Consider a fiber sequence of spaces


$$F \overset{i}{\to} E \to B$$


The cofiber $C(i)$ of the inclusion of the fiber comes with a canonical map $C(i) \to B$. Its possible to show (using some point set topology) that the fiber of that map is homotopy equivalent to $\Omega B \ast F$ hence we obtain a new fiber sequence


$$\Omega B \ast F \to C(i) \to B$$


To demonstrate the power of this statement consider the case of the path space fibration where $B= X , E = PX \cong \ast , F = \Omega X$. The new fiber sequence we obtain would be


$$\Sigma (\Omega X \wedge \Omega X) \cong \Omega X \ast \Omega X \to \Sigma \Omega X \to X$$


This implies immediately that whenever $X$ is $n$-connective for $n\ge 1$ then the canonical map $\Sigma \Omega X \to X$ is ($2n-1$)-connective (which is sort of the Eckman Hilton dual of Freudenthal suspension theorem). 



> 
> **Question:** Is there an Eckman-Hilton dual to this statement?
> 
> 
> 


To be precise, let $X \to Y \overset{\pi}{\to} Z$ be a cofiber sequence and let $F := fib(\pi)$ be the fiber $\pi$. Again there's a canonical (upto homotopy) map $f: X \to F$ and we may take its cofiber to obtain a new cofiber sequence


$$X \to F \to C(f)$$


Is there a "closed formula" for $C(f)$ (and for the maps involved) in terms of $X,Y$ and $Z$ (and the maps) similar to the one from above?



The (fiber of the) cofiber of the fiber of a map of spaces
This sort of question was studied a lot by Ganea (see [here](https://link.springer.com/article/10.1007%2FBF02566956) and [here](https://link.springer.com/article/10.1007%2FBF02564393)). The first piece of bad news is that duality is not perfect: there is no formula for that cofiber that depends only on $\Sigma X$ and $Z$. A counterexample was supplied by Barratt (as Remark 3.5 in the first linked paper above).


Even worse, the homotopy type of that cofiber does not have a formula just in terms of $X$ and $Z$! A counterexample is supplied in the second linked paper as 1.4.


The best we can do is that the *suspension* of the cofiber is given by $\Omega Z \ast X$.


This observation leads to a nice proof of the James splitting, incidentally- with no calculation.


