[ 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. ]




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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.




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**Question:** Could anyone help me complete my proof?



Associativity of consecutive fibrations
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 $G/K$.)


