$\Omega$ of a homotopy cofiber sequence

Question

What is an example of a homotopy cofiber sequence $$ X\to Y\to Z $$ of well-pointed connected CW-complexes such that the associated sequence of loop spaces $$ \Omega X\to \Omega Y\to\Omega Z $$ is not a homotopy cofiber sequence? Are there ''reasonable'' conditions (such as ''simply connected'', etc.) on $X$, $Y$ and $Z$ such that this holds true?

Answer 1

The archetypal example of a homotopy cofiber sequence is a suspension

$$X \to \bullet \to \Sigma X.$$

Taking loops on such a thing gives

$$\Omega X \to \bullet \to \Omega \Sigma X$$

and asking for this to continue to be a homotopy cofiber sequence is equivalent to asking that $\Omega \Sigma X \cong \Sigma \Omega X$; in other words, you're asking for taking loops and taking suspensions to commute (up to homotopy, etc.).

This is almost never true. In particular, $\Omega \Sigma X$ is almost never a suspension. Let's take $n \ge 2$ and $X = S^n$ so that $\Sigma X = S^{n+1}$, which is about as nice as possible and in particular can have arbitrarily high connectivity.

Then $\Omega \Sigma X \cong \Omega S^{n+1}$ turns out to have rational cohomology a polynomial algebra on a generator of degree $n$; in particular it supports nontrivial cup products. But $\Sigma \Omega X \cong \Sigma \Omega S^n$ is a suspension, so the cup product on it vanishes in positive degree.

If you're working with stable objects, e.g. chain complexes or spectra, then homotopy cofiber and fiber sequences agree, and in particular taking loops and taking suspensions are inverses and so $\Omega \Sigma X \cong \Sigma \Omega X \cong X$. In spaces there are theorems such as the Blakers-Massey theorem measuring the extent to which a homotopy cofiber sequence fails to be a homotopy fiber sequence and vice versa.

Answer 2

This is not a complete answer because I can't answer on conditions you ask for, but it's a beginning. You might already know that usually loops is not going to play well with cofiber sequence, homology/cohomology, etc. Loops play well with homotopy, fiber sequences, etc.

A cofiber sequence induces long exact sequences in (co)homology, and in general after looping it once, you can prove it's not a cofiber sequence anymore by showing that you don't have a long exact sequence in homology. One of the easiest example I can think of would be $S^1 \to CS^1 \to S^2$, if you loop it once you get something homotopy equivalent to $\mathbb{Z} \to \ast \to \Omega S^2$ and in homology you don't get a long exact sequence (since $\Omega S^2$ has infinitely many cells and infinite homology). Even that is not a very good example because I cheat by saying that $\Omega S^2$ has infinitely many homology. This is generic to what happens, when you take loops of a space $X$, that is, functions from $S^1$ into your space $X$, this become an "infinite-dimensional thing", it's huge and you loose all control on its homology.

Another way to look at this question is via the Homotopy Excision Thm. This theorem tells you how much a cofiber sequence is not a fiber sequence, or how much the cofiber is not the same as the fiber. A really nice exposition is here http://math.mit.edu/~mbehrens/18.906/notes/lec12.pdf

A criterion for when this would be true is if you work stably, but there it's just tautologically true.