Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $X$, $Y$ be pointed CW complexes, $Y$ connected and $f:X\to Y$ a mapping.

Does the assertion '$\Sigma f:\Sigma X\to\Sigma Y$ is a homotopy equivalence' imply that $f$ is a homotopy equivalence? '$\Sigma$' is the reduced suspension.

If not, is it true with some additional hypotheses on $Y$?

Addition: Does the assertion '$\Omega f:\Omega X\to\Omega Y$ is a homotopy equivalence' imply that $f$ is a homotopy equivalence? '$\Omega$' is the loopspace.

share|improve this question
    
I like this. I think the question can be rephrased as, "Does homotopy the category of topological spaces inject (via the functor $\Sigma^\infty$) into the homotopy category of spectra?" ("Inject" might be the wrong word here, I don't know much category theory.) I'd like the answer to be yes, of course... –  Aaron Mazel-Gee Nov 21 '10 at 19:42
    
Okay. So it's almost true. I guess the fundamental group isn't really a stable notion anyways... –  Aaron Mazel-Gee Nov 23 '10 at 10:33
add comment

1 Answer

up vote 9 down vote accepted

I believe the answer to your first question is no. Let $X$ be any connected acyclic CW-complex with non-trivial fundamental group, for example the space constructed as example 2.38 in Hatcher. Such a space has the property that $H_i(X) = 0$ for $i>0$ and $H_0(X) = \mathbb{Z}$, but $\pi_1(X) \neq 0$. (In particular $\pi_1(X)$ must be perfect). Consider the projection map $f: X \to pt$. By looking $\pi_1$, $f$ cannot be a homotopy equivalence.

However, $\Sigma f: \Sigma X \to \Sigma pt$ is a homotopy equivalence. To see this, note that suspension increases the connectivity, which implies that both spaces are simply connected. Hence the homology Whitehead theorem applies, which says that a map between simply connected CW-complexes is a homotopy equivalence if and only if it induces isomorphisms on all homology groups. Using the suspension axiom in homology, we see that all $H_i(\Sigma X)$ and $H_i(\Sigma pt)$ for all $i>0$ are zero and for $i=0$ are $\mathbb{Z}$. It is then easy to check that $\Sigma f$ is an isomorphism in all degrees.

edit: in your addition, I think the answer is yes, if we replace homotopy equivalence by weak homotopy equivalence. The Whitehead theorem says that $f: X \to Y$ is homotopy equivalence if and only if induces an isomorphism on all $\pi_i$. Because $\Omega X$ and $\Omega Y$ have the homotopy type of CW-complexes, we can replace them by CW-complexes with the price of replacing homotopy equivalence with weak homotopy equivalence. Now note that $Map_+(S^n,\Omega X) \cong Map_+(S^{n+1},X)$ and similarly $Map_+(S^n,\Omega Y) \cong Map_+(S^{n+1},Y)$. Under this isomorphism $(\Omega f)_*$ corresponds to $f_*$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.