Prove that any map \(\mathbb{RP}^2 \to T^2\) must be null-homotopic.
Solution
1. Let \(f:\mathbb{RP}^2\to T^2\). The induced homomorphism\[ f_*:\pi_1(\mathbb{RP}^2)\cong\mathbb Z/2\longrightarrow\pi_1(T^2)\cong\mathbb Z^2 \]is trivial.
The torsion-free group \(\mathbb Z^2\) has no nontrivial element of order dividing \(2\).
2. Therefore \(f\) lifts to the universal cover\[ \widetilde f:\mathbb{RP}^2\to\mathbb R^2. \]
The lifting criterion for the universal covering \(\mathbb R^2\to T^2\) says that a lift exists exactly when \(f_*(\pi_1(\mathbb{RP}^2))\) is trivial.
3. Since \(\mathbb R^2\) is contractible, \(\widetilde f\) is null-homotopic.
Any map into a contractible space is homotopic to a constant.
4. Projecting this homotopy to \(T^2\) shows\[ \boxed{f\simeq\mathrm{const}.} \]
Compose the null-homotopy of \(\widetilde f\) with the covering projection.