Whitehead theorem for maps between CW-complexes

Question

I was wondering : if $f,g : X \rightarrow Y$ are continuous maps between CW-complexes, if they induce the same morphisms on homotopy groups, does that imply that $f$ and $g$ are homotopic? It would seem plausible, similar to a Whitehead theorem.

Answer 1

This is not true.

Take $S^n\times S^n$ and consider the quotient map

$$q_n:S^n\times S^n\rightarrow S^n\wedge S^n\cong S^{2n}$$

where $S^n\wedge S^n=S^n\times S^n/S^n\vee S^n$ is the smash product. Then $q_n$ induces the trivial morphism on homotopy groups in all dimensions, since if $\alpha:S^r\rightarrow S^n\times S^n$ is any map, then by naturality $q_{n*}\alpha$ factors as

$q\circ\alpha:S^r\xrightarrow\Delta S^r\times S^r\xrightarrow{q_{r*}}S^r\wedge S^r\cong S^{2r}\rightarrow S^{2n} $

which is trivial since $\pi_rS^{2r}=0$.

On the other hand $q_n\not\simeq\ast$, since in homology the induced map

$$q_{n*}:H_{2n}(S^n\times S^n)\xrightarrow{}H_{2n}(S^n\wedge S^n)$$

is an isomorphism.