Map of H-spaces inducing zero on homologies

Question

If a map of $H$-spaces $f:X\rightarrow Y$ induces zero on the homology groups at dimensions greater than zero does it necessarily induce zero map on the homotopy groups? It is definitely true for $\pi_1$ as it is Abelian and can be identified with the first homology. The statement is also true rationally since rational homotopy can be identified with primitives of the rational homology.

Answer 1

The answer is no. Let $\pi^k$ denote the kth cohomotopy group, the contravariant functor that measures homotopy classes of maps into the k-sphere. Given a map $[f]:S^n \rightarrow S^k$ we get a natural transformation $\pi^n \rightarrow \pi^k$ (given by postcomposition). Let $\eta$ denote the nontrivial element of $\pi_7(S^3)= \mathbb{Z}_2$. This gives a natural transformation $\pi^7 \rightarrow \pi^3$ that cannot be the 0 natural transformation by Yoneda's lemma. This means there is some map $[g]: S^m \rightarrow S^7$ such that $[\eta \circ g]: S^m \rightarrow S^3$ is not trivial.

We know $[g] \in \pi_m(S^7)$, so we can say that $\eta_*:\pi_m(S^7) \rightarrow \pi_m(S^3)$ is nonzero. Obviously, $\eta$ induces 0 maps on homology. As well, both $S^3$ and $S^7$ are H-spaces.