Prove that every continuous map \(f : \RP^2 \to S^1\) is homotopic to a constant.
Hint: think about covering spaces.
Solution
1. The induced homomorphism\[ f_*:\pi_1(\RP^2,x_0)\to\pi_1(S^1,f(x_0)) \]is trivial.
The fundamental groups are \[ \pi_1(\RP^2,x_0)\cong\mathbb Z/2 \qquad\text{and}\qquad \pi_1(S^1,f(x_0))\cong\mathbb Z. \] Every group homomorphism \[ \mathbb Z/2\to\mathbb Z \] is trivial: if the nonzero element of \(\mathbb Z/2\) maps to \(k\in\mathbb Z\), then \[ 2k=0, \] and torsion-freeness of \(\mathbb Z\) gives \(k=0\). Hence \(f_*=0\).
2. The map \(f\) lifts to the universal covering\[ p:\mathbb R\to S^1, \qquad p(t)=e^{2\pi i t}. \]
Choose \(\widetilde y_0\in\mathbb R\) with \[ p(\widetilde y_0)=f(x_0). \] The covering-space lifting criterion says that a based map \[ f:(\RP^2,x_0)\to(S^1,f(x_0)) \] lifts to \((\mathbb R,\widetilde y_0)\) if and only if \[ f_*\bigl(\pi_1(\RP^2,x_0)\bigr) \subseteq p_*\bigl(\pi_1(\mathbb R,\widetilde y_0)\bigr). \] By 1 the left-hand side is \(0\). Since \(\mathbb R\) is simply connected, the right-hand side is also \(0\). Thus the criterion applies and gives a continuous lift \[ \widetilde f:\RP^2\to\mathbb R \] such that \[ p\circ\widetilde f=f. \]
3. The lift \(\widetilde f\) is homotopic to a constant map.
The real line is contractible. For example, fixing \(c=\widetilde f(x_0)\), define \[ \widetilde H(x,t)=(1-t)\widetilde f(x)+tc. \] Then \(\widetilde H\) is a homotopy from \(\widetilde f\) to the constant map \(x\mapsto c\).
4. Therefore \(f\) is homotopic to a constant map.
Compose the homotopy in 3 with \(p\): \[ H=p\circ\widetilde H:\RP^2\times[0,1]\to S^1. \] At \(t=0\), \[ H(x,0)=p(\widetilde f(x))=f(x), \] while at \(t=1\), \[ H(x,1)=p(c), \] independent of \(x\). Thus \[ \boxed{f\simeq\text{constant}}. \]