# Associativity of consecutive fibrations

[ 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?

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 $$G/K$$.)

• 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)?
– Leo
Mar 30 at 4:52
• 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 $\pi_{2n}$. 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.

• 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)?
– Leo
Mar 30 at 4:42
• I'm interested in your stronger result. How do you show it? (What a pity I can award only one best answer.)
– Leo
Mar 30 at 15:30
• @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 $Y_3 \rightarrow \Omega C \rightarrow B$. Mar 30 at 17:21