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For (nice?) pointed spaces, the reduced suspension $\Sigma$ is left adjoint to the loop space $\Omega$. This adjunction is given by the unit maps

$\eta_X : X \to \Omega \Sigma X$, $x \mapsto (t \mapsto [x,t])$

and the counit maps

$\varepsilon_X : \Sigma \Omega X \to X , [\omega,t] \mapsto \omega(t).$

Question. For which $X$ is $\eta_X$ a homotopy equivalence? For which $X$ is $\varepsilon_X$ a homotopy equivalence?

If this is useful, let's assume that $X$ is sufficiently nice (for example a CW complex). You may also replace "homotopy equivalence" by "homology equivalence" etc., if this yields to interesting statements. If there is no characterization: What are interesting classes of examples? And is there any source in the literature where this sort of question is studied?

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Have you checked Hatcher chapter 4.J? – Simon Markett Jun 13 '12 at 9:03
My guess is - not very often. Try it with the E-M spaces for example. If $X$ if a $K(\mathbb{Z},1)$ then $\Sigma \Omega X = \Sigma \mathbb{C} P^\infty$. Loop spaces are not nice. You probably already know this, but for a topological group $G$ there is a homotopy equivalence $\Omega BG \sim G$ – Juan S Jun 13 '12 at 9:45
I bet that $X\rightarrow \Omega \Sigma X$ is a homotopy equivalence if and only if $X$ is contractible. Certainly such $X$ are simply connected and Sean's point about the tensor algebra suggests they should have trivial homology as well. I didn't think through the details, though. – Dan Ramras Jun 16 '12 at 18:10
Oh, to be clear, my comment assumes $X$ is a CW complex. – Dan Ramras Jun 16 '12 at 18:11
up vote 7 down vote accepted

Hatcher Theorem 4J.1:

The map $J(X)\to \Omega \Sigma X$ is a weak homotopy equivalence for every connected CW complex $X$.

Here $J(X)$ is James' reduced product of $X.

Hatcher Proposition 3.22:

For $n > 0$, $H^*(J(S^n);\mathbb Z)$ consists of a $\mathbb Z$ in each dimension a multiple of $n$.

So your question fails already for spheres.

Edit: Corollary 4.J.3 also tells you how close you can get. The map $X\to \Omega\Sigma X $ factors through $J(X)$ and $(J(X),X)$ is $2n+1$ connected.

Edit2: And in this paper the authors show that the homotopy fibre of the map $S\Omega X\to X$ has the homotopy type of $\Omega X\ast\Omega X$, the join of $\Omega X$ with itself.

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Thank you, this is a good answer as for the unit. – Martin Brandenburg Jun 13 '12 at 16:32

You can build on Simon's answer though. $J(X)$ has a great definition, you should check it out. Then it may be easy to see when it could be true or why it must always fail. (can an algebra ever be isomorhpic as an algebra to the tensor algebra generated by it?)

Another suggestion is that you might try to answer the homology equivalence by yourself using the Serre spectral sequence (using the path-loop space fibration). I bet this will pair nicely with the tensor algebra question.

Good question, but I bet it will never be the case in either situation. Don't be sad though, how often is the unit or counit of the free forgetful adjunction an isomorphism?

Although, there are other adjunctions that behave much better, such as Dold-Kan and the adjunction between $sSets$ and $Top$.

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Thanks. I hoped that this is not just a free-forgetful adjunction. I'm familiar with the two better behaved adjunctions you mention. – Martin Brandenburg Jun 13 '12 at 19:40
James result seems to imply that it is very close to a free forgetful adjunction. I tend to try and think of all adjunctions as being something like a free-forgetful/tensor-hom adjunctions. It is curious to me that both that I mentioned are tensor-hom adjunctions that behave well and your example is one that doesn't... – Sean Tilson Jun 13 '12 at 22:54
J(X) is often called the "free topological monoid generated by $X$". – Dan Ramras Jun 16 '12 at 18:06
Sean: I don't think you mean "never". I bet that $X\rightarrow \Omega \Sigma X$ is a homotopy equivalence if and only if $X$ is contractible. Certainly such $X$ are simply connected and your point about the tensor algebra suggests they should have trivial homology as well. – Dan Ramras Jun 16 '12 at 18:09
Dan: Yeah, you are right since we are requiring the base point be the identity. – Sean Tilson Jun 18 '12 at 17:21

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