $Y_1$.
(ii) They fit into fibrations $A\to Y_2\to B$ and $Y_2\to X\to C$ for some $Y_2$.
Everything is taken up to homotopy equivalence.

I can prove (i) $\Rightarrow$ (ii) as follows. It's not hard to show that we can always find two fibrations $X\stackrel{f}{\to}Y_1$ and $Y_1\stackrel{g}{\to}C$ that are homotopy equivalent to those fibrations in (i) but the two $Y_1$'s here are precisely the same space. Here $X$, $Y_1$, and $C$ are homotopy equivalent to but perhaps different from those spaces above, although I just abuse the same symbols. Then $X\stackrel{h=gf}{\longrightarrow}C$ is also a fibration. Let's pick up an arbitrary $c\in C$. Then $h^{-1}c\stackrel{f}{\to}g^{-1}c$ is clearly again a fibration, whose fiber is the same as a fiber of $X\stackrel{f}{\to}Y_1$ thus homotopy equivalent to $A$. We also note that $g^{-1}c$ is a fiber of $Y_1\stackrel{g}{\to}C$ thus homotopy equivalent to $B$. Therefore, letting $Y_2=h^{-1}c$ proves (ii).

But I found difficulties in proving (ii) $\Rightarrow$ (i). My strategy is to rephrase (ii) in another form similar to (i), i.e.,
(ii) They fit into fibrations $\Omega B\to A\to Y_2$ and $\Omega C\to Y_2\to X$ for some $Y_2$.
Then according to (i) $\Rightarrow$ (ii), we can prove (ii) $\Rightarrow$ (iii) where (iii) is the following.
(iii) They fit into fibrations $\Omega B\to Y_3\to \Omega C$ and $Y_3\to A\to X$ for some $Y_3$.
If $Y_3$ is deloopable, letting $Y_1=BY_3$ proves (i). But I don't know how to argue $Y_3$'s deloopability. I also tried to construct counterexamples, but I still failed. During those trials, I strongly felt that (ii) prohibits me to destroy $Y_3$'s deloopability.

Question: Could anyone help me complete my proof?


2 Answers 2


You are right: they are not equivalent. For an example, choose a group $G$ that has a normal subgroup $H$ in which there is a subgroup $K$ that is normal in $H$ but not in $G$. Take $A$, $B$, and $C$ to be $BK$, $B(H/K)$, and $B(G/H)$. (So $Y_2$ can be $BH$, but what is $Y_1$ going to be? There is no $G/K$.)

  • $\begingroup$ Do you think we can strengthen (ii) in some way (principal fibration, simply-connected spaces, or so on) so that we can prove (ii) $\Rightarrow$ (i)? $\endgroup$
    – Leo
    Mar 30 at 4:52
  • $\begingroup$ Here is another kind of counterexample, with spaces as highly connected as you like. The spaces are rational. $C=S^{2n}_{\mathbb Q}$ is the rational $2n$-sphere. $X$ is contractible. So $Y_2$ is $\Omega C$, which is the product of Eilenberg-Maclane spaces $A\times B$ with $A=K(2n-1,\mathbb Q)$ and $B=K(4n-2,\mathbb Q)$. In the desired fibration sequence $B\to Y_1\to C$, the space $Y_1$ would have to be $K(2n,\mathbb Q)$ (a delooping of $A$); but there is no map $K(2n,\mathbb Q)\to S^{2n}_{\mathbb Q}$ nontrivial on $\pi_{2n}$. $\endgroup$ Apr 1 at 2:17

As Tom Goodwillie has pointed out, your second statement does not imply the first. The best one can generally do is find a (homotopy) fibration sequences of the form $$ Y_3 \rightarrow \Omega C \rightarrow B \text{ and } Y_3 \rightarrow A \rightarrow X,$$ which is a bit stronger than your (iii).

The first bit of what you say can be found in texts like G. Whitehead's Elements of Homotopy Theory, or May's A Concise Course in Algebraic Topology: given $f: X \rightarrow Y_1$ and $g: Y_1 \rightarrow C$, the fibers of $f$, $g \circ f$, and $g$ fit together to form a fibration sequence.

  • $\begingroup$ BTW, do you think we can strengthen (ii) in some way (principal fibration, simply-connected spaces, or so on) so that we can prove (ii) $\Rightarrow$ (i)? $\endgroup$
    – Leo
    Mar 30 at 4:42
  • $\begingroup$ I'm interested in your stronger result. How do you show it? (What a pity I can award only one best answer.) $\endgroup$
    – Leo
    Mar 30 at 15:30
  • 1
    $\begingroup$ @Leo The fiber sequence $A \rightarrow Y_2 \rightarrow B$ maps to the fiber sequence $X = X \rightarrow pt$. Taking fibers gives a fiber sequence $Y_3 \rightarrow \Omega C \rightarrow B$. $\endgroup$ Mar 30 at 17:21

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