Example of an unstable map between finite complexes which is the identity on homotopy but not homotopic to the identity?

Question

Stably, phantom maps (nonzero maps which are zero on homotopy) exist, but it's not known if they exist between finite complexes (Freyd's Generating Hypothesis). Unstably, it's easy to find maps which are the same on homotopy but not homotopic (even between finite complexes), however I don't know an example of a map which is the identity on homotopy but not homotopic to the identity.

I presume I could take $\Omega^\infty(1+f)$ where $f$ is a known stable phantom map -- and I would be interested to see the details worked out -- but known stable phantom maps seem to hinge crucially on $\varprojlim^1$ issues, and so are "inherently infinitary". I'd like to see an unstable example which is not "inherently infinitary"; concretely it would be nice to see an example on a finite complex, but I'm not wedded to this interpretation of "not inherently infinitary". To sum up:

Question: What is an example of a self-map $f: X \to X$ of a (pointed, connected) CW complex which is not homotopic to the identity but such that $\pi_\ast(f): \pi_\ast(X) \to \pi_\ast(X)$ is the identity? Bonus points if $X$ is a finite complex.

An equivalent question is: what's an example of a space $X$ such that $Aut(X)$ doesn't act faithfully on $\pi_\ast(X)$, where $Aut(X)$ is the group of homotopy classes of self-homotopy-equivalences of $X$?

Another way of putting this is: Whitehead's theorem tells us that the functor $\pi_\ast$ reflects isomorphisms, but does it reflect identities, i.e. is it faithful with respect to isomorphisms?

For that matter, I'm having trouble coming up with a space $X$ such that $Aut(X)$ doesn't act faithfully on its homology, either.

Answer 1

Pick a degree $1$ map $h: T^3 \to S^3$ from the $3$-torus to the sphere and define $$f: T^3 \times S^3 \to T^3 \times S^3; \; f(x,y):=(x, yh(x)).$$ This map induces the identity on homotopy groups, but not on homology.

Answer 2

If $X = S^1 \times \Bbb{CP}^\infty$, then there is a map $X \to X$ with the following property. For any space $Y$, $[Y,X] \cong H^1(Y) \times H^2(Y)$, and so the map $X \to X$ classifies the map $$ (a,b) \mapsto (a, a^2 + b). $$ This is clearly not homotopic to the identity self-map of $X$ because it's not the identity natural transformation.

The fact that this is the identity on homotopy groups follows by considering the case $Y = S^n$: in both of those cases, the map $H^1(S^n) \times H^2(S^n)$ given by $(a,b) \mapsto (a,a^2 + b)$ is the same as the identity map $(a,b) \mapsto (a,b)$. [Yes, I neglected the basepoints and they are important but they don't alter this outcome.]

Unfortunately this isn't a finite complex. It also falls into the category of being "$1+f$" for some $f$ which is zero on homotopy, as you suggest, but $f$ in this case is not really a phantom map on its own -- it doesn't restrict to the zero map on the finite subcomplexes.

Answer 3

George Cooke gave an example in Trans, AMS 237 (1978) 391-406. Define a map $h\colon S^n \times S^n \vee S^{2n} \to S^n \times S^n \vee S^{2n}$ by $$ \begin{cases} h|_{S^n \vee S^n \vee S^{2n}} = \mathrm{id} \\ h|_{2n-cell} \ \mathrm{wraps\ non-trivially\ around}\ S^{2n}. \end{cases} $$ This induces the identity on homotopy, but is non-trivial on homology. This in fact gives a homotopy action of $\mathbb{Z}$ on the space and Cooke showed how this could be replaced by a true action on a homotopically equivalent space. As far as the question about $\mathrm{Aut}(X)$, the paper by Pak and me in Topology and its Applications 52 (1993) 11-22 might be relevant. We constructed examples where the representation from components of the homeomorphism group to $\mathrm{Aut}(\pi_*(X))$ is not faithful. This involved interactions between the notions of simple space, the Gottlieb group, principal bundles over tori and the $h$-rank of a space. It all went back to the old question of Gottlieb: is there a finite simple complex with non-trivial fundamental group, but trivial Gottlieb group.