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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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    Oh, to be clear, my comment assumes $X$ is a CW complex.2012-06-16

2 Answers 2

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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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    Tha$n$k you, this is a good a$n$swer as for the u$n$it.2012-06-13
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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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    Dan: Yeah, you are right since we are requiring the base point be the identity.2012-06-18