Finding a counter example of the Whitehead theorem

Question

Let $X$ be a connected (based) CW-complex, $X$ is called simple if $\pi_1(X)$ acts trivially on $\pi_n(X)$ for all $n\geq 1$. The Whitehead theorem states that a self-map of a simple connected CW-complex $X$ is a homotopy equivalence if and only if the induced homological homomorphism $H_n(f)\colon H_n(X)\to H_n(X)$ is an automorphism for all $n\geq 1$.

Q: Is there a CW-complex $X$ satisfying the following conditions:

$(a)$ $X$ is not simple;

$(b)$ For any self-map $f\colon X\to X$, if the induced homological homomorphism $H_n(f)$ is an automorphism for all $n\geq 1$, then $f$ is a homotopy equivalence.

I take it you want the fundamental group to be abelian as well? Otherwise any $K(G,1)$ for $G$ non-abelian works. — Connor Malin, May 27 '19 at 08:18

score 2 · Answer 1 · answered May 27 '19 at 07:56

Let $M(\mathbb{Z}/q,1)$ be the pseudo-projective plane. By an article of P. Olum in 1965, $M(\mathbb{Z}/q,1)$ is not simple and there is a split short exact sequence: $$1\to K\rightarrow \mathcal{E}(M(\mathbb{Z}/q,1))\rightarrow Aut\pi_1\to 1,$$ where $\pi_1=\pi_1(M(\mathbb{Z}/q,1))\cong H_1(M(\mathbb{Z}/q,1))$. It follows the group of self-homotopy equivalences of $M(\mathbb{Z}/q,1)$ are exactly those inducing automorphism on $\pi_1\cong H_1$.

Finding a counter example of the Whitehead theorem

1 Answers1