Let \(p: S^3 \to S^2\) be the Hopf bundle and let \(q: T^3 \to S^3\) be the quotient map collapsing the complement of a ball in the 3-dimensional torus \(T^3 = S^1 \times S^1 \times S^1\) to a point. Show that \(pq: T^3 \to S^2\) induces the trivial map on \(\pi_*\) and \(\tilde{H}_*\), but is not homotopic to a constant map.
Solution
Let \[ f=pq:T^3\xrightarrow{q}S^3\xrightarrow{p}S^2, \] where \(p\) is the Hopf map and \(q\) collapses the complement of an embedded \(3\)-ball.
For homotopy groups, \(\pi_1(S^2)=0\), while \[ \pi_i(T^3)=0\qquad(i>1) \] because the universal cover of \(T^3\) is \(\mathbb R^3\). Hence \(f_*\) is zero on every homotopy group.
For reduced homology, \(q_*\) is zero in degrees \(1\) and \(2\) because \(S^3\) has no homology there, and \[ p_*:H_3(S^3)\to H_3(S^2) \] is zero. Thus \(f_*\) is zero on all reduced homology groups.
To see that \(f\) is nevertheless essential, compare mapping cones. The commutative square \[ \begin{array}{ccc} T^3&\xrightarrow{q}&S^3\\ f\downarrow&&\downarrow p\\ S^2&=&S^2 \end{array} \] induces a map \[ C_f\longrightarrow C_p. \] Contravariantly, this gives \[ H^*(C_p)\longrightarrow H^*(C_f). \] The Hopf mapping cone is \(C_p\simeq\mathbb{CP}^2\). If \(x\in H^2(C_p;\mathbb Z)\) is its standard generator, then \[ x^2\ne0\in H^4(C_p;\mathbb Z). \] The induced map is an isomorphism in degree \(2\), since it is the identity on the \(S^2\) subspace. In degree \(4\), the quotient description of mapping cones identifies the induced map with the suspension of \[ q:T^3\to S^3. \] The map \(q\) has degree \(1\), so this map is also an isomorphism on \(H^4\). Therefore the image \(\bar x\in H^2(C_f)\) satisfies \[ \bar x^2\ne0. \]
If \(f\) were nullhomotopic, then \[ C_f\simeq S^2\vee\Sigma T^3. \] The degree-\(2\) class coming from the \(S^2\)-summand would have square zero, a contradiction. Hence \[ \boxed{pq\text{ is not nullhomotopic}.} \]