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.

Answer 2

In the case that your domain is a suspension or your codomain is a loop space, one may add and subtract maps. In this setting, this means there is an equivalence between your question and the question “Are null homotopic maps the same as maps inducing trivial maps on homotopy groups?”

It tends to be the case in homological algebra as well as homotopy theory that we can phrase similar statements about isomorphisms and zero maps. Your’s is about transferring Whitehead from isomorphisms to zero maps. And there are similar versions of Whitehead in the case of homological algebra.

But these analogous results tend to not to be true in either case