# Iterating the suspension-loop adjunction in two different ways

Let $X$ be a sufficiently nice topological space (i.e. an object of a category of spaces where the reduced suspension-loops, $(\Sigma, \Omega)$, holds.)

There are two directed systems of spaces one can consider:

1) $X\to \Omega\Sigma X \to \Omega^2\Sigma^2X \to \cdots$

Here the first arrow is the unit $\eta_X$ of the adjunction, the second is is $\Omega \eta_{\Sigma X}$, etc. We get that the long composition $X\to \Omega^n\Sigma^nX$ is the unit of the adjunction $(\Sigma^n,\Omega^n)$ obtained by composing $n$ times $(\Sigma, \Omega)$ with itself.

2) $X\to \Omega \Sigma X \to \Omega \Sigma \Omega \Sigma X \to \cdots$

Here each arrow is a unit of the adjunction $(\Omega, \Sigma)$, applied to the spaces $X, \Omega\Sigma X, \dots$.

Now, the category of spaces is cocomplete, so we may take colimits. The first one yields the well-known space $QX$, whose homotopy groups are the stable homotopy groups of $X$.

Question: Is there an interesting description of the colimit of the second system? Or of its homotopy groups?

I have asked a question regarding a generalization of these systems here.

• If any reader feels like the bonus paragraph deserves to be asked as a separate question because there are many interesting things to be said with respect to that, tell me and I will. – Bruno Stonek Oct 13 '15 at 13:54
• I think the second question (the more general one) deserves a post of its own, possibly linking back to this question as motivation. (In general it's inadvisable to ask two different questions in one post). I'm interested to the answers to both, anyway. – Najib Idrissi Oct 13 '15 at 13:58
• @NajibIdrissi: done, it's here – Bruno Stonek Oct 13 '15 at 14:11
• I just realized something one might say about it. If $X$ is connected, then $\Sigma \Omega \Sigma X \simeq \bigvee_n \Sigma X^{\vee n}$. Now iterate. (Hatcher proposition 4.I.2 + 4.J.1) Yikes, ok, this doesn't concatenate correctly, but I leave this here since it might be useful. – Bruno Stonek Oct 13 '15 at 14:19
• You're basically completing along the free $E_1$-algebra monad. I'm don't know much about what kind of object you get though. – Jonathan Beardsley Oct 13 '15 at 14:28