# Is a map a homotopy equivalence if its suspension is so?

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.

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 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

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 $f_*$.