Show that any continuous map \(f : \RP^2 \to S^1 \times S^1\) is necessarily null-homotopic.
Solution
1. The induced homomorphism\[ f_*:\pi_1(\RP^2)\to\pi_1(T^2) \]is zero.
We have \[ \pi_1(\RP^2)\cong\ZZ/2\ZZ, \qquad \pi_1(T^2)\cong\ZZ^2. \] If \(u=f_*([1])\), then \[ 2u=f_*(2[1])=0. \] The group \(\ZZ^2\) is torsion-free, so \(u=0\).
2. The map \(f\) lifts to the universal cover\[ p:\RR^2\to T^2. \]
Choose basepoints. The covering-space lifting criterion requires \[ f_*\pi_1(\RP^2) \subseteq p_*\pi_1(\RR^2). \] Both sides are zero: the left by 1 and the right because \(\RR^2\) is simply connected. Thus there is a continuous lift \[ \widetilde f:\RP^2\to\RR^2 \] such that \[ p\circ\widetilde f=f. \]
3. \(f\) is null-homotopic.
The universal cover \(\RR^2\) is contractible, so \(\widetilde f\) is homotopic to a constant map. Composing this homotopy with \(p\) gives a homotopy from \(f\) to a constant map in \(T^2\).