Two maps inducing the same homotopy groups

Question

Let $X,Y$ be two CW-complexes and $f,g$ two continuous maps from $X$ to $Y$. Suppose that $f_*=g_*:\pi_q(X) \to \pi_q(Y)$ for any $q$, then can we prove that $f$ and $g$ are homotopic?

Answer 1

In addition to the answers given in Dan Rust's link, you can also note that Cohomology Operations are represented by continuous maps between Eilenberg-MacLane spaces $K(G,n) \to K(\pi,m)$ for varying $G,n,\pi, m$. In particular there are non-zero cohomology operations represented by maps where $m>n$, for example the Bockstein operation $\beta\colon K(\mathbb{Z}/2, n) \to K(\mathbb{Z}/2,n+1)$. For degree reasons any map between these CW complexes induces $0$ on all homotopy groups, but the map can't be null-homotopic because there are many spaces which support non-zero Bocksteins.

Answer 2

Here's a little twist: a pair of maps which induce the same on $\pi_*$ but different maps on $H_*$. Hence showing that they are not homotopic is exceedingly simple.

Take the spheres $S^m$, $S^n$ and form the product $S^m\times S^n$. The wedge is the subsapce $S^m\vee S^n=S^m\times e_1\cup e_1\times S^n$ and the smash product is the quotient $S^m\wedge S^n=S^m\times S^n/S^m\vee S^n$. Note that $S^m\wedge S^n\cong S^{m+n}$.

The two maps we will consider will be the pair of the constant map

$$c:S^m\times S^n\rightarrow S^m\wedge S^n,\qquad (x,y)\mapsto e_1\wedge e_1$$

and the quotient map

$$q:S^m\times S^n\rightarrow S^m\wedge S^n,\qquad (x,y)\mapsto x\wedge y.$$

Now clearly the constant map $c$ induces the zero morphism on both homotopy and homology. On the other hand $q$ most certainly does not induce the zero map on $H_*$. Indeed, the exact sequence of the cofibration $S^m\vee S^n\rightarrow S^m\times S^n\rightarrow S^m\wedge S^n$ shows that it induces an isomorphism

$$H_{m+n}(q):H_{m+n}(S^m\times S^n)\xrightarrow{\cong}H_{m+n}(S^{m+n}).$$

Now the map

$$\pi_r(S^m\times S^n)\rightarrow \pi_rS^m\oplus\pi_rS^n,\qquad \alpha\mapsto( pr_1\circ\alpha\oplus pr_2\circ\alpha)$$

is a natural isomorphism. In fact this happens whenever you map into the cartesian product of any pair of spaces. Therefore if $\alpha:S^r\rightarrow S^m\times S^n$ represents a class in $\pi_r(S^m\times S^n)$ it can be written as the composite

$$\alpha:S^r\xrightarrow{\Delta} S^r\times S^r\xrightarrow{\alpha_1\times \alpha_2}S^m\times S^n$$

where $\Delta$ is the diagonal map and $\alpha_i=pr_i\circ\alpha$. Then by inspection we have a strictly commutative diagram $\require{AMScd}$ \begin{CD} S^r\times S^r@>\alpha_1\times \alpha_2>> S^m\times S^n\\ @VV q'V @VV q V\\ S^r\wedge S^r @>\alpha_1\wedge \alpha_2>> S^m\wedge S^n \end{CD}

and since $S^r\wedge S^r\cong S^{2r}$ the composite

$$S^r\xrightarrow{\Delta}S^r\times S^r\xrightarrow{q'}S^{2r}$$

is null-homotopic.

Thus

$$q_*\alpha=0\in\pi_r(S^m\wedge S^n).$$

and we conclude that $q_*$ induces the zero map on all homotopy groups

$$\pi_*(q)=\pi_*(c)=0.$$

Answer 3

Here is a calculation in the spirit of William's answer that I think is accessible without knowing much about the cohomology of Eilenberg-MacLane spaces.

The question reduces to finding a nontrivial cohomology group $H^m(K(G,n),G')$ with $m \neq n$ because cohomology is represented by the Eilenberg-MacLane spaces, and, as William said, dimension reasons imply it is trivial on homotopy groups. You won't have any luck if $m<n$ since you can use the Hurewicz homomorphism and the universal coefficient theorem to deduce all of these are trivial. However, this process tells us $H^m(K(G,n),G')=Hom(G,G')$. For simplicity, pick $G=G'=\mathbb{Z}/2$, so this equals $\mathbb{Z}/2$. Then $H^{m+1}(K(\mathbb{Z}/2,n),\mathbb{Z}/2)=Hom(H_{m+1}(K(\mathbb{Z}/2,n)),\mathbb{Z}/2)\bigoplus Ext(\mathbb{Z}/2,\mathbb{Z}/2)$ which is nontrivial since $Ext(\mathbb{Z}/2,\mathbb{Z}/2)=\mathbb{Z}/2$. So nontrivial elements of this group correspond to nontrivial homotopy classes of maps that induce trivial maps on homotopy.