Proving that a continuous map is homotopic to the constant map

Question

How can I prove that a continuous map $f : \mathbb{R}P^2 \to S^1$ is homotopic to the constant map? I know that in the projective space every point is a line but I do not get why the above has to be true.

Answer 1

Hint:

$\pi(\mathbb{R}P^2)=\mathbb{Z_2}$, $\pi(S^1)=\mathbb{Z}$. So the degree of $f_*:\pi(\mathbb{R}P^2) \rightarrow \pi(S^1)$ can only be zero. So by the lifting property, this map can be lift to $\tilde{f}:\mathbb{R}P^2 \rightarrow \mathbb{R}$.

Answer 2

As $\pi_1(\mathbb{R}P^2)$ is finite, the subgroup $f_*(\pi_1(\mathbb{R}P^2))$ of $\pi_1(S^1)=\mathbb{Z}$ must be trivial: indeed there are no non-trivial finite subgroups of $\mathbb{Z}$. Denote by $p: \mathbb{R} \longrightarrow S^1$ the universal cover of the circle. Using the lifting criterion for covering spaces (because we now have $0=f_*(\pi_1(\mathbb{R}P^2)) \subset p_*(\pi_1(\mathbb{R}))=0$), your map $f$ lifts to a map $\tilde{f}: \mathbb{R}P^2 \rightarrow \mathbb{R}$. In other words $f$ factors through a contractible space (by definition of "lift", $f=p\tilde{f}$). Consequently $f$ is nullhomotopic.