Show that the quotient map \(S^1 \times S^1 \to S^2\) collapsing the subspace \(S^1 \lor S^1\) to a point is not nullhomotopic by showing that it induces an isomorphism on \(H_2\). On the other hand, show via covering spaces that any map \(S^2 \to S^1 \times S^1\) is nullhomotopic.
Solution
Let \[ T^2=S^1\times S^1 \] with its standard CW structure consisting of one \(0\)-cell, two \(1\)-cells, and one \(2\)-cell. Its \(1\)-skeleton is \(S^1\vee S^1\).
1. Collapsing the \(1\)-skeleton gives\[ T^2/(S^1\vee S^1)\cong S^2. \]
The quotient collapses the boundary of the unique \(2\)-cell to a point, so the quotient of its closed characteristic disk is \[ D^2/\partial D^2\cong S^2. \]
2. The quotient map\[ q:T^2\to S^2 \]induces an isomorphism on \(H_2\).
The cellular chain complex of \(T^2\) is \[ 0\to\mathbb Z\xrightarrow{0}\mathbb Z^2\xrightarrow{0}\mathbb Z\to0, \] so the unique \(2\)-cell represents a generator of \(H_2(T^2)\cong\mathbb Z\). The quotient CW structure on \(S^2\) has the same unique \(2\)-cell and no \(1\)-cells. The cellular map induced by \(q\) is the identity on the top cellular chain group \(\mathbb Z\), hence \[ q_*:H_2(T^2)\xrightarrow{\cong}H_2(S^2). \] Therefore \(q\) cannot be nullhomotopic.
3. Every map\[ f:S^2\to T^2 \]is nullhomotopic.
The universal covering map \[ p:\mathbb R^2\to T^2 \] has contractible total space. Since \(S^2\) is simply connected, the lifting criterion gives a lift \[ \widetilde f:S^2\to\mathbb R^2 \] with \(p\widetilde f=f\). The lift is nullhomotopic because \(\mathbb R^2\) is contractible. Composing such a nullhomotopy with \(p\) gives a nullhomotopy of \(f\).