A non-null-homotopic map from $T^{3}$ into $S^{2}$

Question

Suppose we collapse the 2-skeleton of $T^{3}$ and we denote this map by $q$, \begin{equation} q:T^{3}\rightarrow S^{3} \end{equation} Then compositing $q$ with Hopf fibration, \begin{equation} S^{1}\hookrightarrow S^{3}\rightarrow S^{2} \end{equation} we have \begin{equation} f:T^{3}\rightarrow S^{2} \end{equation}

$f$ induces trivial map between homotopy groups and reduced homology groups. But I can not find other invariants to show $f$ is not null-homotopic and do not understand how to solve it.

I will appreciate your help for any suggestion.

Answer 1

Big Hint: Use the fact that the Hopf fibration is a fibration (i.e the homotopy lifting property).