Prove that any continuous map $f\colon RP^3\to S^1$ is homotopic to a constant map?
How to prove it? Thanks in advance.
Prove that any continuous map $f\colon RP^3\to S^1$ is homotopic to a constant map?
How to prove it? Thanks in advance.
Since $\pi_1(\Bbb RP^3)=\Bbb Z_2$ and $\pi_1(\Bbb S^1)=\Bbb Z$ the induced group homomorphism $f_*:\pi_1(\Bbb RP^3)\to \pi_1(\Bbb S^1)$ is the zero homomorphism.
Hence, by the arbitrary map lifting lemma we can lift the map $f$ w.r.t the covering $\Bbb R\ni t\longmapsto \exp(2\pi it)\in \Bbb S^1$. That is there is a map $\widetilde f:\Bbb RP^3\to \Bbb R$ with $\exp\big(2\pi i\widetilde f(x)\big)=f(x)$.
Note that to lift a map, we must ensure that the domain of the map is locally path-connected and connected. But here we are dealing with connected manifolds, so we do not need to worry.
Now, consider the homotopy $H:\Bbb RP^3\times [0,1]\ni (x,t)\longmapsto \exp\big(2\pi i (1-t)\widetilde f(x)\big)\in \Bbb S^1$.