# The (fiber of the) cofiber of the fiber of a map of spaces

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?

• IS there a reason to expect it to be a loop space?! What is the case with the cofibre sequences $X\to *\to \Sigma X$. So, I thought something like $\Omega Map_*(Z,\Sigma X)$ might be a dual object to expect to appear as the cofibre of $X\to F$ ?!? – user51223 May 4 at 14:49

## 1 Answer

This sort of question was studied a lot by Ganea (see here and here). 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.

• Thanks! What a bummer though... :( – Saal Hardali May 7 at 11:58