Prove every map from the projective plane to the circle is nullhomotopic

Prove that every continuous map $f:P^2\to S^1$, where $P^2$ is the projective plane, is nullhomotopic.

I think I need to use the fact that $\pi_1(P^2) = \mathbb{Z}/2\mathbb{Z}$ and covering space theory. The map $f$ induces a map $f_* : \pi_(P^2) = \mathbb{Z}/2\mathbb{Z} \to \pi_1(S^1) = \mathbb{Z}$, but I don't see why this map is trivial.

Solution 1:

As Jim pointed out, it's not enough to show that $f_*$ induces the trivial map on fundamental groups. To finish it, we can proceed as follows. Let $p\colon \mathbb R\to S^1$ be the covering projection. Since $f_*(\pi_1(P^2))=0$ (as Fredrik showed), it's clearly contained in $p_*(\pi_1(\mathbb R))=0$. So, this means there is a map $\tilde f\colon P^2\to\mathbb R$ such that $f=p\tilde f$ (Proposition 1.33 in Hatcher). Now, $\mathbb R$ is contractible so we have a homotopy $H\colon P^2\times I\to \mathbb R$ such that $H_0=\tilde f$ and $H_1$ is constant at some $x_0$. But then $pH$ is a homotopy from $p\tilde f=f$ to $p(x_0)$ meaning that $f$ is null-homotopic.

Solution 2:

I'll assume by $P^2$ you mean the real projective plane and I'll also assume you know that $\pi_1(P^2)=\mathbb{Z}/2$.

Now, every map $f:(X,x_0) \to (Y,y_0)$ of based topological spaces induces a map between their fundamental groups $f_*:\pi_1(X,x_0) \to \pi_1(Y,y_0)$.

Now, $\pi_1(P^2,x_0)=\mathbb{Z}/2$ (and in fact, we may forget about the base point, since $P^2$ is path connected). Also, $\pi_1(S^1, x_0)=\mathbb{Z}$. So a continous map $f:P^2 \to S^1$ induces, by the above paragraph, a map $\mathbb{Z}/2 \to \mathbb{Z}$. Torsion elements are sent to torsion elements, but $\mathbb{Z}$ is torsion-free, so the only possibility is that $\mathbb{Z}/2$ is mapped to zero. So every map $f:P^2 \to S^1$ induces the trivial map on fundamental groups. This means that $f_*$ is homotopic to the null map. EDIT: See SL2's answer below for the conclusion.

(regarding maps $\mathbb{Z}/2 \to \mathbb{Z}$: What happens to $1 \in \mathbb{Z}/2$? The map must satisfy $f(1+1)=f(1)+f(1)$, but $f(1+1)=f(0)=0$ , so $2f(1)=0$, hence $f(1)=0$.)