at.algebraic topology - Maps inducing zero on homotopy groups but are not null-homotopic

Even if you ask that $f$ induces trivial maps on all (singular) homology and cohomology groups, there are still easy manifold examples. (This actually arises as an exercise in Hatcher's AT).

For instance, let $f:T^3rightarrow S^2$ be the composition $T^3rightarrow S^3rightarrow S^2$, where the map from $T^3$ to $S^3$ is simply collapsing the 2-skeleton to a point, and the map from $S^3$ to $S^2$ is the Hopf map.

As others have mentioned, since $T^3$ is a $K(mathbb{Z}^3, 1)$, if follows that $f$ induces trivial maps on homotopy groups.

Since the Hopf map induces trivial maps on homology and cohomology, it follows that $f$ does as well.

Finally, to see that $f$ is NOT nullhomotopic, assume it is. Since the map from $S^3$ to $S^2$ is a fiber bundle, it has the homotopy lifting property. Hence, we can lift the homotopy of $f$ to a homotopy $G:Itimes T^3rightarrow S^3$ where $G_0$ is the above map from $T^3$ to $S^3$ and $G_1$ is is a map from $T^3$ to $S^1subseteq S^3$, the preimage of a point in $S^2$ under the Hopf map.

But $G_0$ has degree 1, while $G_1$ has degree 0, a contradiction.